Solutions - 1 Solutions to Section 1.1 True-False Review:...

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Unformatted text preview: 1 Solutions to Section 1.1 True-False Review: 1. FALSE. A derivative must involve some derivative of the function y = f (x), not necessarily the first derivative. 2. TRUE. The initial conditions accompanying a differential equation consist of the values of y, y , . . . at t = 0. 3. TRUE. If we define positive velocity to be oriented downward, then dv = g, dt where g is the acceleration due to gravity. 4. TRUE. We can justify this mathematically by starting from a(t) = g , and integrating twice to get 1 v (t) = gt + c, and then s(t) = gt2 + ct + d, which is a quadratic equation. 2 5. FALSE. The restoring force is directed in the direction opposite to the displacement from the equilibrium position. 6. TRUE. According to Newton’s Law of Cooling, the rate of cooling is proportional to the difference between the object’s temperature and the medium’s temperature. Since that difference is greater for the object at 100◦ F than the object at 90◦ F , the object whose temperature is 100◦ F has a greater rate of cooling. 7. FALSE. The temperature of the object is given by T (t) = Tm + ce−kt , where Tm is the temperature of the medium, and c and k are constants. Since e−kt = 0, we see that T (t) = Tm for all times t. The temperature of the object approaches the temperature of the surrounding medium, but never equals it. 8. TRUE. Since the temperature of the coffee is falling, the temperature difference between the coffee and the room is higher initially, during the first hour, than it is later, when the temperature of the coffee has already decreased. 9. FALSE. The slopes of the two curves are negative reciprocals of each other. 10. TRUE. If the original family of parallel lines have slopes k for k = 0, then the family of orthogonal tra1 jectories are parallel lines with slope − k . If the original family of parallel lines are vertical (resp. horizontal), then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines. 11. FALSE. The family of orthogonal trajectories for a family of circles centered at the origin is the family of lines passing through the origin. Problems: 1. d2 y dt2 c2 = = g =⇒ 0. dy (0) dt dy dt = gt + c1 =⇒ y (t) = gt2 2 + c1 t + c2 . Now impose the initial conditions. y (0) = 0 =⇒ =⇒ c1 = 0. Hence, the solution to the initial-value problem is: y (t) = ground at time, t0 , when y (t0 ) = 100. Hence 100 = g = 9.8 ms 2. From −2 gt2 0 2, so that t0 = 200 g gt2 2. The object hits the ≈ 4.52 s, where we have taken . d2 y = g , we integrate twice to obtain the general equations for the velocity and the position of the dt2 2 1 dy = gt + c and y (t) = gt2 + ct + d, where c, d are constants of integration. Setting y = 0 dt 2 to be at the top of the boy’s head (and positive direction downward), we know that y (0) = 0. Since the object hits the ground 8 seconds later, we have that y (8) = 5 (since the ground lies at the position y = 5). 5 − 32g From the values of y (0) and y (8), we find that d = 0 and 5 = 32g + 8c. Therefore, c = . 8 (a) The ball reaches its maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore, ball, respectively: t=− 32g − 5 c = ≈ 3.98 s. g 8g (b) To find the maximum height of the tennis ball, we compute y (3.98) ≈ −253.51 feet. So the ball is 253.51 feet above the top of the boy’s head, which is 258.51 feet above the ground. d2 y = g , we integrate twice to obtain the general equations for the velocity and the position of the dt2 1 dy rocket, respectively: = gt + c and y (t) = gt2 + ct + d, where c, d are constants of integration. Setting dt 2 y = 0 to be at ground level, we know that y (0) = 0. Thus, d = 0. 3. From (a) The rocket reaches maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore, the c time that the rocket achieves its maximum height is t = − . At this time, y (t) = −90 (the negative sign g accounts for the fact that the positive direction is chosen to be downward). Hence, −90 = y − c g = 1 c g− 2 g 2 +c − c g = c2 c2 c2 − =− . 2g g 2g √ Solving this for c, we find that c = ± 180g . However, since c represents the initial velocity of the rocket, and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose √ c = − 180g ≈ −42.02 ms−1 , and thus the initial speed at which the rocket must be launched for optimal viewing is approximately 42.02 ms−1 . c −42.02 (b) The time that the rocket reaches its maximum height is t = − ≈ − = 4.28 s. g 9.81 d2 y = g , we integrate twice to obtain the general equations for the velocity and the position of the dt2 dy 1 rocket, respectively: = gt + c and y (t) = gt2 + ct + d, where c, d are constants of integration. Setting dt 2 y = 0 to be at the level of the platform (with positive direction downward), we know that y (0) = 0. Thus, d = 0. 4. From (a) The rocket reaches maximum height at the moment when y (t) = 0. That is, gt + c = 0. Therefore, the c time that the rocket achieves its maximum height is t = − . At this time, y (t) = −85 (this is 85 m above g the platform, or 90 m above the ground). Hence, −85 = y − c g = 1 c g− 2 g 2 +c − c g = c2 c2 c2 − =− . 2g g 2g 3 √ Solving this for c, we find that c = ± 170g . However, since c represents the initial velocity of the rocket, and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose √ c = − 170g ≈ −40.84 ms−1 , and thus the initial speed at which the rocket must be launched for optimal viewing is approximately 40.84 ms−1 . c −40.84 (b) The time that the rocket reaches its maximum height is t = − ≈ − = 4.16 s. g 9.81 5. If y (t) denotes the displacement of the object from its initial position at time t, the motion of the object can be described by the initial-value problem dy d2 y = g, y (0) = 0, (0) = −2. dt2 dt d2 y dy gt2 = g =⇒ = gt + c1 =⇒ y (t) = + c1 t + c2 . Now dt2 dt 2 dy impose the initial conditions. y (0) = 0 =⇒ c2 = 0. (0) = −2 =⇒ c1 = −2. Hence the soution to the dt 2 gt g (10)2 − 2t. We are given that y (10) = h. Consequently, h = − 2 · 10 =⇒ intial-value problem is y (t) = 2 2 −2 h = 10(5g − 2) = 470 m where we have taken g = 9.8 ms . We first integrate this differential equation: 6. If y (t) denotes the displacement of the object from its initial position at time t, the motion of the object can be described by the initial-value problem d2 y dy = g, y (0) = 0, (0) = v0 . dt2 dt d2 y dy gt2 = g =⇒ = gt + c1 =⇒ y (t) = + c1 t + c2 . Now impose dt2 dt 2 dy the initial conditions. y (0) = 0 =⇒ c2 = 0. (0) = v0 =⇒ c1 = v0 . Hence the soution to the intial-value dt gt2 + v0 t. We are given that y (t0 ) = h. Consequently, h = gt2 + v0 t0 . Solving for v0 yields problem is y (t) = 0 2 2h − gt2 0 v0 = . 2t0 We first integrate the differential equation: dy d2 y d2 y = −Aω sin (ωt − φ) =⇒ 2 = −Aω 2 cos (ωt − φ). Hence, 2 + ω 2 y = dt dt dt −Aω 2 cos (ωt − φ) + Aω 2 cos (ωt − φ) = 0. 7. y (t) = A cos (ωt − φ) =⇒ dy d2 y = −c1 ω sin (ωt) + c2 ω cos (ωt) =⇒ = −c1 ω 2 cos (ωt) − dt dt2 d2 y c2 ω 2 sin (ωt) = −ω 2 [c1 cos (ωt) + c2 cos (ωt)] = −ω 2 y. Consequently, + ω 2 y = 0. To determine the dt2 amplitude of the motion we write the solution to the differential equation in the equivalent form: 8. y (t) = c1 cos (ωt) + c2 sin (ωt) =⇒ c1 c2 + c2 1 2 y (t) = c2 1 + c2 2 cos (ωt) + c2 c2 1 + c2 2 sin (ωt) . We can now define an angle φ by cos φ = c1 c2 1 + c2 2 and sin φ = c2 c2 1 + c2 2 . 4 Then the expression for the solution to the differential equation is y (t) = c2 + c2 [cos (ωt) cos φ + sin (ωt) sin φ] = 1 2 c2 + c2 cos (ωt + φ). 1 2 Consequently the motion corresponds to an oscillation with amplitude A = c2 + c2 . 1 2 9. We compute the first three derivatives of y (t) = ln t: d2 y 1 = − 2, dt2 t dy 1 =, dt t d3 y 2 = 3. dt3 t Therefore, 2 dy dt 3 = 2 d3 y = 3, 3 t dt as required. 10. We compute the first two derivatives of y (x) = x/(x + 1): dy 1 = dx (x + 1)2 d2 y 2 =− . dx2 (x + 1)3 and Then y+ d2 y x 2 x3 + 2x2 + x − 2 (x + 1) + (x3 + 2x2 − 3) 1 x3 + 2x2 − 3 = − = = = + , 2 3 3 3 2 dx x + 1 (x + 1) (x + 1) (x + 1) (x + 1) (1 + x)3 as required. 11. We compute the first two derivatives of y (x) = ex sin x: dy = ex (sin x + cos x) dx Then 2y cot x − and d2 y = 2ex cos x. dx2 d2 y = 2(ex sin x) cot x − 2ex cos x = 0, dx2 as required. dT d = −k =⇒ (ln |T − Tm |) = −k . The preceding equation can be integated directly to dt dt yield ln |T − Tm | = −kt + c1 . Exponentiating both sides of this equation gives |T − Tm | = e−kt+c1 , which can be written as T − Tm = ce−kt , 12. (T − Tm )−1 where c = ±ec1 . Rearranging yields T (t) = Tm + ce−kt . 13. After 4 p.m. In the first two hours after noon, the water temperature increased from 50◦ F to 55◦ F, an increase of five degrees. Because the temperature of the water has grown closer to the ambient air temperature, the temperature difference |T − Tm | is smaller, and thus, the rate of change of the temperature of the water grows smaller, according to Newton’s Law of Cooling. Thus, it will take longer for the water temperature to increase another five degrees. Therefore, the water temperature will reach 60◦ F more than two hours later than 2 p.m., or after 4 p.m. 5 14. The object temperature cools a total of 40◦ F during the 40 minutes, but according to Newton’s Law of Cooling, it cools faster in the beginning (since |T − Tm | is greater at first). Thus, the object cooled half-way from 70◦ F to 30◦ F in less than half the total cooling time. Therefore, it took less than 20 minutes for the object to reach 50◦ F. 15. Given a family of curves satisfies: x2 + 4y 2 = c1 =⇒ 2x + 8y dy x dy = 0 =⇒ =− . dx dx 4y Orthogonal trajectories satisfy: dy 4y 1 dy 4 d 4 = =⇒ = =⇒ (ln |y |) = =⇒ ln |y | = 4 ln |x| + c2 =⇒ y = kx4 , where k = ±ec2 dx x y dx x dx x . y(x) 0.8 0.4 x -1.5 -1.0 -0.5 0.5 1.0 1.5 -0.4 -0.8 Figure 0.0.1: Figure for Exercise 15 c dy dy y 16. Given a family of curves satisfies: y = =⇒ x + y = 0 =⇒ =− . x dx dx x Orthogonal trajectories satisfy: x dy d dy = =⇒ y = x =⇒ dx y dx dx 12 y 2 = x =⇒ 12 1 y = x2 + c1 =⇒ y 2 − x2 = c2 , where c2 = 2c1 . 2 2 17. Given family of curves satisfies: y = cx2 =⇒ c = y . Hence, x2 dy y 2y = 2cx = c 2 x = . dx x x Orthogonal trajectories satisfy: dy x dy d2 1 =− =⇒ 2y = −x =⇒ (y ) = −x =⇒ y 2 = − x2 + c1 =⇒ 2y 2 + x2 = c2 , dx 2y dx dx 2 where c2 = 2c1 . y 18. Given family of curves satisfies: y = cx4 =⇒ c = 4 . Hence, x dy y 4y = 4cx3 = 4 4 x3 = . dx x x 6 y(x) 4 2 x -2 -4 4 2 -2 -4 Figure 0.0.2: Figure for Exercise 16 y(x) 2.0 1.6 1.2 0.8 0.4 x -2 -1 1 2 -0.4 -0.8 -1.2 -1.6 -2.0 Figure 0.0.3: Figure for Exercise 17 Orthogonal trajectories satisfy: dy x dy d 1 =− =⇒ 4y = −x =⇒ (2y 2 ) = −x =⇒ 2y 2 = − x2 + c1 =⇒ 4y 2 + x2 = c2 , dx 4y dx dx 2 where c2 = 2c1 . 7 y(x) 0.8 0.4 x -1.5 1.0 -1.0 1.5 -0.4 -0.8 Figure 0.0.4: Figure for Exercise 18 19. Given family of curves satisfies: y 2 = 2x + c =⇒ dy 1 = .Orthogonal trajectories satisfy: dx y dy dy d = −y =⇒ y −1 = −1 =⇒ (ln |y |) = −1 =⇒ ln |y | = −x + c1 =⇒ y = c2 e−x . dx dx dx y(x) 4 3 2 1 x -1 1 2 3 4 -1 -2 -3 -4 Figure 0.0.5: Figure for Exercise 19 20. Given family of curves satisfies: y = cex =⇒ dy 1 dy d = − =⇒ y = −1 =⇒ dx y dx dx 12 y 2 dy = cex = y . Orthogonal trajectories satisfy: dx = −1 =⇒ 12 y = −x + c1 =⇒ y 2 = −2x + c2 . 2 8 y(x) 2 1 x 1 -1 -1 -2 Figure 0.0.6: Figure for Exercise 20 dy 21. y = mx + c =⇒ = m. dx Orthogonal trajectories satisfy: 1 1 dy = − =⇒ y = − x + c1 . dx m m 22. y = cxm =⇒ dy y dy my = cmxm−1 , but c = m so = . Orthogonal trajectories satisfy: dx x dx x dy x dy x d =− =⇒ y = − =⇒ dx my dx m dx 12 y 2 =− x 1 12 1 =⇒ y 2 = − x + c1 =⇒ y 2 = − x2 + c2 . m 2 2m m dy dy mx + 2mx = 0 =⇒ =− . dx dx y Orthogonal trajectories satisfy: 23. y 2 + mx2 = c =⇒ 2y dy y dy 1 d 1 = =⇒ y −1 = =⇒ (ln |y |) = =⇒ m ln |y | = ln |x| + c1 =⇒ y m = c2 x. dx mx dx mx dx mx dy dy m = m =⇒ = . dx dx 2y Orthogonal trajectories satisfy: 24. y 2 = mx + c =⇒ 2y 2x dy 2y dy 2 d 2 2 =− =⇒ y −1 = − =⇒ (ln |y |) = − =⇒ ln |y | = − x + c1 =⇒ y = c2 e− m . dx m dx m dx m m 25. u = x2 + 2y 2 =⇒ 0 = 2x + 4y dy dy x =⇒ =− . dx dx 2y 9 Orthogonal trajectories satisfy: 2y 2 d 2 dy dy = =⇒ y −1 = =⇒ (ln |y |) = =⇒ ln |y | = 2 ln |x| + c1 =⇒ y = c2 x2 . dx x dx x dx x y(x) 2.0 1.6 1.2 0.8 0.4 x -2 1 -1 2 -0.4 -0.8 -1.2 -1.6 -2.0 Figure 0.0.7: Figure for Exercise 25 26. m1 = tan (a1 ) = tan (a2 − a) = tan (a2 ) − tan (a) m2 − tan (a) = . 1 + tan (a2 ) tan (a) 1 + m2 tan (a) Solutions to Section 1.2 True-False Review: 1. FALSE. The order of a differential equation is the order of the highest derivative appearing in the differential equation. 2. TRUE. This is condition 1 in Definition 1.2.11. 3. TRUE. This is the content of Theorem 1.2.15. 4. FALSE. There are solutions to y + y = 0 that do not have the form c1 cos x + 5c2 cos x, such as y (x) = sin x. Therefore, c1 cos x + 5c2 cos x does not meet the second requirement set forth in Definition 1.2.11 for the general solution. 5. FALSE. There are solutions to y + y = 0 that do not have the form c1 cos x + 5c1 sin x, such as y (x) = cos x + sin x. Therefore, c1 cos x + 5c1 sin x does not meet the second requirement set form in Definition 1.2.11 for the general solution. 6. TRUE. Since the right-hand side of the differential equation is a function of x only, we can integrate both sides n times to obtain the formula for the solution y (x). 10 Problems: 1. 2, nonlinear. 2. 3, linear. 3. 2, nonlinear. 4. 2, nonlinear. 5. 4, linear. 6. 3, nonlinear. 7. We can quickly compute the first two derivatives of y (x): y (x) = (c1 +2c2 )ex cos 2x+(−2c1 +c2 )ex sin 2x and y (x) = (−3c1 +4c2 )ex cos 2x+(−4c1 −3c2 )ex sin x. Then we have y − 2y + 5y = [(−3c1 + 4c2 )ex cos 2x + (−4c1 − 3c2 )ex sin x]−2 [(c1 + 2c2 )ex cos 2x + (−2c1 + c2 )ex sin 2x]+5(c1 ex cos 2x+c2 ex sin 2x), which cancels to 0, as required. This solution is valid for all x ∈ R. 8. y (x) = c1 ex + c2 e−2x =⇒ y = c1 ex − 2c2 e−2x =⇒ y = c1 ex + 4c2 e−2x =⇒ y + y − 2y = (c1 ex + 4c2 e−2x ) + (c1 ex − 2c2 e−2x ) − 2(c1 ex + c2 e−2x ) = 0. Thus y (x) = c1 ex + c2 e−2x is a solution of the given differential equation for all x ∈ R. 1 1 1 =⇒ y = − = −y 2 . Thus y (x) = is a solution of the given differential x+4 (x + 4)2 x+4 equation for x ∈ (−∞, −4) or x ∈ (−4, ∞). 9. y (x) = √ √ c1 y 10. y (x) = c1 x =⇒ y = √ = . Thus y (x) = c1 x is a solution of the given differential equation for 2x 2x all x ∈ {x : x > 0}. 11. y (x) = c1 e−x sin (2x) =⇒ y = 2c1 e−x cos (2x)−c1 e−x sin (2x) =⇒ y = −3c1 e−x sin (2x)−4c1 e−x cos (2x) =⇒ y + 2y + 5y = −3c1 e−x sin (2x) − 4c1 e−x cos (2x) + 2[2c1 e−x cos (2x) − c1 e−x sin (2x)] + 5[c1 e−x sin (2x)] = 0. Thus y (x) = c1 e−x sin (2x) is a solution to the given differential equation for all x ∈ R. 12. y (x) = c1 cosh (3x) + c2 sinh (3x) =⇒ y = 3c1 sinh (3x) + 3c2 cosh (3x) =⇒ y = 9c1 cosh (3x) + 9c2 sinh (3x) =⇒ y − 9y = [9c1 cosh (3x) + 9c2 sinh (3x)] − 9[c1 cosh (3x) + c2 sinh (3x)] = 0. Thus y (x) = c1 cosh (3x) + c2 sinh (3x) is a solution to the given differential equation for all x ∈ R. c1 c2 3c1 c2 12c1 2c2 12c1 2c2 + =⇒ y = − 4 − 2 =⇒ y = 5 + 3 =⇒ x2 y + 5xy + 3y = x2 +3+ x3 x x x x x x5 x 3c1 c2 c1 c2 c1 c2 5x − 4 − 2 + 3 3 + = 0. Thus y (x) = 3 + is a solution to the given differential equation x x x x x x for all x ∈ (−∞, 0) or x ∈ (0, ∞). 13. y (x) = √ c1 c1 c1 14. y (x) = c1 x + 3x2 =⇒ y = √ + 6x =⇒ y = − √ + 6 =⇒ 2x2 y − xy + y = 2x2 − √ + 6 − 2x 4 x3 4 x3 11 √ √ c1 √ + 6x + (c1 x +3x2 ) = 9x2 . Thus y (x) = c1 x +3x2 is a solution to the given differential equation 2x for all x ∈ {x : x > 0}. x 15. y (x) = c1 x2 + c2 x3 − x2 sin x =⇒ y = 2c1 x + 3c2 x2 − x2 cos x − 2x sin x =⇒ y = 2c1 + 6c2 x + x2 sin x − 2x cos x − 2x cos −2 sin x. Substituting these results into the given differential equation yields x2 y − 4xy + 6y = x2 (2c1 + 6c2 x + x2 sin x − 4x cos x − 2 sin x) − 4x(2c1 x + 3c2 x2 − x2 cos x − 2x sin x) + 6(c1 x2 + c2 x3 − x2 sin x) = 2c1 x2 + 6c2 x3 + x4 sin x − 4x3 cos x − 2x2 sin x − 8c1 x2 − 12c2 x3 + 4x3 cos x + 8x2 sin x + 6c1 x2 + 6c2 x3 − 6x2 sin x = x4 sin x. Hence, y (x) = c1 x2 + c2 x3 − x2 sin x is a solution to the differential equation for all x ∈ R. 16. y (x) = c1 eax + c2 ebx =⇒ y = ac1 eax + bc2 ebx =⇒ y = a2 c1 eax + b2 c2 ebx . Substituting these results into the differential equation yields y − (a + b)y + aby = a2 c1 eax + b2 c2 ebx − (a + b)(ac1 eax + bc2 ebx ) + ab(c1 eax + c2 ebx ) = (a2 c1 − a2 c1 − abc1 + abc1 )eax + (b2 c2 − abc2 − b2 c2 + abc2 )ebx = 0. Hence, y (x) = c1 eax + c2 ebx is a solution to the given differential equation for all x ∈ R. 17. y (x) = eax (c1 + c2 x) =⇒ y = eax (c2 ) + aeax (c1 + c2 x) = eax (c2 + ac1 + ac2 x) =⇒ y = eaax (ac2 ) + aeax (c2 + ac1 + ac2 x) = aeax (2c2 + ac1 + ac2 x). Substituting these into the differential equation yields y − 2ay + a2 y = aeax (2c2 + ac1 + ac2 x) − 2aeax (c2 + ac1 + ac2 x) + a2 eax (c1 + c2 x) = aeax (2c2 + ac1 + ac2 x − 2c2 − 2ac1 − 2ac2 x + ac1 + ac2 x) = 0. Thus, y (x) = eax (c1 + c2 x) is a solution to the given differential eqaution for all x ∈ R. 18. y (x) = eax (c1 cos bx + c2 sin bx) so, y = eax (−bc1 sin bx + bc2 cos bx) + aeax (c1 cos bx + c2 sin bx) = eax [(bc2 + ac1 ) cos bx + (ac2 − bc1 ) sin bx] so, y = eax [−b(bc2 + ac1 ) sin bx + b(ac2 + bc1 ) cos bx] + aeax [(bc2 + ac1 ) cos bx + (ac2 + bc1 ) sin bx] = eax [(a2 c1 − b2 c1 + 2abc2 ) cos bx + (a2 c2 − b2 c2 − abc1 ) sin bx]. Substituting these results into the differential equation yields y − 2ay + (a2 + b2 )y = (eax [(a2 c1 − b2 c1 + 2abc2 ) cos bx + (a2 c2 − b2 c2 − abc1 ) sin bx]) − 2a(eax [(bc2 + ac1 ) cos bx + (ac2 − bc1 ) sin bx]) + (a2 + b2 )(eax (c1 cos bx + c2 sin bx)) = eax [(a2 c1 − b2 c1 + 2abc2 − 2abc2 − 2a2 c1 + a2 c1 + b2 c1 ) cos bx 2 2 2 2 2 + (a c2 − b c2 − 2abc1 + 2abc1 − 2a c2 + a c2 + b c2 ) sin bx] =0 Thus, y (x) = eax (c1 cos bx + c2 sin bx) is a solution to the given differential equation for all x ∈ R. 19. y (x) = erx =⇒ y = rerx =⇒ y = r2 erx . Substituting these results into the given differential equation yields erx (r2 + 2r − 3) = 0, so that r must satisfy r2 + 2r − 3 = 0, or (r + 3)(r − 1) = 0. Consequently r = −3 and r = 1 are the only values of r for which y (x) = erx is a solution to the given differential equation. The corresponding solutions are y (x) = e−3x and y (x) = ex . . 12 20. y (x) = erx =⇒ y = rerx =⇒ y = r2 erx . Substitution into the given differential equation yields erx (r2 − 8r + 16) = 0, so that r must satisfy r2 − 8r + 16 = 0, or (r − 4)2 = 0. Consequently the only value of r for which y (x) = erx is a solution to the differential equation is r = 4. The corresponding solution is y (x) = e4x . 21. y (x) = xr =⇒ y = rxr−1 =⇒ y = r(r − 1)xr−2 . Substitution into the given differential equation yields xr [r(r − 1) + r − 1] = 0, so that r must satisfy r2 − 1 = 0. Consequently r = −1 and r = 1 are the only values of r for which y (x) = xr is a solution to the given differential equation. The corresponding solutions are y (x) = x−1 and y (x) = x. 22. y (x) = xr =⇒ y = rxr−1 =⇒ y = r(r − 1)xr−2 . Substitution into the given differential equation yields xr [r(r − 1) + 5r + 4] = 0, so that r must satisfy r2 + 4r + 4 = 0, or equivalently (r + 2)2 = 0. Consequently r = −2 is the only value of r for which y (x) = xr is a solution to the given differential equation. The corresponding solution is y (x) = x−2 . 23. y (x) = 1 x(5x2 − 3) = 1 (5x3 − 3x) =⇒ y = 1 (15x2 − 3) =⇒ y = 15x. Substitution into the Legendre 2 2 2 equation with N = 3 yields (1 − x2 )y − 2xy + 12y = (1 − x2 )(15x) + x(15x2 − 3) + 6x(5x2 − 3) = 0. Consequently the given function is a solution to the Legendre equation with N = 3. 24. y (x) = a0 + a1 x + a2 x2 =⇒ y = a1 +2a2 x =⇒ y = 4a2 . Substitution into the given differential equation yields (1 − x2 )(2a2 ) − x(a1 +2a2 x)+4(a0 + a1 x + a2 x2 ) = 0 =⇒ 3a1 x +2a2 +4a0 = 0. For this equation to hold for all x we require 3a1 = 0, and 2a2 + 4a0 = 0. Consequently a1 = 0, and a2 = −2a0 . The corresponding solution to the differential equation is y (x) = a0 (1 − 2x2 ). Imposing the normalization condition y (1) = 1 requires that a0 = −1. Hence, the required solution to the differential equation is y (x) = 2x2 − 1. 25. x sin y − ex = c =⇒ x cos y 26. xy 2 + 2y − x = c =⇒ 2xy dy dy ex − sin y + sin y − ex = 0 =⇒ = . dx dx x cos y dy dy 1 − y2 dy + y2 + 2 − 1 = 0 =⇒ = . dx dx dx 2(xy + 1) dy dy 1 − yexy + y ] − 1 = 0 =⇒ xexy + yexy = 1 =⇒ . Given y (1) = 0 =⇒ dx dx xexy ln x e0(1) − 1 = c =⇒ c = 0. Therefore, exy − x = 0, so that y = . x 27. exy + x = c =⇒ exy [x x y/x 28. e 2 y/x + xy − x = c =⇒ e dy −y dy dy x2 (1 − y 2 ) + yey/x dx . + 2xy + y 2 − 1 = 0 =⇒ = 2 x dx dx x(ey/x + 2x2 y ) dy dy cos x − 2xy 2 1 + 2xy 2 − cos x = 0 =⇒ = . Since y (π ) = , then dx dx 2x2 y π 2 1 1 + sin x 1 π2 − sin π = c =⇒ c = 1. Hence, x2 y 2 − sin x = 1 so that y 2 = . Since y (π ) = , take the 2 π x π √ 1 + sin x branch of y where x < 0 so y (x) = . x 29. x2 y 2 − sin x = c =⇒ 2x2 y 30. dy = sin x =⇒ y (x) = − cos x + c for all x ∈ R. dx 13 31. dy = x−1/2 =⇒ y (x) = 2x1/2 + c for all x > 0. dx 32. d2 y dy = xex =⇒ = xex − ex + c1 =⇒ y (x) = xex − 2ex + c1 x + c2 for all x ∈ R. 2 dx dx d2 y = xn , where n is an integer. dx2 dy If n = −1 then = ln |x| + c1 =⇒ y (x) = x ln |x| + c1 x + c2 for all x ∈ (−∞, 0) or x ∈ (0, ∞). dx dy If n = −2 then = −x−1 + c1 =⇒ y (x) = c1 x + c2 − ln |x| for all x ∈ (−∞, 0) or x ∈ (0, ∞). dx dy xn+1 xn+2 If n = −1 and n = −2 then = + c1 =⇒ + c1 x + c2 for all x ∈ R. dx n+1 (n + 1)(n + 2) 33. dy = ln x =⇒ y (x) = x ln x − x + c. Since, y (1) = 2 =⇒ 2 = 1(0) − 1 + c =⇒ c = 3. Thus, dx y (x) = x ln x − x + 3. 34. d2 y dy = cos x =⇒ = sin x + c1 =⇒ y (x) = − cos x + c1 x + c2 . 2 dx dx Thus, y (0) = 1 =⇒ c1 = 1, and y (0) = 2 =⇒ c2 = 3. Thus, y (x) = 3 + x − cos x. 35. d2 y dy d3 y = 6x =⇒ 2 = 3x2 + c1 =⇒ = x3 + c1 x + c2 =⇒ y = 1 x4 + 1 c1 x2 + c2 x + c3 . 4 2 3 dx dx dx Thus, y (0) = 4 =⇒ c1 = 4, and y (0) = −1 =⇒ c2 = −1, and y (0) = 1 =⇒ c3 = 1. Thus, y (x) = 14 2 4 x + 2x − x + 1. 36. 37. y = xex =⇒ y = xex − ex + c1 =⇒ y = xex − 2ex + c1 x + c2 . Thus, y (0) = 4 =⇒ c1 = 5, and y (0) = 3 =⇒ c2 = 5. Thus, y (x) = xex − 2ex + 5x + 5. 38. Starting with y (x) = c1 ex + c2 e−x , we find that y (x) = c1 ex − c2 e−x and y (x) = c1 ex + c2 e−x . Thus, y − y = 0, so y (x) = c1 ex + c2 e−x is a solution to the differential equation on (−∞, ∞). Next we establish that every solution to the differential equation has the form c1 ex + c2 e−x . Suppose that y = f (x) is any solution to the differential equation. Then according to Theorem 1.2.15, y = f (x) is the unique solution to the initial-value problem y − y = 0, y (0) = f (0), y (0) = f (0). However, consider the function y (x) = f (0) + f (0) x f (0) − f (0) −x e+ e. 2 2 This is of the form y (x) = c1 ex + c2 e−x , where c1 = f (0)+f (0) and c2 = f (0)−f (0) , and therefore solves the 2 2 differential equation y − y = 0. Furthermore, evaluation this function at x = 0 yields y (0) = f (0) and y (0) = f (0). Consequently, this function solves the initial-value problem above. However, by assumption, y (x) = f (x) solves the same initial-value problem. Owing to the uniqueness of the solution to this initial-value problem, it follows that these two solutions are the same: f (x) = c1 ex + c2 e−x . 14 Consequently, every solution to the differential equation has the form y (x) = c1 ex + c2 e−x , and therefore this is the general solution on any interval I . dy d2 y = e−x =⇒ = −e−x + c1 =⇒ y (x) = e−x + c1 x + c2 . Thus, y (0) = 1 =⇒ c2 = 0, and 2 dx dx y (1) = 0 =⇒ c1 = − 1 . Hence, y (x) = e−x − 1 x. e e 39. d2 y dy = −6 − 4 ln x =⇒ = −2x − 4x ln x + c1 =⇒ y (x) = −2x2 ln x + c1 x + c2 . Since, y (1) = 0 =⇒ dx2 dx 2e2 2e2 c1 + c2 = 0, and since, y (e) = 0 =⇒ ec1 + c2 = 2e2 . Solving this system yields c1 = , c2 = − . e−1 e−1 2 2e (x − 1) − 2x2 ln x. Thus, y (x) = e−1 40. 41. y (x) = c1 cos x + c2 sin x (a) y (0) = 0 =⇒ 0 = c1 (1) + c2 (0) =⇒ c1 = 0. y (π ) = 1 =⇒ 1 = c2 (0), which is impossible. No solutions. (b) y (0) = 0 =⇒ 0 = c1 (1) + c2 (0) =⇒ c1 = 0. y (π ) = 0 =⇒ 0 = c2 (0), so c2 can be anything. Infinitely many solutions. 42-47. Use some kind of technology to define each of the given functions. Then use the technology to simplify the expression given on the left-hand side of each differential equation and verify that the result corresponds to the expression on the right-hand side. 48. (a) Use some form of technology to substitute y (x) = a + bx + cx2 + dx3 + ex4 + f x5 where a, b, c, d, e, f are constants, into the given Legendre equation and set the coefficients of each power of x in the resulting equation to zero. The result is: e = 0, 20f + 18d = 0, e + 2c = 0, 3d + 14b = 0, c + 15a = 0. 9 Now solve for the constants to find: a = c = e = 0, d = − 14 b, f = − 10 d = 3 corresponding solution to the Legendre equation is: y (x) = bx 1 − (−1)k 2 k=0 (k !) ∞ x 2 2k = 1 − 1 x2 + 4 14 64 x Consequently the 14 2 21 4 x+ x . 3 5 Imposing the normalization condition y (1) = 1 requires 1 = b(1 − 1 required solution is y (x) = 8 x(15 − 70x2 + 63x4 ). 49. (a) J0 (x) = 21 5 b. 14 3 + 21 5) =⇒ b = 15 8. Consequently the + ... (b) A Maple plot of J (0, x, 4) is given in the acompanying figure. From this graph, an approximation to the first positive zero of J0 (x) is 2.4. Using the Maple internal function BesselJZeros gives the approximation 2.404825558. (c) A Maple plot of the functions J0 (x) and J (0, x, 4) on the interval [0,2] is given in the acompanying figure. We see that to the printer resolution, these graphs are indistinguishable. On a larger interval, for example, [0,3], the two gaphs would begin to differ dramatically from one another. (d) By trial and error, we find the smallest value of m to be m = 11. A plot of the functions J (0, x) and J (0, x, 11) is given in the acompanying figure. 15 J(0, x, 4) 1 0.8 0.6 Approximation to the first positive zero of J0(x) 0.4 0.2 0 1 2 3 4 x –0.2 Figure 0.0.8: Figure for Exercise 49(b) J0(x), J(0, x, 4) 1 0.8 0.6 0.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x Figure 0.0.9: Figure for Exercise 49(c) Solutions to Section 1.3 True-False Review: 1. TRUE. This is precisely the remark after Theorem 1.3.2. 2. FALSE. For instance, the differential equation in Example 1.3.6 has no equilibrium solutions. 3. FALSE. This differential equation has equilibrium solutions y (x) = 2 and y (x) = −2. 4. TRUE. For this differential equation, we have f (x, y ) = x2 + y 2 . Therefore, any equation of the form x2 + y 2 = k is an isocline, by definition. 5. TRUE. Equilibrium solutions are always horizontal lines. These are always parallel to each other. 6. TRUE. The isoclines have the form is valid. x2 +y 2 2y = k , or x2 + y 2 = 2ky , or x2 + (y − k )2 = k 2 , so the statement 16 J(0, x), J(0, x, 11) 1 0.8 0.6 0.4 0.2 0 2 4 6 8 10 x J(0, x) –0.2 –0.4 J(0, x, 11) Figure 0.0.10: Figure for Exercise 49(d) 7. dy dx TRUE. An equilibrium solution is a solution, and two solution curves to the differential equation = f (x, y ) do not intersect. Problems: 1. y = cx−1 =⇒ c = xy . Hence, 2. y = cx2 =⇒ c = dy y = −cx−2 = −(xy )x−2 = − . dx x y dy y 2y . Hence, = 2cx = 2 2 x = . 2 x dx x x x2 + y 2 dy x2 + y 2 dy x2 + y 2 = c. Hence, 2x + 2y = 2c = , so that, y = − x. 2x dx x dx 2x 2 2 dy y −x = . Consequently, dx 2xy 3. x2 + y 2 = 2cx =⇒ 4. y 2 = cx =⇒ c = y2 dy dy c y . Hence, 2y = c, so that, = = . x dx dx 2y 2x 5. 2cy = x2 − c2 =⇒ c2 + 2cy − x2 = 0 =⇒ c = 6. y 2 − x2 = c =⇒ 2y −2y ± 4y 2 + 4x2 = −y ± 2 x2 + y 2 . dy dy x − 2x = 0 =⇒ =. dx dx y x2 + y 2 . Differentiating the given 2(x + y ) dy x2 + y 2 x2 + y 2 dy equation yields 2(x − c) + 2(y − c) = 0, so that 2 x − +2 y− = 0, that is dx 2(x + y ) 2(x + y ) dx dy x2 + 2xy − y 2 =− 2 . dx y + 2xy − x2 7. (x − c)2 + (y − c)2 = 2c2 =⇒ x2 − 2cx + y 2 − 2cy = 0 =⇒ c = 17 8. x2 + y 2 = c =⇒ 2x + 2y dy x dy = 0 =⇒ =− . dx dx y y(x) 2.0 1.6 1.2 0.8 0.4 x -2 -1 1 -0.4 2 -0.8 -1.2 -1.6 -2.0 Figure 0.0.11: Figure for Exercise 8 dy y 3y = 3cx2 = 3 3 x2 = . The initial condition y (2) = 8 =⇒ 8 = c(2)3 =⇒ c = 1. Thus the dx x x unique solution to the initial value problem is y = x3 . 9. y = cx3 =⇒ y(x) 8 (2, 8) 4 x -2 2 -4 -8 Figure 0.0.12: Figure for Exercise 9 18 dy dy y2 dy = c =⇒ 2y = =⇒ = y 2x =⇒ 2x · dy − y · dx = 0. The initial condition dx dx x dx y (1) = 2 =⇒ c = 4, so that the unique solution to the initial value problem is y 2 = 4x. 10. y 2 = cx =⇒ 2y y(x) 3 (1, 2) 1 x -3 -2 -1 1 2 3 -1 -3 Figure 0.0.13: Figure for Exercise 10 11. (x − c)2 + y 2 = c2 =⇒ x2 − 2cx + c2 + y 2 = c2 , so that x2 − 2cx + y 2 = 0. (0.0.1) Differentiating with respect to x yields 2x − 2c + 2y But from (0.0.1), c = dy = 0. dx x2 + y 2 which, when substituted into (0.0.2), yields 2x − 2x (0.0.2) x2 + y 2 x + 2y dy = 0, dx dy y 2 − x2 = . Imposing the initial condition y (2) = 2 =⇒ from (0.0.1) c = 2, so that the unique dx 2xy solution to the initial value problem is y = + x(4 − x). that is, 12. Let f (x, y ) = x sin (x + y ), which is continuous for all x, y ∈ R. ∂f = x cos (x + y ), which is continuous for all x, y ∈ R. ∂y dy By Theorem 1.3.2, = x sin (x + y ), y (x0 ) = y0 has a unique solution for some interval I ∈ R. dx dy x =2 (y 2 − 9), y (0) = 3. dx x +1 x (y 2 − 9), which is continuous for all x, y ∈ R. f (x, y ) = 2 x +1 ∂f 2xy =2 , which is continuous for all x, y ∈ R. ∂y x +1 13. 19 y(x) 3 (2, 2) 2 1 x 1 2 3 4 5 6 -1 -2 -3 Figure 0.0.14: Figure for Exercise 11 So the IVP stated above has a unique solution on any interval containing (0, 3). By inspection we see that y (x) = 3 is the unique solution. 14. The IVP does not necessarily have a unique solution since the hypothesis of the existence and uniqueness ∂f 1 theorem are not satisfied at (0,0). This follows since f (x, y ) = xy 1/2 , so that = 2 xy −1/2 which is not ∂y continuous at (0, 0). ∂f = −4xy . Both of these functions are continuous for all (x, y ), and therefore ∂y the hypothesis of the uniqueness and existence theorem are satisfied for any (x0 , y0 ). 1 2x (b) y (x) = 2 =⇒ y = − 2 = −2xy 2 . x +c (x + c)2 1 (c) y (x) = 2 . x +c 1 1 (i) y (0) = 1 =⇒ 1 = =⇒ c = 1. Hence, y (x) = 2 . The solution is valid on the interval (−∞, ∞). c x +1 1 1 (ii) y (1) = 1 =⇒ 1 = =⇒ c = 0. Hence, y (x) = 2 . This solution is valid on the interval (0, ∞). 1+c x 1 1 (iii) y (0) = −1 =⇒ −1 = =⇒ c = −1. Hence, y (x) = 2 . This solution is valid on the interval c x −1 (−1, 1). (d) Since, by inspection, y (x) = 0 satisfies the given IVP, it must be the unique solution to the IVP. 15. (a) f (x, y ) = −2xy 2 =⇒ ∂f = 2y − 1 are continuous at all points (x, y ). Consequently, the ∂y hypothesis of the existence and uniqueness theorem are satisfied by the given IVP for any x0 , y0 . (b) Equilibrium solutions: y (x) = 0, y (x) = 1. 16. (a) Both f (x, y ) = y (y − 1) and 20 y(x) 1.2 0.8 0.4 x -2 2 Figure 0.0.15: Figure for Exercise 15c(i) y(x) 8 6 4 2 x 1 2 3 4 5 6 Figure 0.0.16: Figure for Exercise 15c(ii) d2 y dy = (2y − 1) = (2y − 1)y (y − 1). Hence the 2 dx dx 1 solution curves are concave up for 0 < y < 2 , and y > 1, and concave down for y < 0, and 1 < y < 1. 2 (d) The solutions will be bounded provided 0 ≤ y0 ≤ 1. (c) Differentiating the given differential equation yields 17. y = 4x. There are no equalibrium solutions. The slope of the solution curves is positive for x > 0 and is negative for x < 0. The isoclines are the lines x = k . 4 Slope of Solution Curve -4 -2 0 2 4 Equation of Isocline x = −1 x = −1/2 x=0 x = 1/2 x=1 21 y(x) -1.0 -0.5 0.5 1.0 x -1 -2 -3 Figure 0.0.17: Figure for Exercise 15c(iii) y(x) 2 1 –2 –1 0 1 2 x –1 –2 Figure 0.0.18: Figure for Exercise 16(d) 1 18. y = x . There are no equalibrium solutions. The slope of the solution curves is positive for x > 0 and increases without bound as x → 0+ . The slope of the curve is negative for x < 0 and decreases without 1 bound as x → 0− . The isoclines are the lines x = k . 22 y(x) 1.5 1 0.5 –1.5 –1 0 –0.5 0.5 1 x 1.5 –0.5 –1 –1.5 Figure 0.0.19: Figure for Exercise 17 Slope of Solution Curve ±4 ±2 ±1/2 ±1/4 ±1/10 Equation of Isocline x = ±1/4 x = ±1/2 x = ±2 x = ±4 x = ±10 y(x) 2 1 –2 –1 0 1 2 x –1 –2 Figure 0.0.20: Figure for Exercise 18 19. y = x + y . There are no equalibrium solutions. The slope of the solution curves is positive for y > −x, and negative for y < −x. The isoclines are the lines y + x = k . Slope of Solution Curve −2 −1 0 1 2 Equation of Isocline y = −x − 2 y = −x − 1 y = −x y = −x + 1 y = −x + 2 23 Since the slope of the solution curve along the isocline y = −x − 1 coincides with the slope of the isocline, it follows that y = −x − 1 is a solution to the differential equation. Differentiating the given differential equation yields: y = 1 + y = 1 + x + y . Hence the solution curves are concave up for y > −x − 1, and concave down for y < −x − 1. Putting this information together leads to the slope field in the acompanying figure. y(x) 3 2 1 x -3 -2 1 -1 2 3 -1 -2 -3 Figure 0.0.21: Figure for Exercise 19 20. y = x . There are no equalibrium solutions. The slope of the solution curves is zero when x = 0. The y solution has a vertical tangent line at all points along the x-axis (except the origin). Differentiating the 1 x 1 x2 1 differential equation yields: y = − 2 y = − 3 = 3 (y 2 − x2 ). Hence the solution curves are concave y y y y y up for y > 0 and y 2 > x2 ; y < 0 and y 2 < x2 and concave down for y > 0 and y 2 < x2 ; y < 0 and y 2 > x2 . The isoclines are the lines x = k . y Slope of Solution Curve ±2 ±1 ±1/2 ±1/4 ±1/10 Equation of Isocline y = ±x/2 y = ±x y = ±2x y = ±4x y = ±10x Note that y = ±x are solutions to the differential equation. x 21. y = − 4y . Slope is zero when x = 0 (y = 0). The solutions have a vertical tangent line at all points x along the x-axis(except the origin). The isoclines are the lines − 4y = k . Some values are given in the table below. 24 y(x) 2 1 –2 –1 0 1 2 x –1 –2 Figure 0.0.22: Figure for Exercise 20 Slope of Solution Curve ±1 ±2 ±3 Equation of Isocline y = ±4x y = ±2x y = ±4x/3 4 4xy 4 16x2 4(y 2 + 4x2 ) + 2 =− − 3 =− . y y y y y Consequently the solution curves are concave up for y < 0, and concave down for y > 0. Putting this information together leads to the slope field in the acompanying figure. Differentiating the given differential equation yields: y = − 22. y = x2 y . Equalibrium solution: y (x) = 0 =⇒ no solution curve can cross the x-axis. Slope: zero when x = 0 or y = 0. Positive when y > 0 (x = 0), negative when y < 0 (x = 0). Differentiating the given dy d2 y = 2xy + x2 = 2xy + x4 y = xy (2+ x3 ). So, when y > 0, the solution curves differential equation yields: dx2 dx are concave up for x ∈ (−∞, (−2)1/3 ), and for x > 0, and are concave down for x ∈ ((−2)1/3 , 0). When y < 0, the solution curves are concave up for x ∈ ((−2)1/3 , 0), and concave down for x ∈ (−∞, (−2)1/3 ) and for x > 0. The isoclines are the hyperbolas x2 y = k . Slope of Solution Curve ±2 ±1 ±1/2 ±1/4 ±1/10 0 Equation of Isocline y = ±2/x2 y = ±1/x2 y = ±1/(2x)2 y = ±1/(4x)2 y = ±1/(10x)2 y=0 23. y = x2 cos y . The slope is zero when x = 0. There are equalibrium solutions when y = (2k + 1) π . The 2 π slope field is best sketched using technology. The accompanying figure gives the slope field for − π < y < 32 . 2 24. y = x2 + y 2 . The slope of the solution curves is zero at the origin, and positive at all the other points. There are no equilibrium solutions. The isoclines are the circles x2 + y 2 = k . 25 y(x) 4 3 2 1 x -2 -1 2 1 Figure 0.0.23: Figure for Exercise 21 y(x) 2 1 –2 –1 0 1 2 x –1 –2 Figure 0.0.24: Figure for Exercise 22 Slope of Solution Curve 1 2 3 4 5 Equation of Isocline x = ±1/4 x = ±1/2 x = ±2 x = ±4 x = ±10 26 y(x) 5 4 3 2 1 –3 –2 0 –1 1 2 x 3 –1 Figure 0.0.25: Figure for Exercise 23 y(x) 2 1 –2 –1 0 1 2 x –1 –2 Figure 0.0.26: Figure for Exercise 24 1 = − 80 (T − 70). Equilibrium solution: T (t) = 70. The slope of the solution curves is positive for d2 T 1 dT 1 T > 70, and negative for T < 70. =− = (T − 70). Hence the solution curves are concave 2 dt 80 dt 6400 1 up for T > 70, and concave down for T < 70. The isoclines are the horizontal lines − 80 (T − 70) = k . 25. dT dt Slope of Solution Curve −1/4 1/5 0 1/5 1/4 26. y = −2xy . 27. y = x sin x . 1 + y2 Equation of Isocline T = 90 T = 86 T = 70 T = 54 T = 50 27 T(t) 100 80 60 40 20 0 20 40 60 80 t 100 120 140 Figure 0.0.27: Figure for Exercise 25 y(x) 2 1 –2 –1 0 1 2 x –1 –2 Figure 0.0.28: Figure for Exercise 26 y(x) 2 1 –2 –1 0 1 2 x –1 –2 Figure 0.0.29: Figure for Exercise 27 28. y = 3x − y . 28 y(x) 2 1 –2 0 –1 1 2 x –1 –2 Figure 0.0.30: Figure for Exercise 28 29. y = 2x2 sin y . y(x) 3 2 1 –2 0 –1 1 2 x –1 –2 –3 Figure 0.0.31: Figure for Exercise 29 30. y = 2 + y2 . 3 + 0.5x2 31. y = 1 − y2 . 2 + 0.5x2 32.(a) Slope field for the differential equation y = x−1 (3 sin x − y ). (b) Slope field with solution curves included. The figure suggests that the solution to the differential equation are unbounded as x → 0+ . 6 (c) Slope field with solution curve corresponding to the initial condition y ( π ) = π . 2 This solution curve is bounded as x → 0+ . (d) In the accompanying figure we have sketched several solution curves on the interval (0,15]. The figure suggests that the solution curves approach the x-axis as x → ∞. 29 y(x) 2 1 –2 –1 0 1 x 2 –1 –2 Figure 0.0.32: Figure for Exercise 30 y(x) 2 1 –2 –1 0 1 x 2 –1 –2 Figure 0.0.33: Figure for Exercise 31 y(x) 4 2 0 2 4 6 8 10 x –2 –4 Figure 0.0.34: Figure for Exercise 32(a) 33.(a) Differentiating the given equation gives dy y = 2kx = 2 . Hence the differential equation of the dx x 30 y(x) 4 2 0 2 4 6 8 10 x –2 –4 Figure 0.0.35: Figure for Exercise 32(b) y(x) 4 2 0 2 4 6 8 10 x –2 –4 Figure 0.0.36: Figure for Exercise 32(c) y(x) 4 2 0 2 4 6 8 10 12 14 x –2 –4 Figure 0.0.37: Figure for Exercise 32(d) othogonal trajectories is dy x =− . dx 2y 31 y(x) 4 2 –4 –2 0 2 x 4 –2 –4 Figure 0.0.38: Figure for Exercise 33(a) (b) The orthogonal trajectories appear to be ellipses. This can be verified by intergrating the differential equation derived in (a). b 34. If a > 0, then as illustrated in the following slope field (a = 0.5, b = 1), it appears that limt→∞ i(t) = a . y(x) 4 2 –3 –2 –1 0 1 2 3 x –2 –4 Figure 0.0.39: Figure for Exercise 34 when a > 0 If a < 0, then as illustrated in the following slope field (a = −0.5, b = 1) it appears that i(t) diverges as t → ∞. If a = 0 and b = 0, then once more i(t) diverges as t → ∞. The acompanying figure shows a representative case when b > 0. Here we see that limt→∞ i(t) = +∞. If b < 0, then limt→∞ i(t) = −∞. If a = b = 0, then the general solution to the differential equation is i(t) = i0 where i0 is a constant. Solutions to Section 1.4 32 i(t) 4 2 –3 –2 –1 0 1 2 3 t –2 –4 Figure 0.0.40: Figure for Exercise 34 when a < 0 y(x) 4 2 –3 –2 –1 0 1 2 3 x –2 –4 Figure 0.0.41: Figure for Exercise 34 when a = 0 True-False Review: dy 1. TRUE. The differential equation dx = f (x)g (y ) can be written according to Definition 1.4.1, for a separable differential equation. 1 dy g (y ) dx = f (x), which is the proper form, 2. TRUE. A separable differential equation is a first-order differential equation, so the general solution contains one constant. The value of that constant can be determined from an initial condition, as usual. 3. TRUE. Newton’s Law of Cooling is usually expressed as dT dt = −k (T − Tm ), and this can be rewritten as 1 dT = −k, T − Tm dt and this form shows that the equation is separable. 4. FALSE. The expression x2 + y 2 cannot be separated in the form f (x)g (y ), so the equation is not separable. 5. FALSE. The expression x sin(xy ) cannot be separated in the form f (x)g (y ), so the equation is not separable. 33 dy 6. TRUE. We can write the given equation as e−y dx = ex , which is the proper form for a separable equation. dy 7. TRUE. We can write the given equation as (1 + y 2 ) dx = equation. x+4y 4x+y 8. FALSE. The expression 3 1 x2 , which is the proper form for a separable cannot be separated in the form f (x)g (y ), so the equation is not separable. 22 y 9. TRUE. We can write x x2+x y = xy , so we can write the given differential equation as +xy is the proper form for a separable equation. 1 dy y dx = x, which Problems: 1. Separating the variables and integrating yields dy =2 y 2 xdx =⇒ ln |y | = x2 + c1 =⇒ y (x) = cex . 2. Separating the variables and integrating yields y −2 dy = dx 1 =⇒ y (x) = − . −1 +1 tan x + c x2 3. Separating the variables and integrating yields ey dy = e−x dx = 0 =⇒ ey + e−x = c =⇒ y (x) = ln (c − e−x ). 4. Separating the variables and integrating yields dy = y (ln x)−1 dx =⇒ y (x) = c ln x. x 5. Separating the variables and integrating yields dx = x−2 dy =⇒ ln |x − 2| − ln |y | = c1 =⇒ y (x) = c(x − 2). y 6. Separating the variables and integrating yields dy = y−1 7. y − x 2x dx =⇒ ln |y − 1| = ln |x2 + 3| + c1 =⇒ y (x) = c(x2 + 3) + 1. x2 + 3 dy dy dy = 3 − 2x2 =⇒ x(2x − 1) = (3 − y ). Separating the variables and integrating yields dx dx dx − dy = y−3 dx dx 2 =⇒ − ln |y − 3| = − + dx x(2x − 1) x 2x − 1 =⇒ − ln |y − 3| = − ln |x| + ln |2x − 1| + c1 x cx − 3 =⇒ = c2 =⇒ y (x) = . (y − 3)(2x − 1) 2x − 1 dy cos (x − y ) dy cos x cos y 8. = − 1 =⇒ = =⇒ dx sin x sin y dx sin x sin y cos y = c csc x. sin y dy = cos y cos x dx =⇒ − ln | cos y | = ln | sin x| + c1 =⇒ cos y 34 dy x(y 2 − 1) = =⇒ dx 2(x − 2)(x − 1) 9. − 1 2 dy 1 + y+1 2 dy 1 = (y + 1)(y − 1) 2 dy 1 = y−1 2 2 dx − x−2 dx x−1 xdx , y = ±1. Thus, (x − 2)(x − 1) =⇒ − ln |y + 1|+ln |y − 1| = 2 ln |x − 2|−ln |x − 1|+c1 y−1 (x − 2)2 (x − 1) + c(x − 2)2 =c =⇒ y (x) = . By inspection we see that y (x) = 1, and y (x) = −1 y+1 x−1 (x − 1) − c(x − 2)2 are solutions of the given differential equation. The former is included in the above solution when c = 0. =⇒ x2 y − 32 dy x2 16 dy 1+ 2 dx =⇒ ln |y − 2| = = + 2 =⇒ = dx =⇒ ln |y − 2| = − 2 dx 16 − x y−2 16 − x2 x − 16 dx dx dx −x − 16 =⇒ ln |y − 2| = −x − 16 − 1 +1 =⇒ ln |y − 2| = −x + 2 ln |x + 4| − 8 x2 − 16 x+4 8 x−4 2 x+4 2 ln |x − 4| + c1 =⇒ y (x) = 2 + c e−x . x−4 10. dy dy dx dy 1 1 1 −(y −c) = 0 =⇒ = =⇒ = − dx =⇒ dx y−c (x − a)(x − b) y−c a−b x−a x−b 1/(a−b) 1/(a−b) 1/(a−b) x−b x−a x−a =⇒ (y − c) = c1 =⇒ y c = c2 =⇒ ln |y − c| = ln c1 x−b x−a x−b 11. (x−a)(x−b) y (x) = c + c2 x−a x−b 1/(a−b) . dy dy + y 2 = −1 =⇒ =− dx 1 + y2 π 1−x Thus, tan−1 y = tan−1 x + or y (x) = . 4 1+x 12. (x2 + 1) dx π =⇒ tan−1 y = tan−1 x + c, but y (0) = 1 so c = . 2 1+x 4 dy 2x dy 13. (1 − x2 ) + xy = ax =⇒ = −1 − dx =⇒ − ln |a − y | = − 1 ln |1 − x2 | + c1 =⇒ y (x) = 2 2 dx a−y 1 − x2 √ √ 2 , but y (0) = 2a so c = a and therefore, y (x) = a(1 + 2 ). a+c 1−x 1−x sin (x + y ) dy sin y sin x dy 14. = 1− =⇒ = − tan x cot y =⇒ − dy = dx =⇒ − ln | cos x cos y | = c, but dx sin x sin y dx cos y cos x π π y ( ) = so c = ln (2). Hence, − ln | cos x cos y | = ln (2) =⇒ y (x) = cos−1 1 sec x . 2 4 4 dy dy 1 = y 3 sin x =⇒ = sin xdx for y = 0. Thus − 2 = − cos x + c. However, we cannot impose the dx y3 2y initial condition y (0) = 0 on the last equation since it is not defined at y = 0. But, by inspection, y (x) = 0 is a solution to the given differential equation and further, y (0) = 0; thus, the unique solution to the initial value problem is y (x) = 0. 15. dy dy = 2 dx if y = 1 =⇒ 2(y − 1)1/2 = 2 x + c but y (1) = 1 so = 2 (y − 1)1/2 =⇒ 3 3 3 dx (y −√ 1/2 1) √ c = − 2 =⇒ 2 y − 1 = 2 x − 2 =⇒ y − 1 = 1 (x − 1). This does not contradict the Existance-Uniqueness 3 3 3 3 theorem because the hypothesis of the theorem is not satisfied when x = 1. 16. 35 m dv = mg − kv 2 =⇒ dv = dt. If we let a = mg then the preceding equation can k dt k [(mg/k ) − v 2 ] 1 m be written as dv = dt which can be integrated directly to obtain k a2 − v 2 17.(a) m m ln 2ak a+v a−v = t + c, that is, upon exponentiating both sides, 2ak a+v = c1 e m t . a−v Imposing the initial condition v (0) = 0, yields c = 0 so that 2ak a+v = e m t. a−v Therefore, v (t) = a e 2akt m −1 e 2akt m +1 which can be written in the equivalent form v (t) = a tanh gt . a (b) No. As t → ∞, v → a and as t → 0+ , v → 0. dy (c) v (t) = a tanh gt =⇒ = a tanh gt =⇒ a tanh a a dt a2 y (0) = 0 then y (t) = ln cosh ( gt ) . a g gt a dt =⇒ y (t) = a2 ln(cosh ( gt )) + c1 and if a g dy x = − , y (0) = 1 . Separating the variables in 2 dx 4y x2 the differential equation yields 4y −1 dy = −1dx, which can be integrated directly to obtain 2y 2 = − + c. 2 1 Imposing the initial condition we obtain c = 2 , so that the solution curve has the equation 2y 2 = −x2 + 1 , 2 or equivalently, 4y 2 + 2x2 = 1. 18. The required curve is the solution curve to the IVP dy 19. The required curve is the solution curve to the IVP = ex−y , y (3) = 1. Separating the variables in the dx differential equation yields ey dy = ex dx, which can be integrated directly to obtain ey = ex + c. Imposing the initial condition we obtain c = e − e3 , so that the solution curve has the equation ey = ex + e − e3 , or equivalently, y = ln(ex + e − e3 ). dy = x2 y 2 , y (−1) = 1. Separating the variables dx 1 in the differential equation yields y12 dy = x2 dx, which can be integrated directly to obtain − y = 1 x3 + c. 3 2 1 Imposing the initial condition we obtain c = − 3 , so that the solution curve has the equation y = − 1 x3 − 2 , 20. The required curve is the solution curve to the IVP or equivalently, y = 3 2−x3 . 3 3 36 1 dv = −dt. Integrating we 1 + v2 −1 −1 obtain tan (v ) = −t + c. The initial condition v (0) = v0 implies that c = tan (v0 ), so that tan−1 (v ) = −t + tan−1 (v0 ). The object will come to rest if there is time t, at which the velocity is zero. To determine tr , we set v = 0 in the previous equation which yields tan−1 (0) = tr +tan−1 (v0 ). Consequently, tr = tan−1 (v0 ). dv The object does not remain at rest since we see from the given differential equation that < 0 att = tr , dt and so v is decreasing with time. Consequently v passes through zero and becomes negative for t < tr . dv dx dv dv (b) From the chain rule we have = . Then = v . Substituting this result into the differential dt dt dx dx dv v equation (1.4.17) yields v = −(1 + v 2 ). We now separate the variables: dv = −dx. Integrating we dx 1 + v2 2 obtain ln (1 + v 2 ) = −2x + c. Imposing the initial condition v (0) = v0 , x(0) = 0 implies that c = ln (1 + v0 ), 2 2 so that ln (1 + v ) = −2x + ln (1 + v0 ). When the object comes to rest the distance traveled by the object 2 is x = 1 ln (1 + v0 ). 2 21.(a) Separating the variables in the given differential equation yields dv = −kv n =⇒ v −n dv = −kdt. dt 1 1 n = 1 :=⇒ v 1−n = −kt + c. Imposing the initial condition v (0) + v0 yields c = v 1−n , so that 1−n 1−n 0 1 v = [v0 −n + (n − 1)kt]1/(1−n) . The object comes to rest in a finite time if there is a positive value of t for which v = 0. n = 1 :=⇒ Integratingv −n dv = −kdt and imposing the initial conditions yields v = v0 e−kt , and the object does not come to rest in a finite amount of time. dx 1 (b) If n = 1, 2, then = [v0 −n + (n − 1)kt]1/(1−n) , where x(t) denotes the distanced traveled by the object. dt 1 Consequently, x(t) = − [v 1−n + (n − 1)kt](2−n)/(1−n) + c. Imposing the initial condition x(0) = 0 k (2 − n) 0 1 1 1 v 2−n , so that x(t) = − [v 1−n + n(n − 1)kt](2−n)/(1−n) + v 2−n . For yields c = k (2 − n) 0 k (2 − n) o k (2 − n) 0 2−n 1 < 0, so that limt→∞ x(t) = . Hence the maximum distance that the 1 < n < 2, we have 1−n k (2 − n) 1 object can travel in a finite time is less than . k (2 − n) v0 If n = 1, then we can integrate to obtain x(t) = (1 − e−kt ), where we have imposed the initial condition k v0 x(0) = 0. Consequently, limt→∞ x(t) = . Thus in this case the maximum distance that the object can k v0 travel in a finite time is less than . k 1 1 (c) If n > 2, then x(t) = − [v 1−n + n(n − 1)kt](2−n)/(1−n) + v 2−n is still valid. However, k (2 − n) o k (2 − n) 0 2−n in this case > 0, and so limt→∞ x(t) = +∞. COnsequently, there is no limit to the distance that the 1−n object can travel. dx 1 − If n = 2, then we return to v = [v0 −n + (n − 1)kt]1/(1−n) . In this case = (v0 1 + kt)−1 , which can dt 1 be integrated directly to obtain x(t) = ln (1 + v0 kt), where we have imposed the initial condition that k x(0) = 0. Once more we see that limt→∞ x(t) = +∞, so that there is no limit to the distance that the object can travel. 22.(a) 37 ρ 1/γ p ) . Consequently the given differential equation can be written as dp = −gρ0 ( )1/γ dy , ρ0 p0 gρ0 γp(γ −1)/γ gρ0 y or equivalently, p−1/γ dp = − 1/γ dy . This can be integrated directly to obtain = − 1/γ + c. At γ−1 p0 p0 the center of the Earth we have p = p0 . Imposing this initial condition on the preceding solution gives (γ −1)/γ γp0 c= . Substituting this value of c into the general solution to the differential equation we find, γ−1 (γ −1)/γ (γ − 1)ρ0 gy (γ − 1)ρ0 gy (γ −1)/γ after some simplification, p(γ −1)/γ = p0 1− , so that p = p0 1 − . γp0 γp0 23. Solving p = p0 ( dT dT dT = −k (T − Tm ) =⇒ = −k (T − 75) =⇒ = −kdt =⇒ ln |T − 75| = −kt + c1 =⇒ T (t) = dt dt T − 75 75 + ce−kt . T (0) = 135 =⇒ c = 60 so T = 75 + 60e−kt . T (1) = 95 =⇒ 95 = 75 + 60e−k =⇒ k = ln 3 =⇒ T (t) = 75 + 60e−t ln 3 . Now if T (t) = 615 then 615 = 75 + 60−t ln 3 =⇒ t = −2h. Thus the object was placed in the room at 2p.m. 24. dT = −k (T − 450) =⇒ T (t) = 450 + Ce−kt .T (0) = 50 =⇒ C = −400 so T (t) = 450 − 400e−kt and dt 1 T (20) = 150 =⇒ k = ln 4 ; hence, T (t) = 450 − 400( 3 )t/20 . 4 20 3 3 (i) T (40) = 450 − 400( 4 )2 = 225◦ F. 20 ln 4 (ii) T (t) = 350 = 450 − 400( 3 )t/20 =⇒ ( 3 )t/20 = 1 =⇒ t = ≈ 96.4 minutes. 4 4 4 ln(4/3) 25. dT dT = −k (T − 34) =⇒ = −kdt =⇒ T (t) = 34 + ce−kt . T (0) = 38 =⇒ c = 4 so that dt T − 34 T (t) = 34 + 4e−kt .T (1) = 36 =⇒ k = ln 2; hence, T (t) = 34 + 4e−t ln 2 . Now T (t) = 98 =⇒ T (t) = 34 + 4e−kt = 98 =⇒ 2−t = 16 =⇒ t = −4h. Thus T (−4) = 98 and Holmes was right, the time of death was 10 a.m. 26. 27. T (t) = 75 + ce−kt . T (10) = 415 =⇒ 75 + ce−10k = 415 =⇒ 340 = ce−10k and T (20) = 347 =⇒ 1 75 + ce−20k = 347 =⇒ 272 = ce−20k . Solving these two equations yields k = 10 ln 5 and c = 425; hence, 4 T = 75 + 425( 4 )t/10 5 (a) Furnace temperature: T (0) = 500◦ F. 10 ln 17 4 (b) If T (t) = 100 then 100 = 75 + 425( 5 )t/10 =⇒ t = ≈ 126.96 minutes. Thus the temperature of ln 5 4 the coal was 100◦ F at 6:07 p.m. dT dT dT = −k (T − 72) =⇒ = −kdt =⇒ T (t) = 72 + ce−kt . Since = −20, −k (T − 72) = −20 or dt T − 72 dt 10 −10/39 10/39 =⇒ c = 78e ; consequently, T (t) = 72 + 78e10(1−t)/39 . k = 39 . Since T (1) = 150 =⇒ 150 = 72 + ce (i) Initial temperature of the object: t = 0 =⇒ T (t) = 72 + 78e10/30 ≈ 173◦ F dT (ii)Rate of change of the temperature after 10 minutes: T (10) = 72 + 78e−30/13 so after 10 minutes, = dt 10 dT 260 −30/13 − (72 + 78e−30/13 − 72) =⇒ =− e ≈ 2◦ F per minute. 39 dt 13 28. Solutions to Section 1.5 38 True-False Review: 1. TRUE. The differential equation for such a population growth is dP = kP , where P (t) is the population dt as a function of time, and this is the Malthusian growth model described at the beginning of this section. 2. FALSE. The initial population could be greater than the carrying capacity, although in this case the population will asymptotically decrease towards the value of the carrying capacity. 3. TRUE. The differential equation governing the logistic model is (1.5.2), which is certainly separable as D dP = r. P (C − P ) dt Likewise, the differential equation governing the Malthusian growth model is 1 as P dP = k . dt dP dt = kP , and this is separable 4. TRUE. As (1.5.3) shows, as t → ∞, the population does indeed tend to the carrying capacity C independently of the initial population P0 . As it does so, its rate of change dP slows to zero (this is best dt seen from (1.5.2) with P ≈ C ). 5. TRUE. Every five minutes, the population doubles (increase 2-fold). Over 30 minutes, this population will double a total of 6 times, for an overall 26 = 64-fold increase. 6. TRUE. An 8-fold increase would take 30 years, and a 16-fold increase would take 40 years. Therefore, a 10-fold increase would take between 30 and 40 years. 7. FALSE. The growth rate is constant. dP dt = kP , and so as P changes, dP dt changes. Therefore, it is not always 8. TRUE. From (1.5.2), the equilibrium solutions are P (t) = 0 and P (t) = C , where C is the carrying capacity of the population. 9. FALSE. If the initial population is in the interval ( C , C ), then although it is less than the carrying 2 capacity, its concavity does not change. To get a true statement, it should be stated instead that the initial population is less than half of the carrying capacity. 10. TRUE. Since P (t) = kP , then P (t) = kP (t) = k 2 P > 0 for all t. Therefore, the concavity is always positive, and does not change, regardless of the initial population. Problems: dP 1. = kP =⇒ P (t) = P0 ekt . Since P (0) = 10, then P = 10ekt . Since P (3) = 20, then 2 = e3k =⇒ k = dt ln 2 . Thus P (t) = 10e(t/3) ln 3 . Therefore, P (24) = 10e(24/3) ln 3 = 10 · 28 = 2560 bacteria. 3 2. Using P (t) = P0 ekt we obtain P (10) = 5000 =⇒ 5000 = p0 e10k and P (12) = 6000 =⇒ 6000 = P0 e12k 6 which implies that e2k = 5 =⇒ k = 1 ln 6 . Hence, P (0) = 5000( 5 )5 = 2009.4. Also, P = 2P0 when 2 5 6 2 ln 2 1 t = 2 ln 2 = ≈ 7.6h. ln 6 5 3. From P (t) = P0 ekt and P (0) = 2000 it follows that P (t) = 2000ekt . Since td = 4, k = P = 2000et ln 2/4 . Therefore, P (t) = 106 =⇒ 106 = 2000et ln 2/4 =⇒ t ≈ 35.86h. 4. 1 4 ln 2 so dP = kP =⇒ P (t) = P0 ekt . Since, P (0) = 10000 then P (t) = 10000ekt . Since P (5) = 20000 then dt 39 20000 = 10000ekt =⇒ k = 1 ln 2. Hence P (t) = 10000e(t ln 2)/5 . 5 (a) P (20) = 10000e4 ln 2 = 160000. 5 ln 100 (b) 1000000 = 10000e(t ln 2)/5 =⇒ 100 = e(t ln 2)/5 =⇒ t = ≈ 33.22yrs. ln 2 50C . In formulas (1.5.5) and (1.5.6) we have P0 = 500, P1 = 800, P2 = 1000, t1 = 5, 50 + (C − 50)e−rt (1000)(300) 1 800[(800)(1500) − 2(500)(1000)] 1 and t2 = 10. Hence, r = ln = ln 3, C = ≈ 1142.86, so 5 (500)(200) 5 8002 − (500)(1000) 1142.86)(500) 571430 that P (t) = ≈ . Inserting t = 15 into the preceding formula −0.2t ln 3 500 + 642.86e 500 + 642.86e−0.2t ln 3 yields P (15) = 1091. 5. P (t) = 50C In formulas (1.5.5) and (1.5.6) we have P0 = 50, P1 = 62, P2 = 76, t1 = 2, 50 + (C − 50)e−rt 1 (76)(12) 62[(62)(126) − 2(50)(76)] and t2 = 2t1 = 4. Hence, r = ln ≈ 0.132, C = ≈ 298.727, so that 2 (50)(14) 622 − (50)(76) 14936.35 P (t) = . Inserting t = 20 into the preceding formula yields P (20) ≈ 221. 50 + 248.727e−0.132t 6. P (t) = P2 (P1 − P0 ) > 1. Rearranging the terms in this inequality and P0 (P2 − P1 ) 2P0 P2 P1 (P0 + P2 ) − 2P0 P2 using the fact that P2 > P1 yields P1 > . Further, C > 0 requires that > 0. 2 P0 + P2 P1 − P0 P2 2P0 P2 From P1 > we see that the numerator in the preceding inequality is positive, and therefore the P0 + P2 2P0 P2 2 denominator must also be positive. Hence in addition to P1 > , we must also have P1 > P0 P2 . P0 + P2 7. From equation (1.5.5) r > 0 requires dy = dt ky (1500 − y ), y (0) = 5, y (1) = 10, where k is a positive constant. Separating the differential equation 1 and integrating yields dy = k dt. Using a partial fraction decomposition on the left-hand y (1500 − y ) 1 1 1 y side gives + dy = kt + c, so that ln = kt + c, which upon 1500y 1500(1500 − y ) 1500 1500 − y y 1 exponentiation yields = c1 e1500kt . Imposing the initial condition y (0) = 5, we find that c1 = . 1500 − y 299 y 1 1500kt 10 1 1500k Hence, = e . The further condition y (1) = 10 requires = e . solving 1500 − y 299 1490 299 1 299 y 1 t ln (299/149) for k gives k = ln . Therefore, = e . Solving algebraically for y we find 1500 149 1500 − y 299 1500et ln (299/149) 1500 1500 y (t) = = . Hence, y (14) = = 1474. 299 + et ln (299/149) 1 + 299e−t ln (299/149) 1 + 299e−14 ln (299/149) 8. Let y (t) denote the number of passengers who have the flu at time t. Then we must solve 9. (a) Equilibrium solutions: P (t) = 0, P (t) = T . dP dP Slope: P > T =⇒ > 0, 0 < P < T =⇒ < 0. dt dt 40 Isoclines: r(P − T ) = k =⇒ P 2 − T P − k 1 = 0 =⇒ P = r 2 T± rT 2 + 4k . We see that slope of the r −rT 2 . 4 2 dP dP Concavity: = r(2P − T ) = r2 (2P − T )(P − T )P . Hence, the solution curves are concave up for dt2 dt T t P > , and are concave down for 0 < P < . 2 2 (b) See accompanying figure. solution curves satisfies k ≥ P(t) T 0 t Figure 0.0.42: Figure for Exercise 9(b) (c) For 0 < P0 < T , the population dies out with time. For P0 > T , there is a population growth. The term threshold level is appropriate since T gives the minimum value of P0 above which there is a population growth. 1 dP = r, which can be written P (P − T ) dt 1 1 dP 1 P −T in the equivalent form − = r. Integrating yields ln = rt + c, so that T (P − T ) T P dt T P P −T P0 − T P −T P0 − T = c1 eT rt . The initial condition P (0) = P0 requires = c1 , so that = erT t . P P0 P P0 T P0 . Solving algebraically for P yields P (t) = P0 − (P0 − T )erT t T P0 (b) If P0 < T , then the denominator in is positive, and increases without bound as P0 − (P0 − T )erT t t → ∞. Consequently limt→∞ P (t) = 0. In this case the population dies out as t increases. T P0 (c) If P0 > T , then the denominator of vanishes when (P0 − T )erT t = P0 , that is when P0 − (P0 − T )erT t 1 P0 t= ln . This means that within a finite time the population grows without bound. We can rT P0 − T interpret this as a mathematical model of a population explosion. 10. (a) Separating the variables in differential equation (1.5.7) gives dP = r(C − P )(P − T )P, P (0) = P0 , r > 0, 0 < T < C . dt Equilibrium solutions: P (t) = 0, P (t) = T, P (t) = C . The slope of the solution curves is negative for 0 < P < T , and for P > C . It is positive for T < P < C . 11. 41 Concavity: d2 P = r2 [(C − P )(P − T ) − (P − T )P + (C − P )P ](C − P )(P − T )P , which simplifies to dt2 d2 P = r2 (−3P 2 + 2P T + 2CP − CT )(C − P )(P − T ). Hence changes in concavity occur when P = 1 (C + 3 dt2 √ T ± C 2 − CT + T 2 ). A representative slope field with some solution curves is shown in the acompanying figure. We see that for 0 < P0 < T the population dies out, whereas for T < P0 < C the population grows and asymptotes to the equilibrium solution P (t) = C . If P0 > C , then the solution decays towards the equilibrium solution P (t) = C . P(t) C T t 0 Figure 0.0.43: Figure for Exercise 11 dP = rP (ln C − ln P ), P (0) = P0 , and r, C , and P0 are positive constants. dt Equilibrium solutions: P (t) = C . The slope of the solution curves is positive for 0 < P < C , and negative for P > C . d2 P C dP C C Concavity: = r ln −1 = r2 ln − 1 P ln . Hence, the solution curves are concave dt2 P dt P P C C and P > C . They are concave down for < P < C . A representative slope field with up for 0 < P < e e some solution curves are shown in the acompanying figure. 12. 25 20 15 P(t) 10 5 0 1 2 t 3 4 5 Figure 0.0.44: Figure for Exercise 12 42 dP 1 = r which can be integrated directly to P (ln C − ln P ) dt obtain − ln (ln C − ln P ) = rt + c so that ln ( C ) = c1 e−rt . The initial condition P (0) = P0 requires that P −rt C C ln ( P0 ) = c1 . Hence, ln ( C ) = e−rt ln ( P0 ) so that P (t) = Celn (P0 /k)e . Since limt→∞ e−rt = 0, it follows P that limt→∞ P (t) = C . 13. Separating the variables in (1.5.8) yields dP = kP , which is easily integrated to obtain P (t) = P0 ekt . dt The initial condition P (0) = 400 requires that P0 = 400, so that P (t) = 400ekt . We also know that 1 17 P (30) = 340. This requires that 340 = 400e30k so that k = ln . Consequently, 30 20 14. Using the exponential decay model we have P (t) = 400e 30 ln( 20 ) . t 17 (0.0.3) 17 (a) From (0.0.3), P (60)400e2 ln( 20 ) = 289. 10 17 (b) From (0.0.3), P (100) = 400e 3 ln( 20 ) ≈ 233 (c) From (0.0.3), the half-life, tH , is determine from tH 200 = 400e 30 ln( 17 ) 20 =⇒ tH = 30 ln 2 ≈ 128 days. ln 20 17 15. (a) More. dP = kP , which is easily integrated to obtain P (t) = P0 ekt . dt The initial condition P (0) = 100, 000 requires that P0 = 100, 000, so that P (t) = 100, 000ekt . We also know 4 1 that P (10) = 80, 000. This requires that 100, 000 = 80, 000e10k so that k = ln . Consequently, 10 5 (b) Using the exponential decay model we have P (t) = 100, 000e 10 ln( 5 ) . t 4 (0.0.4) Using (0.0.4), the half-life is determined from tH 50, 000 = 100, 000e 10 ln( 4 ) 5 =⇒ tH = 10 ln 2 ≈ 31.06 min. ln 5 4 (c) Using (0.0.4) there will be 15,000 fans left in the stadium at time t0 , where t0 15, 000 = 100, 000e 10 ln( 5 ) =⇒ t0 = 10 4 ln ln 3 20 4 15 ≈ 85.02 min. dP 16. Using the exponential decay model we have = kP , which is easily integrated to obtain P (t) = P0 ekt . dt Since the half-life is 5.2 years, we have 1 ln 2 P0 = P0 e5.2k =⇒ k = − . 2 5.2 Therefore, ln 2 P (t) = P0 e−t 5.2 . 43 Consequently, only 4% of the original amount will remain at time t0 where ln 2 4 ln 25 P0 = P0 e−t0 5.2 =⇒ t0 = 5.2 ≈ 24.15 years. 100 ln 2 17. Maple, or even a TI 92 plus, has no problem in solving these equations. 18. (a) Malthusian model is P (t) = 151.3ekt . Since P (1) = 179.4, then 179.4 = 151.3e10k =⇒ k = t ln (179.4/151.1) 10 1 10 ln 179.4 . 151.3 Hence, P (t) = 151.3e . 151.3C (b)P (t) = . Imposing the initial conditions P (10) = 179.4 and P (20) = 203.3 151.3 + (C − 151.3)e−rt 20 ln (179.4/151.1) 10 ln (179.4/151.1) 10 10 and 203.3 = 151.3e whose solution is gives the pair of equations 179.4 = 151.3e 39935.6 C ≈ 263.95, r ≈ 0.046. Using these values for C and r gives P (t) = . 151.3 + 112.65e−0.046t (c) Malthusian model: P (30) ≈ 253 million; P (40) ≈ 300 million. P(t) 320 300 280 260 240 220 200 180 160 0 10 20 30 40 t Figure 0.0.45: Figure for Exercise 18(c) Logistics model: P (30) ≈ 222 million; P (40) ≈ 236 million. The logistics model fits the data better than the Malthusian model, but still gives a significant underestimate of the 1990 population. 50C . Imposing the conditions P (5) = 100, P (15) = 250 gives the pair of equations 50 + (C − 50)e−rt 50C 50C 100 = and 250 = whose positive solutions are C ≈ 370.32, r ≈ 0.17. 50 + (C − 50)e−5r 50 + (C − 50)e−15r 18500 Using these values for C and r gives P (t) = . From the figure we see that it will take 50 + 18450e−0.17t approximately 52 years to reach 95% of the carrying capacity. 19. P (t) = Solutions to Section 1.6 True-False Review: 1. FALSE. Any solution to the differential equation (1.6.7) serves as an integrating factor for the differential 44 P(t) Carrying Capacity 350 300 250 200 150 100 50 0 t 10 20 30 40 50 60 70 Figure 0.0.46: Figure for Exercise 19 equation. There are infinitely many solutions to (1.6.7), taking the form I (x) = c1 e arbitrary constant. p(x)dx , where c1 is an 2. TRUE. Any solution to the differential equation (1.6.7) serves as an integrating factor for the differential equation. There are infinitely many solutions to (1.6.7), taking the form I (x) = c1 e p(x)dx , where c1 is an arbitrary constant. The most natural choice is c1 = 1, giving the integrating factor I (x) = e p(x)dx . 3. TRUE. Multiplying y + p(x)y = q (x) by I (x) yields y I + pIy = qI . Assuming that I = pI , the requirement on the integrating factor, we have y I + I y = qI , or by the product rule, (I · y ) = qI , as requested. 4. FALSE. Before determining an integrating factor, the equation must be rewritten as dy − x2 y = sin x, dx and with p(x) = −x2 , we have integrating factor I (x) = e −x2 dx , not e x2 dx . 5. TRUE. Rewriting the differential equation as dy 1 + y = x, dx x 1 we have p(x) = x , and so an integrating factor must have the form I (x) = e Since 5x does indeed have this form, it is an integrating factor. Problems: In this section the function I (x) = e of the form y + p(x)y = q (x). 1. y − y = e2x . I (x) = e− dx p(x)dx = e−x =⇒ p(x)dx =e dx x = eln x+c = ec x. will represent the integrating factor for a differential equation d(e−x y ) = ex =⇒ e−x y = ex + c =⇒ y (x) = ex (ex + c). dx 4 2. x2 y − 4xy = x7 sin x, x > 0 =⇒ y − x y = x5 sin x. I (x) = x−4 =⇒ sin x − x cos x + c =⇒ y (x) = x4 (sin x − x cos x + c). d(x−4 y ) = x sin x =⇒ x−4 y = dx 45 3. y + 2xy = 2x3 . I (x) = e2 xdx 2 = ex =⇒ 2 2 d x2 2 2 (e y ) = 2ex x3 =⇒ ex y = 2 dx 2 2 ex x3 dx =⇒ ex y = ex (x2 − 1) + c =⇒ y (x) = x2 − 1 + ce−x . 2x 1 d y = 4x, −1 < x < 1. I (x) = =⇒ 2 2 1−x 1−x dx c =⇒ y (x) = (1 − x2 )[− ln (1 − x2 )2 + c]. 4. y + 5. y + 4 2x 4 y= . I (x) = e 1 + x2 (1 + x2 )2 2x 1+x2 dx y 1 − x2 = 1 + x2 =⇒ = 4x y =⇒ = − ln(1 − x2 )2 + 2 1−x 1 − x2 4 d [(1 + x2 )y ] = =⇒ (1 + x2 )y = dx (1 + x2 )2 dx 1 =⇒ (1 + x2 )y = 4 tan−1 x + c =⇒ y (x) = (4 tan−1 x + c). 1 + x2 1 + x2 dy sin 2x 1 d + y sin 2x = 4 cos4 x, 0 ≤ x ≤ π =⇒ y + y = 2 cos2 x. I (x) = =⇒ (y sec x) = 2 2x dx 2 cos cos x dx cos x =⇒ y (x) = cos x(2 sin x + c) =⇒ y (x) = sin 2x + c cos x. 6. 2 cos2 x dx d 1 y = 9x2 . I (x) = e x ln x = ln x =⇒ (y ln x) = 9 x2 ln xdx =⇒ y ln x = 3x3 ln x − x3 + c =⇒ x ln x dx x3 (3 ln x − 1) 3x3 ln x − x3 + c but y (e) = 2e3 so c = 0; thus, y (x) = . y (x) = ln x ln x 7. y + d (y cos x) = 8 cos x sin3 x =⇒ y cos x = 8 cos x sin3 xdx + c =⇒ 8. y − y tan x = 8 sin3 x. I (x) = cos x =⇒ dx 1 y cos x = 2 sin4 x + c =⇒ y (x) = (2 sin4 x + c). cos x dx 2 4et + 2x = 4et =⇒ x + x = . I (x) = e2 dt t t 4et (t − 1) + c . t2 x = 4et (t − 1) + c =⇒ x(t) = t2 9. t 10. dt t = t2 =⇒ d2 (t x) = 4tet =⇒ t2 x = 4 tet dt + c =⇒ dt y = (sin x sec x)y − 2 sin x =⇒ y − (sin x sec x)y = −2 sin x. I (x) = cos x =⇒ −2 sin x cos x =⇒ y cos x = −2 sin x cos xdx + c = 1 1 cos 2x + c =⇒ y (x) = 2 cos x 1 cos 2x + c . 2 d (y sec x) = dx 2 sec xdx + c =⇒ y sec x = tan x + c =⇒ y (x) = cos x(tan x + c) =⇒ y (x) = sin x + c cos x. 11. (1 − y sin x)dx − cos xdy = 0 =⇒ y + (sin x sec x)y = sec x. I (x) = e sec2 x =⇒ y sec x = d (y cos x) = dx sin x sec xdx = sec x =⇒ 1 d −1 12. y − x−1 y = 2x2 ln x. I (x) = e− x dx = x−1 =⇒ (x y ) = 2x ln x =⇒ x−1 y = 2 x ln xdx + c =⇒ dx 1 1 x−1 y = x2 (2 ln x − 1) + c. Hence, y (x) = x3 (2 ln x − 1) + cx. 2 2 d αx (e y ) = e(α+β )x =⇒ eαx y = e(α+β )x dx + c. If α + β = 0, dx e(α+β )x eβx then eαx y = x + c =⇒ y (x) = e−αx (x + c). If α + β = 0, then eαx y = + c =⇒ y (x) = + ce−αx . α+β α+β 13. y + αy = eβx . I (x) = eα dx = eαx =⇒ 46 dm m y = ln x. I (x) = xm =⇒ (x y ) = xm ln x =⇒ xm y = xm ln xdx + c. If m = −1, then x dx (ln x)2 xm+1 xm+1 (ln x)2 xm y = + c =⇒ y (x) = x + c . If m = −1, then xm y = ln x − + c =⇒ y (x) = 2 2 m+1 (m + 1)2 x x c ln x − + m. m+1 (m + 1)2 x 14. y + 2 15. y + y = 4x. I (x) = e2 x dx x = e2 ln x = x2 =⇒ but y (1) = 2 so c = 1; thus, y (x) = x4 + 1 . x2 d2 (x y ) = 4x3 =⇒ x2 y = 4 dx x3 dx + c =⇒ x2 y = x4 + c, d (y csc x) = 2 csc x cos x =⇒ dx π y csc x = 2 ln (sin x) + c, but y ( 2 ) = 2 so c = 2; thus, y (x) = 2 sin x[ln (sin x) + 1]. 16. y sin x − y cos x = sin 2x =⇒ y − y cot x = 2 cos x. I (x) = csc x =⇒ dt 2 d x = 5. I (t) = e2 4−t = e−2 ln(4−t) = (4 − t)t−2 =⇒ ((4 − t)−2 x) = 5(4 − t)−2 =⇒ (4 − t)−2 x = 4−t dt 5 (4 − t)−2 dt + c =⇒ (4 − t)−2 x = 5(4 − t)−1 + c, but x(0) = 4 so c = −1; thus, x(t) = (4 − t)2 [5(4 − t)−1 − 1] or x(t) = (4 − t)(1 + t). 17. x + e2x dx 18. (y − e−x )dx + dy = 0 =⇒ y + y = ex . I (x) = ex =⇒ (e y ) = e2x =⇒ ex y = + c, but y (0) = 1 so dx 2 1 1 c = ; thus, y (x) = (ex + e−x ) = cosh x. 2 2 19. y + y = f (x), y (0) = 3, f (x) = 1, if x ≤ 1, 0, if x > 1. dx x (e y ) = ex f (x) =⇒ [ex y ]x = 0 ex f (x)dx =⇒ ex y − y (0) = 0 dx x x ex y − 3 = 0 ex dx =⇒ y (x) = e−x 3 + 0 ex f (x)dx . xx xx x If x ≤ 1, 0 e f (x)dx = 0 e dx = e − 1 =⇒ y (x) = e−x (2 + ex ) x x If x > 1, 0 ex f (x)dx = 0 ex dx = e − 1 =⇒ y (x) = e−x (2 + e). I (x) = e dx = ex =⇒ xx e f (x)dx 0 =⇒ 20. y − 2y = f (x), y (0) = 1, f (x) = 1 − x, if x < 1, 0, if x ≥ 1. d −2x x (e y ) = e−2x f (x) =⇒ [e−2x y ]x = 0 e−2x f (x)dx =⇒ e−2x y − y (0) = 0 dx x x −2x x e f (x)dx =⇒ e−2x y − 1 = 0 e−2x f (x) =⇒ y (x) = e2x 1 + 0 e−2x f (x)dx . 0 1 1 x x If x < 1, 0 e−2x f (x)dx = 0 e−2x (1 − x)dx = e−2x (2x−1+e2x ) =⇒ y (x) = e2x 1 + e−2x (2x − 1 + e2x ) = 4 4 1 2x (5e + 2x − 1). 4 1 1 1 x x If x ≥ 1, 0 e−2x f (x)dx = 0 e−2x (1 − x)dx = (1 + e−2 ) =⇒ y (x) = e2x 1 + (1 + e−2 ) = e2x (5 + e−2 ). 4 4 4 I (x) = e− 2dx = e−2x =⇒ 21. On (−∞, 1), y − y = 1 =⇒ I (x) = e−x =⇒ y (x) = c1 ex − 1. Imposing the initial condition y (0) = 0 requires c1 = 1, so that y (x) = ex − 1, for x < 1. 47 d −x (e y ) = (2 − x)e−x =⇒ y (x) = x − 1 + c2 e−x . dx Continuity at x = 1 requires that limx→1 y (x) = y (1). Consequently we must choose c2 to satisfy c2 e = e − 1, so that c2 = 1 − e−1 . Hence, for x ≥ 1, y (x) = x − 1 + (1 − e−1 )ex . On [1, ∞), y − y = 2 − x =⇒ I (x) = e−x =⇒ 1 dy dy du d2 y du 1 d2 y + = 9x, x > 0. Let u = so = 2 . The first equation becomes + u = 9x which is dx2 x dx dx dx dx dx x d first-order linear. An integrating factor for this is I (x) = x so (xu) = 9x2 =⇒ xu = x2 dx + c =⇒ xu = dx dy dy 3x3 + c1 =⇒ u = 3x2 + c1 x−1 , but u = so = 3x2 + c1 x−1 =⇒ y = (3x2 + c1 x−1 )dx + c2 =⇒ y (x) = dx dx x3 + c1 ln x + c2 . 22. 23. The differential equation for Newton’s Law of Cooling is dT = −k (T − Tm ). We can re-write this dt equation in the form of a first-order linear differential equation: dT + kT = kTm . An integrating factor for dt d k dt kt kt this differential equation is I = e (T e ) = kTm ekt . Integrating both sides, we get = e . Thus, dt T ekt = Tm ekt + c, and hence, T = Tm + ce−kt , which is the solution to Newton’s Law of Cooling. dTm dT dT = α =⇒ Tm = αt + c1 so = −k (T − αt − c1 ) =⇒ + kT = k (αt + c1 ). An integrating dt dt dt d kt factor for this differential equation is I = ek dt = ekt . Thus, (e T ) = kekt (αt + c1 ) =⇒ ekt T = dt α α 1 ekt (αt − + c1 ) + c2 =⇒ T = αt − + c1 + c2 e−kt =⇒ T (t) = α(t − ) + β + T0 e−kt where β = c1 and k k k T0 = c2 . 24. dTm dT = 10 =⇒ Tm = 10t + c1 but Tm = 65 when t = 0 so c1 = 65 and Tm = 10t + 65. = dt dt dT 1 dT −k (T − Tm ) =⇒ = −k (T − 10t − 65), but (1) = 5, so k = . The last differential equation can be dt dt 8 dT d kt 2 2 13 kt written + kT = k (10t + 65) =⇒ (e T ) = 5ke (2t + 13) =⇒ ekt T = 5kekt t− 2 + + c =⇒ dt dt k k k 1 2 1 t T = 5(2t − + 13) + ce−kt , but k = so T (t) = 5(2t − 3) + ce− 8 . Since T (1) = 35, c = 40e 8 . Thus, k 8 1 T (t) = 10t − 15 + 40e 8 (1−t) . 25. dT 1 26. (a) In this case, Newton’s law of cooling is = − (t − 80e−t/20 ). This linear differential equation dt 40 dT 1 has standard form + T = 2e−t/20 , with integrating factor I (t) = et/40 . Consequently the differential dt 40 d t/40 equation can be written in the integrable form (e T ) = 2e−t/40 , so that T (t) = −80e−t/20 + ce−t/40 . dt Then T (0) = 0 =⇒ c = 80, so that T (t) = 80(e−t/40 − e−t/20 ). (b) We see that limt→∞ = 0. This is a reasonable result since the temperature of the surrounding medium also approaches zero as t → ∞. We would expect the temperature of the object to approach to the temperature of the surrounding medium at late times. dT 1 1 (c) T (t) = 80(e−t/40 − e−t/20 ) =⇒ = 80 − e−t/40 + e−t/20 . So T (t) has only one critical point dt 40 20 1 1 when 80 − e−t/40 + e−t/20 = 0 =⇒ t = 40 ln 2. Since T (0) = 0, and limt→∞ T (t) = 0 the function 40 20 assumes a maximum value at tmax = 40 ln 2. T (tmax ) = 80(e− ln 2 − e−2 ln 2 ) = 20◦ F, Tm (tmax ) = 80e−2 ln 2 = 48 20◦ F. (d) The behavior of T (t) and Tm (t) is given in the accompanying figure. T(t), Tm(t) 80 Tm(t) 60 40 T(t) 20 t 0 0 20 40 60 80 100 120 Figure 0.0.47: Figure for Exercise 26(d) 27. (a) The temperature varies from a minimum of A − B at t = 0 to a maximum of A + B when t = 12. T A+B A-B t 5 10 15 20 Figure 0.0.48: Figure for Exercise 27(a) dT + k1 T = k1 (A − B cos ωt) + T0 . Multiplying dt reduces this differential equation to the integrable form (b) First write the differential equation in the linear form by the integrating factor I = ek1 t d k1 t (e T ) = k1 ek1 t (A − B cos ωt) + T0 ek1 t . dt Consequently, ek1 t T (t) = Aek1 t − Bk1 ek t cos ωtdt + 1 T 0 k1 t e +c k1 49 so that T (t) = A + Bk1 T0 −2 (k1 cos ωt + ω sin ωt) + ce−k1 t . k1 k1 + ω 2 This can be written in the equivalent form T (t) = A + T0 − k1 Bk1 2 k1 + ω 2 cos (ωt − α) + ce−k1 t for an approximate phase constant α. 28. (a) c1 e− dy dy + p(x)y = 0 =⇒ = −p(x)dx =⇒ dx y p(x)dx dy = − p(x)dx =⇒ ln |y | = − p(x)dx + c =⇒ yH = y . dv du dy = u + v . Substituting this last (b) Replace c1 in part (a) by u(x) and let v = e− p(x)dx . y = uv =⇒ dx dx dx dy dv du result into the original differential equation, + p(x) = q (x), we obtain u + v + p(x)y = q (x), but since dx dx dx dv du −1 = −vp, the last equation reduces to v = q (x) =⇒ du = v (x)q (x)dx =⇒ u = v −1 (x)q (x)dx + c. dx dx Substituting the values for u and v into y = uv , we obtain y = e− p(x)dx e p(x)dx q (x)dx + c . dy + x−1 y = 0, with solution yH = cx−1 . According to problem dx 28 we determine the function u(x) such that y (x) = x−1 u(x) is a solution to the given differential equation. du dy du 1 dy We have = x−1 − x−2 u. Substituting into + x−1 y = cos x yields x−1 − 2 u + x−1 (x−1 u) = cos x, dx dx dx dx x du so that = x cos x. Integrating we obtain u = x sin x + cos x + c, so that y (x) = x−1 (x sin x + cos x + c). dx 29. The associated homogeneous equation is dy + y = 0, with solution yH = ce−x . According to problem dx 28 we determine the function u(x) such that y (x) = e−x u(x) is a solution to the given differential equation. dy du −x dy du −x We have = e − e−x u. Substituting into + y = e−2x yields e − e−x u + e−x u(x) = e−2x , so dx dx dx dx du that = e−x . Integrating we obtain u = −e−x + c, so that y (x) = e−x (−e−x + c). dx 30. The associated homogeneous equation is dy + cot x · y = 0, with solution yH = c · csc x. According to dx problem 28 we determine the function u(x) such that y (x) = csc x · u(x) is a solution to the given differential dy du dy equation. We have = csc x · − csc x · cot x · u. Substituting into + cot x · y = 2 cos x yields dx dx dx du du csc x · − csc x · cot x · u + csc x · cot x · u = cos x, so that = 2 cos x sin x. Integrating we obtain dx dx 2 2 u = sin x + c, so that y (x) = csc x(sin x + c). 31. The associated homogeneous equation is dy 1 − y = 0, with solution yH = cx. We determine the dx x dy du function u(x) such that y (x) = xu(x) is a solution of the given differential equation. We have = x + u. dx dx dy 1 du Substituting into − y = x ln x and simplifying yields = ln x, so that u = x ln x − x + c. Consequently, dx x dx 32. The associated homogeneous equation is 50 y (x) = x(x ln x − x + c). Problems 33 - 39 are easily solved using a differential equation solver such as the dsolve package in Maple. Solutions to Section 1.7 True-False Review: 1. TRUE. Concentration of chemical is defined as the ratio of mass to volume; that is, c(t) = Therefore, A(t) = c(t)V (t). A(t) V (t) . 2. FALSE. The rate of change of volume is “rate in” − “rate out”, which is r1 − r2 , not r2 − r1 . 3. TRUE. This is reflected in the fact that c1 is always assumed to be a constant. 4. FALSE. The concentration of chemical leaving the tank is c2 (t) = can be nonconstant, c2 (t) can also be nonconstant. A(t) V (t) , and since both A(t) and V (t) 5. FALSE. Kirchoff’s second law states that the sum of the voltage drops around a closed circuit is zero, not that it is independent of time. 6. TRUE. This is essentially Ohm’s law, (1.7.10). 7. TRUE. Due to the negative exponential in the formula for the transient current, iT (t), it decays to zero as t → ∞. Meanwhile, the steady-state current, iS (t), oscillates with the same frequency ω as the alternating current, albeit with a phase shift. 8. TRUE. The amplitude is given in (1.7.19) as A = gets smaller. √ E0 , R2 +ω 2 L2 and so as ω gets larger, the amplitude A Problems: dV = 1 =⇒ V (t) = 1. Given V (0) = 10, A(0) = 20, c1 = 4, r1 = 2, and r2 = 1. Then ∆V = r1 ∆t − r2 ∆t =⇒ dt dA A A dA 1 t + 10 since V (0) = 10. ∆A ≈ c1 r1 ∆t − c2 r2 ∆t =⇒ = 8 − c2 = 8 − = 8 − =⇒ + A= dt V t + 10 dt t + 10 4 8 =⇒ (t + 10)A = 4(t + 10)2 + c1 . Since A(0) = 20 =⇒ c1 = −200 so A(t) = [(t + 10)2 − 50]. Therefore, t + 10 A(40) = 196 g. A(60) . ∆ V = r1 ∆ t − V (60) dV dA r2 ∆t =⇒ = 3 =⇒ V (t) = 3(t + 200) since V (0) = 600. ∆A ≈ c1 r1 ∆t − c2 r2 ∆t =⇒ = 30 − 3c2 = dt dt A A 30 − 3 = 30 − =⇒ (t + 200)A = 15(t + 200)2 + c. Since A(0) = 1500, c = −300000 and therefore V t + 200 15 A(60) 596 A(t) = [(t + 200)2 − 20000]. Thus = g/L. t + 200 V (60) 169 2. Given V (0) = 600, A(0) = 1500, c1 = 5, r1 = 6, and r2 = 3. We need to find dV = 2 =⇒ V = dt 2(t + 10) since V (0) = 20. Thus V (t) = 40 for t = 10, so we must find A(10).∆A ≈ c1 r1 ∆t − c2 r2 ∆t =⇒ dA 2A A dA 1 d = 40 − 2c2 = 40 − = 40 − =⇒ + A = 40 =⇒ [(t + 10)A] = 40(t + 10)dt =⇒ dt V t + 10 dt t + 10 dt 3. Given V (0) = 20, A(0) = 0, c1 = 10, r1 = 4, and r2 = 2. Then ∆V = r1 ∆t − r2 ∆t =⇒ 51 20 (t + 10)A = 20(t + 10)2 + c. Since A(0) = 0 =⇒ c = −2000 so A(t) = [(t + 10)2 − 100] and A(10) = 300 t + 10 g. dV 4. Given V (0) = 100, A(0) = 100, c1 = 0.5, r1 = 6, and r2 = 4. Then ∆V = r1 ∆t − r2 ∆t =⇒ = 2 =⇒ dt dA 4A dA 2A d V (t) = 2(t + 50) since V (0) = 100. Then + = 3 =⇒ + = 3 =⇒ [(t + 50)2 A] = dt 2(t + 50) dt t + 50 dt 125000 3(t +50)2 =⇒ (t +50)2 A = (t +50)3 + c but A(0) = 100 so c = 125000 and therefore A(t) = t +50+ . (t + 50)2 The tank is full when V (t) = 200, that is when 2(t+50) = 200 so that t = 50 min. Therefore the concentration 9 A(50) just before the tank overflows is: = g/L. V (50) 16 5. Given V (0) = 10, A(0) = 0, c1 = 0.5, r1 = 3, r2 =, and A(5) = 0.2. V (5) dV (a) ∆V = r1 ∆t − r2 ∆t =⇒ = 1 =⇒ V (t) = t + 10 since V (0) = 10. Then ∆A ≈ c1 r1 ∆t − c2 r2 ∆t =⇒ dt dA A 2A dA 2dt = −2c2 = −2 = − =⇒ =− =⇒ ln |A| = −2 ln |t + 10| + c =⇒ A = k (t +10)−2 . Then dt V t + 10 A t + 10 A(5) 675 A(5) = 3 since V (5) = 15 and = 0.2. Thus, k = 675 and A(t) = . In particular, A(0) = 6.75 V (5) (t + 10)2 g. A(t) 675 A(t) 675 (b) Find V (t) when = 0.1. From part (a) A(t) = and V (t) = t + 10 =⇒ = . V (t) (t + 10)2 V (t) (t + 10)3 √ √ A(t) = 0.1 =⇒ (t + 10)3 = 6750 =⇒ t + 10 = 15 3 2 so V (t) = t + 10 = 15 3 2 L. Since V (t) dV = 1 =⇒ V = t + 20 since dt 2 d (20 + t)3 + c dA + A = 3 =⇒ [(t + 20)2 A] = 3(t + 20)2 =⇒ A(t) = and since V (0) = 20. Then dt t + 20 dt (t + 20)2 3 3 (t + 20) − 20 A(0) = 0, c = −203 which means that A(t) = . (t + 20)2 A(t) A(t) (b) The concentration of chemical in the tank, c2 , is given by c2 = or c2 = so from part (a), V (t) t + 20 √ (t + 20)3 − 203 1 (t + 20)3 − 203 c2 = . Therefore c2 = df rac12 g/l when = =⇒ t = 20( 3 2 − 1)minutes. 3 3 (t + 20) 2 (t + 20) 6. Given V (0) = 20, c1 = 1, r1 = 3, and r2 = 2. Then ∆V = r1 ∆t − r2 ∆t =⇒ dV 7. Given V (0) = w, c1 = k, r1 = r, r2 = r, and A(0) = A0 . Then ∆V = r1 ∆r − r2 ∆t =⇒ = dt dA A A 0 =⇒ V (t) = V (0) = w for all t. Then ∆A = c1 r1 ∆t − c2 r2 ∆t =⇒ = kr − r = kr − r = dt V V r dA r d kr − A =⇒ + A = kr =⇒ (e−rt/w A) = kre−rt/w =⇒ A(t) = kw + ce−rt/w . Since A(0) = A0 so w dt w dt c = A0 − kw =⇒ A(t) = e−rt/w [kw(ert/w − 1) + A0 ]. A(t) e−rt/w A0 (b) limt→∞ = limt→∞ [kw(ert/w − 1) + A0 ] = limt→∞ [k + − k e−rt/w ] = k . This is V (t) w w reasonable since the volume remains constant, and the solution in the tank is gradually mixed with and replaced by the solution of concentration k flowing in. 52 dA1 A1 (t) dA1 dA1 = c1 r1 − c2 r2 =⇒ = c1 r1 − r2 =⇒ = c1 r1 − dt dt V1 (t) dt r2 dA1 r2 A1 (t) =⇒ + A1 = c1 r1 . (r1 − r2 )t + V1 dt (r1 − r2 )t + v1 dA2 dA2 A1 A2 (t) dA2 For the bottom tank we have: = c2 r2 − c3 r3 =⇒ = r2 − r3 =⇒ = dt dt (r1 − r2 )t + V1 V2 (t) dt dA2 r3 r2 A1 A1 A2 (t) − r3 =⇒ + A2 = . r2 (r1 − r2 )t + V1 (r2 − r3 )t + V2 dt (r1 − r2 )t + V2 (r1 − r2 )t + V1 dA1 r2 dA1 4 dA1 2 (b) From part (a) + A1 = c1 r1 =⇒ + A1 = 3 =⇒ + A1 = 3 =⇒ dt (r1 − r2 )t + v1 dt 2t + 40 dt t + 20 c d [(t + 20)2 A] = 3(t + 20)2 =⇒ A1 = t + 20 + but A1 (0) = 4 so c = −6400. Consequently dt (t + 20)2 6400 dA2 3 2 6400 dA2 3 A1 (t) = t + 20 − . Then + A2 = t + 20 − =⇒ + A2 = (t + 20)2 dt t + 20 t + 20 (t + 20)2 dt t + 20 2[(t + 20)3 − 6400] d 2[(t + 20)3 − 6400] t + 20 12800t =⇒ [(t + 20)3 A2 ] = (t + 20)3 =⇒ A2 (t) = − + (t + 20)3 dt (t + 20)3 2 (t + 20)3 k t + 20 12800t 80000 but A2 (0) = 20 so k = 80000. Thus A2 (t) = − + and in particular 3 3 (t + 20) 2 (t + 20) (t + 20)3 119 ≈ 13.2 g. A2 (10) = 9 8. (a) For the top tank we have: di R 1 di d 40t + i = E (t) =⇒ + 40i = 200 =⇒ (e i) = dt L L dt dt −40t −40t =⇒ i(t) = 5 + ce . But i(0) = 0 =⇒ c = −5. Consequently i(t) = (1 − e ). 9. Let E (t) = 20, R = 4 and L = 200e40t 1 10 . Then dq 1 E dq d + q= =⇒ +10q = 20 =⇒ (qe10t ) = 20e10t =⇒ dt RC R dt dt q (t) = 2 + ce−10t . But q (0) = 0 =⇒ c = −2 so q (t) = 2(1 − e−40t ). 10. Let R = 5, C = 1 50 and E (t) = 100. Then di R 1 di + i = E (t) =⇒ + 3i = 15 sin 4t =⇒ dt L L dt 3t d 3t 3e 3 4 (e i) = 15e3t sin 4t =⇒ e3t i = (3 sin 4t − 4 cos 4t) + c =⇒ i = 3 sin 4t − cos 4t + ce−3t , but dt 5 5 5 12 3 i(0) = 0 =⇒ c = so i(t) = (3 sin 4t − 4 cos 4t + 4e−3t ). 5 5 11. Let R = 2, L = 12. Let R = 2, C = 1 8 2 3 and E (t) = 10 sin 4t. Then and E (t) = 10 cos 3t. Then dq 1 E dq d + q= =⇒ + 4q = 5 cos 3t =⇒ (e4t q ) = dt RC R dt dt e4t 5e4t cos 3t =⇒ e4t q = (4 cos 3t + 3 sin 3t) + c =⇒ q (t) = 1 (4 cos 3t + 3 sin 3t) + ce4t , but q (0) = 1 =⇒ c = 5 5 1 1 −4t dq 1 (4 cos 3t + 3 sin 3t) + e and i(t) = = (9 cos 3t − 12 sin 3t − 4e−4t ). 5 5 dt 5 dq 1 E 13. In an RC circuit for t > 0 the differential equation is given by + q = . If E (t) = 0 then dt RC R dq 1 d t/RC t/RC −t/RC + q = 0 =⇒ (e q ) = 0 =⇒ q = ce and if q (0) = 5 then q (t) = 5e . Then limt→∞ q (t) = dt RC dt 0. Yes, this is reasonable. As the time increases and E (t) = 0, the charge will dissipate to zero. 14. In an RC circuit the differential equation is given by i + 1 E0 q= . Differentiating this equation with RC R 53 1 dq dq di 1 d di + = 0, but = i so + i = 0 =⇒ (et/RC i) = 0 =⇒ i(t) = ce−t/RC . dt RC dt dt dt RC dt E0 E0 −t/RC Since q (0) = 0, i(0) = and so i(t) = e =⇒ d = E0 k so q (t) = E0 k (1 − e−t/RC ). Then R R limt→∞ q (t) = E0 k , and lim t → ∞i(t) = 0. respect to t we obtain q(t) E0k t Figure 0.0.49: Figure for Exercise 14 di R E (t) di R E0 15. In an RL circuit, + i= and since E (t) = E0 sin ωt, then + i= sin ωt =⇒ dt L L dt L L E0 Rt/L E0 d Rt/L (e i) = e sin ωt =⇒ i(t) = 2 [R sin ωt − ωL cos ωt] + Ae−Rt/L . We can write this dt L R + L2 ω 2 E0 R ωL √ as i(t) = √ sin ωt − √ cos ωt + Ae−Rt/L . Defining the phase φ by 2 + L2 ω 2 2 + L2 ω 2 2 + L2 ω 2 R R R R ωL E0 cos φ = √ , sin φ = √ , we have i(t) = √ [cos φ sin ωt−sin φ cos ωt]+Ae−Rt/L . 2 + L2 ω 2 2 + L2 ω 2 2 + L2 ω 2 R R R E0 sin (ωt − φ) + Ae−Rt/L . That is, i(t) = √ 2 + L2 ω 2 R Transient part of the solution: iT (t) = Ae−Rt/L . E0 sin (ωt − φ). Steady state part of the solution: iS (t) = √ 2 + L2 ω 2 R di E0 R + ai = , i(0) = 0, where a = , and E0 denotes the dt L L constant EMF. An integrating factor for the differential equation is I = eat , so that the differential equation d at E0 at E0 can be written in the form (e i) = e . Integrating yields i(t) = + c1 e−at . The given initial dt L aL E0 E0 E0 E0 condition requires c1 + = 0, so that c1 = − . Hence i(t) = (1 − e−at ) = (1 − e−at ). aL aL aL R 16. We must solve the initial value problem dq 1 E (t) dq 1 E0 −at d E0 (1/RC −a)t + q= =⇒ + q= e =⇒ (et/RC q ) = e =⇒ dt RC R dt RC R dt R E0 C E0 C q (t) = e−t/RC e(1/RC −a)t + k =⇒ q (t) = e−at + ke−t/RC . Imposing the initial condition 1 − aRC 1 − aRC 17. 54 q (0) = 0 (capacitor initially uncharged) requires k = − Thus i(t) = E0 C dq = dt 1 − aRC E0 C E0 C , so that q (t) = (e−at − e−t/RC ). 1 − aRC 1 − aRC 1 −t/RC e − ae−at . RC 1 di 1 dq 1 12 d2 q + q = 0 =⇒ i + q = 0, since i = . Then idi = − qdq =⇒ i2 = − q + k but dt2 LC dq LC dt LC LC 2 2 q0 − q 2 q0 − q 2 q2 12 q2 dq q (0) = q0 and i(0) = 0 so k = 0 =⇒ i2 = − q + 0 =⇒ i = ± √ =⇒ =± √ =⇒ LC LC LC dt LC LC dq dt t t + k1 =⇒ q = q0 sin ± √ + k1 but q (0) = q0 so q0 = = ±√ =⇒ sin−1 ( qq0 ) = ± √ 2 − q2 LC LC LC qo 1 π t q0 sin k1 =⇒ k1 = π + 2nπ where n is an integer =⇒ q = q0 sin ± √ + =⇒ q (t) = q0 cos √ 2 2 LC LC dq q0 t and i(t) = = −√ sin √ . dt LC LC 18. d2 q 1 E0 dq d2 di dq di + q= . Since i = then 2 = = i . Hence the original equation can be written 2 dt LC L dt dt dq dt dq 1 E0 1 E0 i2 q2 E0 q di + q= =⇒ idi + qdq = dq or + = + A. Since i(0) = 0 and q (0) = q0 as i dq LC L LC L 2 2LC L 1/2 2 q0 E0 q 0 i2 q2 E0 q 2E0 q q2 then A = − . From + = + A we get that i = 2A + − =⇒ i = 2LC L 2 2LC L L LC 19. 2 1/2 1/2 (2E0 C )2 (q − E0 C )2 (E0 C )2 dq q − E0 C √ 2A + − and we let D2 = 2A + then i =D 1− =⇒ LC LC LC dt D LC √ √ q − E0 C q − E0 C √ √ LC sin−1 = t + B . Then since q (0) = 0 so B = LC sin−1 and therefore D LC D LC √ t+B t+B q − E0 C t+B dq √ = sin √ =⇒ q (t) = D LC sin √ + E0 c =⇒ i = = D cos √ . Since dt D LC LC LC LC 2 2 q E0 q 0 |q0 − E0 C | 2A + (E0 C ) D2 = and A = 0 − we can substitute to eliminate A and obtain D = ± √ . LC 2LC L LC t+B Thus q (t) = ±|q0 − E0 C | sin √ + E0 c. LC Solutions to Section 1.8 True-False Review: 1. TRUE. We have f (tx, ty ) = 2(xt)(yt) − (xt)2 2xyt2 − x2 t2 2xy − x2 = = = f (x, y ), 2(xt)(yt) + (yt)2 2xyt2 + y 2 t2 2xy + y 2 so f is homogeneous of degree zero. 2. FALSE. We have f (tx, ty ) = (yt)2 y 2 t2 y2 t y2 = = = , 2 2 t2 2t (xt) + (yt) xt + y x+y x + y2 55 so f is not homogeneous of degree zero. 3. FALSE. Setting f (x, y ) = 1+xy 2 1+x2 y , we have f (tx, ty ) = 1 + (xt)(yt)2 1 + xy 2 t3 = = f (x, y ), 1 + (xt)2 (yt) 1 + x2 yt3 so f is not homogeneous of degree zero. Therefore, the differential equation is not homogeneous. 4. TRUE. Setting f (x, y ) = x2 y 2 x4 +y 4 , f (tx, ty ) = we have x2 y 2 t4 x2 y 2 (xt)2 (yt)2 = 44 =4 = f (x, y ). (xt)4 + (yt)4 x t + y 4 t4 x + y4 Therefore, f is homogeneous of degree zero, and therefore, the differential equation is homogeneous. 5. TRUE. This is verified in the calculation leading to Theorem 1.8.5. 6. TRUE. This is verified in the calculation leading to (1.8.12). 7. TRUE. We can rewrite the equation as y− √ xy = √ xy 1/2 , √ √ which is the proper form for a Bernoulli equation, with p(x) = − x, q (x) = x, and n = 1/2. 8. FALSE. The presence of an exponential exy involving y prohibits this equation from having the proper form for a Bernoulli equation. 9. TRUE. This is a Bernoulli equation with p(x) = x, q (x) = x2 , and n = 2/3. Unless otherwise indicated in this section v = y dy dv =v+x and t > 0. , x dx dx Problems: (tx)2 − (ty )2 x2 − y 2 = = f (x, y ). Thus f is homogeneous of degree zero. f (x, y ) = (tx)(ty ) xy y2 1 − (x) x2 − y 2 1 − v2 = = = F (v ). y xy v x 1. f (tx, ty ) = 2. f (tx, ty ) = (tx) − (ty ) = t(x − y ) = tf (x, y ). Thus f is homogeneous of degree one. ty (tx) sin ( tx ) − (ty ) cos ( tx ) ty 3. f (tx, ty ) = sin x − y zero. f (x, y ) = y x y x y x y cos x = = y x sin x − y cos x y 5. f (tx, ty ) = = f (x, y ). Thus f is homogeneous of degree 1 sin v − v cos v = F (v ). v (tx)2 + (ty )2 |t| x2 + y 2 = = tx − ty t(x − y ) 1 + v2 . 1−v 4. f (tx, ty ) = √ zero. F (v ) = y ty . Thus f is not homogeneous. tx − 1 x2 + y 2 = f (x, y ). Thus f is homogeneous of degree x−y 56 5(ty ) + 9 3(tx) + 5(ty ) 3x + 5y tx − 3 + = = = f (x, y ). Thus f is homogeneous of degree ty 3(ty ) 3(ty ) 3y y 3 + 5x 3x + 5y 3 + 5v zero. f (x, y ) = = = = F (v ). y 3y 3x 3v 6. f (tx, ty ) = (tx)2 + (ty )2 x2 + y 2 = = f (x, y ). Thus f is homogeneous of degree zero. f (x, y ) = tx x y2 √ |x| 1 + ( x ) y2 x2 + y 2 =− 1+ = − 1 + v 2 = F (v ). = x x x 7. f (tx, ty ) = (tx)2 + 4(ty )2 − (tx) + (ty ) x2 + 4y 2 − x + y = f (x, y ). Thus f is homogeneous of = x + 3y (tx) + 3(ty ) √ y y 1 + 4( x )2 − 1 + x 1 + 4v 2 − 1 + v x2 + 4y 2 − x + y = F (v ). = = degree zero. f (x, y ) = y 1 + 3v x + 3y 1 + 3x 8. f (tx, ty ) = dy y dy y dv dv 3v = 3y =⇒ (3 − 2 ) = 3 =⇒ (3 − 2v ) v + x = 3v =⇒ x = − v =⇒ dx x dx x dx dx 3 − 2v dx 3 3x y 3x 3 − 2v dv = =⇒ − − ln |v | = ln |x| + c1 =⇒ − − ln | | = ln |x| + c1 =⇒ ln y = − + c2 =⇒ 2v 2 x 2v 2y x 2y y 2 = ce−3x/y . 9. (3x − 2y ) dy (x + y )2 dy 1 y = =⇒ = 1+ dx 2x2 dx 2 x y 1 1 ln |x| + c =⇒ tan−1 = ln |x| + c. 2 x 2 10. =⇒ v + x dv 1 = (1 + v )2 =⇒ dx 2 dv = v2 + 1 dx =⇒ tan−1 v = x dy y x dy y y dv − y = x cos =⇒ sin − = cos =⇒ sin v v + x −v = dx x y dx x x dx sin v dx y dv cos v =⇒ sin v x = cos v =⇒ dv = =⇒ − ln | cos v | = ln |x| + c1 =⇒ x cos = c2 =⇒ dx cos v x x c y (x) = x cos−1 . x 11. sin 12. y x 2 dy = dx x 16x2 − y 2 + y dy =⇒ = x dx y 16 − ( x )2 + y x =⇒ v + x √ dv = 16 − v 2 + v =⇒ dx √ dv = 16 − v 2 dx =⇒ sin−1 ( v ) = ln |x| + c =⇒ sin−1 ( 4y ) = ln |x| + c. 4 x x 13. We first rewrite the given differential equation in the equivalent form y = (9x2 + y 2 ) + y . Factoring x y |x| 9 + ( x )2 + y . Since we are told to solve the differential x y y equation on the interval x > 0 we have |x| = x, so that y = 9 + ( x )2 + x , which we recognize as being homogeneous. We therefore let y = xV , so that y = xV + V . Substitution into the preceding differential √ √ equation yields xV + V = 9 + V 2 + V , that is xV = 9 + V 2 . Separating the variables in this equation √ 1 1 we obtain √ dV = dx. Integrating we obtain ln (V + 9 + V 2 ) = ln c1 x. Exponentiating both sides 2 x 9+V √ y yields V + 9 + V 2 = c1 x. Substituting = V and multiplying through by x yields the general solution x out an x2 from the square root yields y = 57 9x2 + y 2 = c1 x2 . y+ 14. The given differential equation can be written in the equivalent form dy y (x2 − y 2 ) = , dx x(x2 + y 2 ) which we recognize as being first order homogeneous. The substitution y = xv yields v+x so that dv v (1 − v 2 ) dv 2v 3 = =⇒ x =− , dx 1 + v2 dx 1 + v2 dx v −2 =⇒ − + ln |v | = −2 ln |x| + c1 . x 2 1 + v2 dv = −2 v3 Consequently, − x2 + ln |xy | = c1 . 2y 2 dy dy yy dv + y ln x = y ln y =⇒ = ln =⇒ v + x = v ln v =⇒ dx dx xx dx y ln x − 1 = c =⇒ y (x) = xe1+cx . ln |x| + c1 =⇒ x 15. x dv v (ln v − 1) dx =⇒ ln | ln v − 1| = x dy y 2 + 2xy − 2x2 dv v 2 + 2v − 2 dv −v 3 + 2v 2 + v − 2 v2 − v + 1 =2 =⇒ v +x = =⇒ x = =⇒ dv = 2 2 2−v+1 3 − 2v 2 − v + 2 dx x − xy + y dx 1−v+v dx v v v2 − v + 1 dx 1 1 1 dx dx =⇒ dv = − =⇒ − + dv = − =⇒ − x (v − 1)(v + 2)(v + 1) x v − 2 2(v − 1) 2(v + 1) x 1 1 (v − 2)2 (v + 1) (y − 2x)2 (y + x) ln |v − 2|− ln |v − 1|+ ln |v + 1| = − ln |x|+c1 =⇒ ln = −2 ln x+c2 =⇒ = 2 2 v−1 y−x c. 16. 17. 2xydy − (x2 e−y 2v 2 ) = 0 =⇒ 2vx x2 ln (ln (cx)). dy 18. x2 = y2 dx dv = (v + 1)2 1 −x 1 + ln (cx) 2 /x2 y dy y2 dv 2 2 2 − e−y /x + 2 = 0 =⇒ 2v v + x − (e−v + x dx x dx dx 2 2 2 2 ev (2vdv ) = =⇒ ev = ln |x| + c1 =⇒ ey /x = ln (cx) =⇒ y 2 = x + 2y 2 )dx = 0 =⇒ 2 dv 2 = e−v =⇒ dx dv dy dv y y + 3xy + x2 =⇒ = ( x )2 + 3 x + 1 =⇒ v + x = v 2 + 3v + 1 =⇒ x = (v + 1)2 =⇒ dx dx dx dx 1 1 y 1 =⇒ − = ln |x| + c1 =⇒ − y = ln |x| + c1 =⇒ =− =⇒ y (x) = x v+1 x ln (cx) x +1 . √ √ y 1 + ( x )2 − 1 x2 + y 2 − x dy dv 1 + v2 − dv 1 + v2 − x =⇒ =⇒ = =⇒ v + x = =⇒ x = y y dx dx v dx v x dx v c √ dv = =⇒ ln |1 − u| = ln |x| + c1 =⇒ |x(1 − u)| = c2 =⇒ 1 − u = =⇒ u2 = 2 − 1 − v2 x x 1+v c2 c c2 c − 2 + 1 =⇒ v 2 = 2 − 2 =⇒ y 2 = c2 − 2cx. 2 x x x x 19. dy = dx 58 dy y dy dv dv y y = y (4x − y ) =⇒ 2 +2 = x (4 − x ) =⇒ 2(v + 2) v + x = v (4 − v ) =⇒ 2x = dx x dx dx dx 3v 2 v+2 dx 4 − =⇒ 2 dv = −3 =⇒ 2 ln |v | − v = −3 ln |x| + c1 =⇒ y 2 = cxe4x/y . v+2 v2 x 20. 2x(y + 2x) dy dv dv y = x tan ( x ) + y =⇒ v + x = tan v + v =⇒ x = tan v =⇒ dx dx dx −1 −1 ln |x| + c1 =⇒ sin v = cx =⇒ v = sin (cx) =⇒ y (x) = x sin (cx). 21. x cot vdv = dx =⇒ ln | sin v | = x 2 x x2 + y 2 + y 2 dy x y dv dv dy 1 = =⇒ = +1 + =⇒ v + x = ( v )2 + 1 + v =⇒ = dx xy dx y x dx dx √ √ dv dx y y 1 = =⇒ ln |v + 1 + v 2 | = ln |x| + c =⇒ v + 1 + v 2 = cx =⇒ x + 1 + ( x )2 = ( v )2 + 1 =⇒ x 12 (v) 22. y xc =⇒ 2( x ) + 1 = (cx)2 =⇒ y 2 = x2 [(cx)2 − 1] . 2 23. The given differential equation can be written as (x−4y )dy = (4x+y )dx. Converting to polar coordinates we have x = r cos θ =⇒ dx = cos θdr − r sin θdθ, and y = r sin θdr + r cos θdθ. Substituting these results into the preceding differential equation and simplifying yields the separable equation 4r−1 dr = dθ which can be integrated directly to yield 4 ln r = θ + c, so that r = c1 eθ/4 . x dy 2(2y − x) dy y +1 24. . = . Since x = 0, divide the numerator and denominator by y yields = 2(2 − x ) dx x+y dx y dx dv dv v+1 2(2 − v ) dy =⇒ v + y = =⇒ dv = =⇒ Now let v = x so that = v+y y dy dy dy 2(2 − v ) 2v 2 − 3v + 1 y dv dv (v − 1)2 −6 +2 = ln |y | + c1 =⇒ ln = ln (c2 |y |) =⇒ (x − y )2 = c(y − 2x)3 . Since y (0) = 2 2v − 1 v−1 |2v − 1|3 1 1 then c = . Thus, (x − y )2 = (y − 2x)3 . 2 2 y 2− x dy 2x − y dy dv 2−v dv 2 − 2v − 4v 2 1 1 + 4v =⇒ v +x = =⇒ = = =⇒ x = =⇒ dv = y 2+v−1 dx x + 4y dx 1 + 4x dx 1 + 4v dx 1 + 4v 2 2v dx 1 1 1 − =⇒ ln |2v 2 + v − 1| = − ln |x| + c =⇒ ln |x2 (2v 2 + v − 1)| = c =⇒ ln |2y 2 + yx − x2 | = c, but x 2 2 2 1 1 1 2 2 y (1) = 1 so c = ln 2. Thus ln |2y + yx − x | = ln 2 and since y (1) = 1 it must be the case that 2 2 2 2y 2 + yx − x2 = 2. 25. √ dy y dv dv y − x2 + y 2 dy y = − 1 + v 2 =⇒ √ =− = =⇒ = − 1 + ( )2 =⇒ x x dx dx x dx x 1 + v2 √ |x| ln (v + 1 + v 2 ) = − ln |x| + c1 =⇒ y + x2 + y 2 = c2 . Since y (3) = 4 then c2 = 9. Then take x since we must have y (3) = 4; thus y + x2 + y 2 = 9. 26. √ y2 dv dy y =⇒ v + x = v + 4 − v 2 =⇒ − = 4− x dx dx x y sin−1 = ln x + c since x > 0. 2x 27. √ dv = 4 − v2 dx =⇒ x |x| =1 x dx v =⇒ sin−1 = ln |x| + c =⇒ x 2 59 dy x + ay dv 1 + v2 = . Substituting y = xv and simplifying yields x = . Separating the variables dx ax − y dx a−v 1 y and integrating we obtain a tan−1 v − ln (1 + v 2 ) = ln x + ln c or equivalently, a tan−1 x − 1 ln (x2 + y 2 ) = 2 2 ln c. Substituting for x = r cos θ, y = r sin θ yields aθ − ln r √ ln c. Exponentiating then gives r = keaθ . = (b) The initial condition y (1) = 1 corresponds to r( π ) = 2. Imposing this condition on the polar form 4 √ of the solution obtained in (a) yields k = 2e−π/8 . Hence, the solution to the initial value problem is √ 1 dy 2x + y = . Consequently every solution r = 2e(θ−π/4)/2 . When a = , the differential equation is 2 dx x − 2y x curve has a vertical tangent line at points of intersection with the line y = . The maximum interval of 2 x existence for the solution of the initial value problem can be obtained by determining where y = intersects 2 √ (θ−π/4)/2 x 1 the curve r = 2e . The line y = has a polar equation tan θ = . The corresponding values of θ 2 2 1 −1 are θ = θ1 = tan ≈ 0.464, θ = θ2 = θ1 + π ≈ 3.61. Consequently, the x-coordinates of the intersection 2 √ √ points are x1 = r cos θ1 = 2e(θ1 −π/4)/2 cos θ1 ≈ 1.08, x2 = r cos θ2 = 2e(θ2 −π/4)/2 cos θ2 ≈ −5.18. Hence the maximum interval of existence for the solution is approximately (−5.18, 1.08). (c) See the accompanying figure. 28. (a) y(x) 2 1 -6 -5 -4 -3 -2 -1 1 0 x -1 -2 -3 -4 Figure 0.0.50: Figure for Exercise 28(c) x2 + y 2 dy dy dy . Hence 2x + 2y = 2c =⇒ = 2y dx dx dx 2 2 x 2xy dy y −x dy dv =2 . Orthogonal trajectories satisfies: = . Let y = vx so that = v+x . 2 c−y x −y dx 2xy dx dx dv v2 + 1 1 Substituting these results into the last equation yields x =− =⇒ ln |v + 1| = − ln |x| + c1 =⇒ dx 2v 29. Given family of curves satisfies: x2 + y 2 = 2cy =⇒ c 60 c2 y2 +1= =⇒ x2 + y 2 = 2kx. 2 x x y(x) x Figure 0.0.51: Figure for Exercise 29 x2 + y 2 dy . Hence 2(x − c)+2(y − c) = 2(x + y ) dx 2 2 2 2 y − 2xy − x dy y + 2xy − x c−x =2 . Orthogonal trajectories satisfies: =2 . Let y = vx so 0 =⇒ 2 y−c y + 2xy − x dx x + 2xy − y 2 dv dv v 2 + 2v − 1 dy = v + x . Substituting these results into the last equation yields v + x = =⇒ that dx dx dx 1 + 2v − v 2 2 1 + 2v − v 1 1 2v 1 dv = dx =⇒ −2 dv = dx =⇒ x2 + y 2 = 2k (x − y ) =⇒ (x − k )2 +(y + k )2 = 3 − v2 + v − 1 v x v−1 v +1 x 2k 2 . 30. Given family of curves satisfies: (x − c)2 +(y − c)2 = 2c2 =⇒ c = 31. (a) Let r represent the radius of one of the circles with center at (a.ma) and passing through (0, 0). √ r =√ (a − 0)2 + (ma − 0)2 = |a| 1 + m2 . Thus, the circle’s equation can be written as (x−a)2 +(y −ma)2 = (|a| 1 + m2 )2 or (x − a)2 + (y − ma)2 = a2 (1 + m2 ). x2 + y 2 (b) (x − a)2 + (y − ma)2 = a2 (1 + m2 ) =⇒ a = . Differentiating the first equation with respect 2(x + my ) dy a−x dy y 2 − x2 − 2mxy x and solving we obtain = . Substituting for a and simplifying yields = . dx y − ma dx my 2 − mx2 + 2xy y2 y 2 2 m − m( ) − 2 x dy mx − my − 2xy dy Orthogonal trajectories satisfies: =2 =⇒ = y2 x y . Let y = vx so that 2 − 2mxy dx y −x dx ( x ) − 1 − 2m x dy dv dv m − mv 2 − 2v xdv = v + x . Substituting these results into the last equation yields v + x =2 =⇒ = dx dx dx v − 1 − 2mv dx 2 2 (m − v )(1 + v ) v − 2mv − 1 dx dv 2v dx =⇒ dv = =⇒ − dv = =⇒ ln |v − m| − v 2 − 2mv − 1 (m − v )(1 + v 2 ) x v−m 1 + v2 x 61 y(x) x Figure 0.0.52: Figure for Exercise 30 ln (1 + v 2 ) = ln |x| + c1 =⇒ v − m = c2 x(1 + v 2 ) =⇒ y − mx = c2 x2 + c2 y 2 =⇒ x2 + y 2 + cmx − cy = 0. Completing the square we obtain (x + cm/2)2 + (y − c/2)2 = c2 /4(m2 + 1). Now letting b = c/2, the last equation becomes (x + bm)2 + (y − b)2 = b2 (m2 + 1) which is a family or circles lying on the line y = −my and passing through the origin. (c) See the accompanying figure. y(x) x Figure 0.0.53: Figure for Exercise 31(c) 62 m2 − tan ( π ) dy x −x/y − 1 x+y 4 = − = m2 . m1 = = . Let y = vx so that π= dx y 1 + m2 tan ( 4 ) 1 − x/y x−y dy dv dv 1+v 1−v = v + x . Substituting these results into the last equation yields v + x = =⇒ dv = dx dx dx 1−v 1 + v2 v 1 dx 1 dx =⇒ − +2 dv = =⇒ − ln (1 + v 2 )+tan−1 v = ln |x| + c1 =⇒ Oblique trajectories: 2 x 1+v v +1 x 2 ln (x2 + y 2 ) − 2 tan−1 (y/x) = c2 . 32. x2 + y 2 = c =⇒ m2 − tan ( π ) 6y − x dy 6y/x − 1 4 = 6y/x = m2 . m1 = = = . Let y = vx so that π dx 1 + m2 tan ( 4 ) 1 + 6y/x 6y + x dv dv 6v − 1 dv dy = v + x . Substitute these results into the last equation yields v + x = =⇒ x = dx dx dx 6v + 1 dx (3v − 1)(1 − 2v ) 9 8 dx =⇒ − dv = =⇒ 3 ln |3v − 1| − 4 ln |2v − 1| = ln |x| + c1 =⇒ 6v + 1 3v − 1 2v − 1 x Oblique trajectories (3y − x)3 = k (2y − x)4 . 33. y = cx6 =⇒ 2 34. x2 + y 2 = 2cx =⇒ c = y 2 − x2 − 2xy . y 2 − x2 + 2xy x +y 2x 2 2 and 2 dy y −x = = m2 . m1 = dx 2xy m2 − tan ( π ) 4 1 + m2 tan ( π ) 4 y 2 − x2 −1 2xy = = y 2 − x2 1+ 2xy dy dv = v + x . Substituting these results into the last equadx dx dv v 2 − 2v − 1 dv −v 3 − v 2 − v − 1 −v 2 − 2v + 1 dx tion yields v + x =2 =⇒ x = =⇒ 3 dv = =⇒ dx v + 2v − 1 dx v 2 + 2v − 1 v + v2 + v + 1 x 1 2v dx − dv = =⇒ ln |v + 1| − ln (v 2 + 1) = ln |x| + c1 =⇒ ln |y + x| = ln |y 2 + x2 | + c1 =⇒ v + 1 v2 + 1 x Oblique trajectories: x2 + y 2 = 2k (x + y ) or, equivalently, (x − k )2 + (y − k )2 = 2k 2 . Let y = vx so that y m2 − tan α0 −y/x − tan α0 dy = −cx−2 = − . m1 = = . Let y = vx so dx x 1 + m2 tan α0 1 − y/x tan α0 dy dv dv tan α0 + v that = v + x . Substituting these results into the last equation yields v + x = =⇒ dx dx dx v tan α0 − 1 2v tan α0 − 2 2dx dv = − =⇒ ln |v 2 tan α0 − 2v − tan α0 | = −2 ln |x| + c1 =⇒ (y 2 − x2 ) tan α0 − 2 tan α − 2v − tan alpha v x 0 0 2xy = k . (b) See the accompanying figure. 35. (a) y = cx−1 =⇒ dy x m2 − tan α0 −x/y − m x + my dy = − . m1 = = = . Let y = vx so that = dx y 1 + m2 tan α0 1 − (x/y )m mx − y dx dv dv 1 + mv dv 1 + v2 v + x . Substituting these results into the last equation yields v + x = =⇒ x = =⇒ dx dx m−v dx m−v v−m dx v m dx 1 dv = − =⇒ − dv = − =⇒ ln (1 + v 2 ) − m tan v = − ln |x| + c1 . In polar 2 2 2 1+v x 1+v 1+v x 2 coordinates, r = x2 + y 2 and θ = tan−1 y/x, so this result becomes ln r − mθ = c1 =⇒ r = emθ where k is an arbitrary constant. (b) See the accompanying figure. 36. (a) x2 + y 2 = c =⇒ 37. dy 1 dy 1 2 − y = 4x2 y −1 cos x. This is a Bernoulli equation. Multiplying both sides y results in y −y= dx x dx x 63 y(x) x Figure 0.0.54: Figure for Exercise 35(b) y(x) x Figure 0.0.55: Figure for Exercise 36(b) du dy dy 1 du dy 1 4x2 cos x. Let u = y 2 so = 2y or y = . Substituting these results into y − y 2 = 4x2 cos x dx dx dx 2 dx dx x du 2 d −2 yields − u = 8x2 cos x which has an integrating factor I (x) = x−2 =⇒ (x u) = 8 cos x =⇒ x−2 u = dx x dx 8 cos xdx + c =⇒ x−2 u = 8 sin x + c =⇒ u = x2 (8 sin x + c) =⇒ y 2 = x2 (8 sin x + c). dy 1 du dy 38. y −3 + y −2 tan x = 2 sin x. This is Bernoulli equation. Let u = y −2 so = −2y −3 or dx 2 dx dx dy 1 du du y −3 = . Substituting these results into the last equation yields −u tan x = −4 sin x. An integrating dx 2 dx dx 64 d (u cos x) = −4 cos x sin x =⇒ u cos x = 4 cos x sin xdx =⇒ factor for this equation is I (x) = cos x. Thus, dx 1 c u(x) = (cos2 x + c) =⇒ y −2 = 2 cos x + . cos x cos x dy 3 1 dy 3 2/3 dy 2 dy − y = 6y 1/3 x2 ln x or 1/3 −y = 6x2 ln x. Let u = y 2/3 =⇒ = y −1/3 . Substituting dx 2x dx 2x dx 3 dx y 1 dy 3 2/3 du 1 these results into 1/3 − y = 6x2 ln x yields − u = 4x2 ln x. An integrating factor for this dx 2x dx x y 1 d −1 equation is I (x) = so (x u) = 4x ln x =⇒ x−1 u = 4 x ln xdx + c =⇒ x−1 u = 2x2 ln x − x2 + c =⇒ x dx u(x) = x(2x2 − x2 + c) =⇒ y 2/3 = x(2x2 − x2 + c). 39. √ √ dy 2 dy 2 du dy + y = 6 1 + x2 y 1/2 or y −1/2 + y 1/2 = 6 1 + x2 . Let u = y 1/2 =⇒ 2 = y −1/2 . dx x dx x dx dx √ √ du dy 2 1 Substituting these results into y −1/2 + y 1/2 = 6 1 + x2 yields + u = 3 1 + x2 . An integrating dx x dx x √ √ d 2= factor for this equation is I (x) = x so (xu) = 3x 1 + x ⇒ xu = x 1 + x2 dx + c =⇒ xu = dx 1 c 1 c 2 3/2 2 3 /2 (1 + x ) + c =⇒ u = (1 + x ) + =⇒ y 1/2 = (1 + x2 )3/2 + . x x x x 40. dy 2 dy 2 du dy + y = 6y 2 x4 or y −2 + y −1 = 6x4 . Let u = y −1 =⇒ − = y −2 . Substituting these results dx x dx x dx dx 2 −1 du 2 −2 dy 4 4 into y + y = 6x yields − u = −6x . An integrating factor for this equation is I (x) = x−2 so dx x dx x d −2 1 (x u) = −6x2 =⇒ x−2 u = −2x3 + c =⇒ u = −2x5 + cx2 =⇒ y −1 = −2x5 + cx2 =⇒ y (x) = 2 . dx x (c − 2x3 ) 41. dy 1 −2 1 du dy + y = −x−2 . Let u = y −2 =⇒ − = y −3 . Substituting dx 2x 2 dx dx 1 −2 du 1 −2 2 −3 dy + y = −x yields − u = 2x . An integrating factor for this equation is these results into y dx 2x dx x d −1 1 (x u) = 2x =⇒ x−1 u = x2 + c =⇒ u = x3 + cx =⇒ y −2 = x3 + cx. I (x) = so x dx 42. 2x dy + y 3 x2 dx + y = 0 or y −3 dy 2(b − a) du − y 1/2 = 1. Let u = y 1/2 =⇒ 2 = dx (x − a)(x − b) dx dy dy 2(b − a) du (b − a) 1 y −1/2 . Substituting these results into y −1/2 − y 1/2 = 1 yields − u= . dx dx (x − a)(x − b) dx (x − 1)(x − b) 2 x−a d x−a x−a x−a 1 An integrating factor for this equation is I (x) = so u= =⇒ u = [x +(b − x−b dx x − b 2(x − b) x−b 2 2 x−b 1 x−b a) ln |x − b| + c] =⇒ y 1/2 = [x +(b − a) ln |x − b| + c] =⇒ y (x) = [x +(b − a) ln |x − b| + c]2 . 2(x − a) 4 x−a 43. (x − a)(x − b) dy − y 1/2 dx = 2(b − a)y or y −1/2 dy 6 cos x dy 6 cos x du dy + y = 3y 2/3 or y −2/3 + y 1/3 = 3 . Let u = y 1/3 =⇒ 3 = y −2/3 . Substituting dx x x dx x x dx dx dy 6 1/3 cos x du 2 cos x these results into y −2/3 +y =3 yields + u= . An integrating factor for this equation dx x x dx x x d2 cos x + x sin x + c is I (x) = x2 so (x u) = x cos x =⇒ x2 u = cos x + x sin x + c =⇒ y 1/3 = . dx x2 44. 65 dy du dy dy + 4xy = 4x3 y 1/2 or y −1/2 + 4xy 1/2 = 4x3 . Let u = y 1/2 =⇒ 2 = y −1/2 . Substituting these dx dx dx dx dy du results into y −1/2 + 4xy 1/2 = 4x3 yields + 2xu = 2x3 . An integrating factor for this equation is I (x) = dx dx d x2 2 2 x2 x2 3 x2 x2 e so (e u) = 2e x =⇒ e u = e (x2 − 1)+ c =⇒ y 1/2 = x2 − 1+ ce−x =⇒ y (x) = [(x2 − 1)+ ce−x ]2 . dx 45. dy 1 dy 1 1 du dy − = 2xy 3 or y −3 − y −2 = 2x. Let u = y −2 =⇒ = y −3 . Substituting these dx 2x ln x dx 2x ln x 2 dx dx 1 du 1 dy results into y −3 − y −2 = 2x yields + u = −4x. An integrating factor for this equation is dx 2x ln x dx x ln x d ln x I (x) = ln x so (u ln x) = −4x ln x =⇒ u ln x = x2 − 2x2 ln x + c =⇒ y 2 = 2 . dx x (1 − 2 ln x) + c 46. dy 1 3 dy 1 3x 1 du dy − y= xy π or y −π − y 1−π = . Let u = y 1−π =⇒ = y −π . dx (π − 1)x (1 − π ) dx (π − 1)x 1−π 1 − π dx dx dy 1 3x du 1 Substituting these results into y −π − y 1−π = yields + u = 3x. An integrating factor for dx (π − 1)x 1−π dx x 1/(1−π ) d x3 + c x3 + c this equation is I (x) = x so (xu) = 3x2 =⇒ xu = x3 + c =⇒ y 1−π = =⇒ y (x) = . dx x x 47. dy du dy dy + y cot x = 8y −1 cos3 x or 2y + y 2 cot x = 8 cos2 x. Let u = y 2 =⇒ = 2y . Substituting these dx dx dx dx du dy results into 2y + y 2 cot x = 8 cos2 x yields + u sec x = sec x. An integrating factor for this equation is dx dx d −2 cos4 x + c 3 (u sin x) = 8 cos x sin x =⇒ u sin x = −2 cos4 x + c =⇒ y 2 = . I (x) = sin x so dx sin x 48. 2 √ √ dy √ √ √ dy √ du 49. (1 − 3) + y sec x = y 3 sec x or (1 − 3)y − 3 + y 1− 3 sec x = sec x. Let u = y 1− 3 =⇒ = (1 − dx dx dx √ dy √ dy √ √ −3 √ du . Substituting these results into (1 − 3)y − 3 + y 1− 3 sec x = sec x yields + u sec x = sec x. 3)y dx dx dx d An integrating factor for this equation is I (x) = sec x +tan x so [(sec x +tan x)u] = sec x(sec x +tan x) =⇒ dx √ 1/(1− 3) √ 1 c . (sec x + tan x)u = tan x + sec x + c =⇒ y 1− 3 = 1 + =⇒ y (x) = 1 + sec x + tan x sec x + tan x dy 2x 1 dy 2x 1 du dy + y = xy 2 or 2 + = x. Let u = y −1 so = −y −2 . Substituting these dx 1 + x2 y dx 1 + x2 y dx dx 1 dy 2x 1 du 2x results into 2 + = x yields − u = −x. An integrating factor for this equation is y dx 1 + x2 y dx 1 + x2 1 d u x u xdx u 1 I (x) = so =− =⇒ =− + c =⇒ = − ln (1 + x2 )+ c =⇒ 1 + x2 dx 1 + x2 1 + x2 1 + x2 1 + x2 1 + x2 2 1 1 u = (1 + x2 ) − ln (1 + x2 ) + c =⇒ y −1 = (1 + x2 ) − ln (1 + x2 ) + c . Since y (0) = 1 =⇒ c = 1 so 2 2 1 1 = (1 + x2 ) − ln (1 + x2 ) + 1 . y 2 50. 51. dy dy 1 du dy + y cot x = y 3 sin3 x or y −3 + y −2 cot x = sin3 x. Let u = y −2 =⇒ − = y −3 . Substituting dx dx 2 dx dx 66 du dy + y −2 cot x = sin3 x yields − 2u cot x = −2 sin3 x. An integrating factor for this dx dx d equation is I (x) = csc2 x so (u csc2 x) = − sin x =⇒ u csc2 x = 2 cos x + c. Since y (π/2) = 1 =⇒ c = 1. dx 1 Thus y 2 = . sin2 x(2 cos x + 1) these results into y −3 52. 1 b 53. dv dy dy dv dy dy − a =⇒ = = F (ax + by + c). Let v = ax + by + c so that = a+b =⇒ b = dx dx dx dx dx dx dv dv dv dv = dx. − a = F (v ) =⇒ − a = bF (v ) =⇒ = bF (v ) + a =⇒ dx dx dx bf (v ) + a dy dy dv dv = (9x − y )2 . Let v = 9x − y so that = 9− =⇒ = 9 − v 2 =⇒ dx dx dx dx dv = 9 − v2 dx =⇒ dv = +4 dx =⇒ 1 tanh−1 (v/3) = x + c1 but y (0) = 0 so c = 0. Thus, tanh−1 (3x − y/3) = 3x or y (x) = 3(3x − tanh 3x). 3 54. dy dv dy dv = (4x + y + 2)2 . Let v = 4x + y + 2 so that = 4+ =⇒ 2 = dx =⇒ dx dx dx v +4 v2 1 tan−1 v/2 = x + c1 =⇒ tan−1 (2x + y/2 + 1) = 2x + c =⇒ y (x) = 2[tan (2x + c) − 2x − 1]. 2 dy 1 dv 1 dv dy = sin2 (3x − 3y + 1). Let v = 3x − 3y + 1 so that 1− =⇒ 1 − = sin2 v =⇒ 55. dx dx 3 dx 3 dx dv = 3 cos2 v =⇒ sec2 vdv = 3 dx =⇒ tan v = 3x + c =⇒ tan (3x − 3y + 1) = 3x + c =⇒ y (x) = dx 1 [3x − tan−1 (3x + c) + 1]. 3 56. V = xy =⇒ V = xy + y =⇒ y = (V − y )/x. Substitution into the differential equation yields dV 1 1 =. (V − y )/x = yF (V )/x =⇒ V = y [F (V ) + 1] =⇒ V = V [F (V ) + 1]/x, so that V [F (V ) + 1] dx x dV 1 1 1 1 = for F (V ) = ln V − 1 yields dV = dx =⇒ ln ln V = V [F (V ) + 1] dx x V ln V x 1 =⇒ y (x) = ecx . x 57. Substituting into ln cx =⇒ V = ecx dy dv 58. (a) x = u − 1, y = v + 1 =⇒ = . Substitution into the given differential equation yields dx du dv u + 2v . = du 2u − v (b) The differential equation obtained in (a) is first order homogeneous. We therefore let W = v/u, and 1 + 2W 1 + W2 substitute into the differential equation to obtain W u + W = =⇒ W u = . Separating the 2−W 2−W 2 W 1 variables yields − dW = du. This can be integrated directly to obtain 2 tan−1 W − 1 + W2 1 + W2 u 1 −1 −1 ln (1 + W 2 ) = ln u + ln c. Simplifying we obtain cu2 (1 + W 2 ) = e4 tan W =⇒ c(u2 + v 2 ) = etan (v/u) . 2 −1 Substituting back in for x and y yields c[(x + 1)2 + (y − 1)2 ] = etan [(y−1)/(x+1)] . 59. (a) y = Y (x) + v −1 (x) =⇒ y = Y (x) − v −2 (x)v (x). Now substitute into the given differential 67 equation and simplify algebraically to obtain Y (x) + p(x)Y (x) + q (x)Y 2 (x) − v −2 (x)v (x) + v −1 (x)p(x) + q (x)[2Y (x)v −1 (x) + v −2 (x)] = r(x). We are told that Y (x) is a particular solution to the given differential equation, and therefore Y (x) + p(x)Y (x) + q (x)Y 2 (x) = r(x). Consequently the transformed differential equation reduces to −v −2 (x)v (x) + v −1 p(x) + q (x)[2Y (x)v −1 (x) + v −2 (x)] = 0, or equivalently v − [p(x) + 2Y (x)q (x)]v = q (x). (b) The given differential equation can be written as y − x−1 y − y 2 = x−2 , which is a Riccati differential equation with p(x) = −x−1 , q (x) = −1, and r(x) = x−2 . Since y (x) = −x−1 is a solution to the given differential equation, we make a substitution y (x) = −x−1 + v −1 (x). According to the result from part (a), the given differential equation then reduces to v − (−x−1 +2x−1 )v = −1, or equivalently v − x−1 v = −1. This d −1 linear differential equation has an integrating factor I (x) = x−1 , so that v must satisfy (x v ) = −x−1 =⇒ dx 1 1 1 1 v (x) = x(c−ln x). Hence the solution to the original equation is y (x) = − + = −1 . x x(c − ln x) x c − ln x 60. (a) If y = axr , then y = arxr−1 . Substituting these expressions into the given differential equation yields arxr−1 + 2axr−1 − a2 x2r = −2x−2 . For this to hold for all x > 0, the powers of x must match up on either side of the equation. Hence, r = −1. Then a is determined from the quadratic −a + 2a − a2 = −1 ⇐⇒ a2 − a − 2 = 0 ⇐⇒ (a − 2)(a + 1) = 0. Consequently, a = 2, −1 in order for us to have a solution to the given differential equation. Therefore, two solutions to the differential equation are y1 (x) = 2x−1 , y2 (x) = −x−1 . (b) Taking Y (x) = 2x−1 and using the result from problem 59(a), we now substitute y (x) = 2x−1 + v −1 into the given Riccati equation. The result is (−2x−2 − v −2 v ) + 2x−1 (2x−1 + v −1 ) − (4x−2 + 4x−1 v −1 + v −2 ) = −2x−2 . Simplifying this equation yields the linear equation v + 2x−1 v = −1. Multiplying by the integrating −1 d2 factor I (x) = e 2x dx = x2 results in the integrable differential equation (x v ) = −x2 . Integrating this dx 1 1 differential equation we obtain v (x) = x2 − x3 + c = x2 (c1 − x3 ). Consequently, the general solution 3 3 2 3 . to the Riccati equation is y (x) = + 2 x x (c1 − x3 ) 61. (a) y = x−1 + w(x) =⇒ y = −x−2 + w . Substituting into the given differential equation yields (−x−2 + w ) + 7x−1 (x−1 + w) − 3(x−2 + 2x−1 w + w2 ) = 3x−2 which simplifies to w + x−1 w − 3w2 = 0. (b) The preceding equation can be written in the equivalent form w−2 w + x−1 w−1 = 3. We let u = w−1 , so that u = −w−2 w . Substitution into the differential equation gives, after simplification, u − x−1 u = −3. An integrating factor for this linear differential equation is I (x) = x−1 , so that the differential equation d −1 can be written in the integrable form (x u) = −3x−1 . Integrating we obtain u(x) = x(−3 ln x + dx 1 c), so that w(x) = . Consequently the solution to the original Riccati equation is y (x) = x(c − 3 ln x) 1 1 1+ . x c − 3 ln 3 62. y −1 dy du 1 dy du + p(x) ln y = q (x). If we let u = ln y , then = and the given equation becomes + dx dx y dx dx p(x)u = q (x) which is a first order linear and has a solution of the form u = e− Substituting ln y = e− where I (x) = e p(t)dt p(x)dx e p(x)dx p(x)dx e p(x)dx I q (x)dx + c into u = ln y we obtain y (x) = e and c is an arbitrary constant. −1 q (x)dx + c . I (t)q (t)dt+c 68 2 1 − 2 ln x du 2 dy − ln y = . Let u = ln y so using the technique of the preceding problem: − u= dx x x dx x dx dx 2 −2 1 − 2 ln x 1 − 2 ln x 1 − 2 ln x x e x and u = e dx + c1 = x2 dx + c1 = ln x + cx2 , x x x3 63. y −1 2 and since u = ln y, ln y = ln x + cx2 . Now y (1) = e so c = 1 =⇒ y (x) = xex . du dy dy du = f (y ) and the given equation f (y ) + p(x)f (y ) = q (x) becomes + dx dx dx dx − p(x)dx p(x)dx p(x)u = q (x) which has a solution of the form u(x) = e e q (x)dx + c . Substituting 64. If u = f (y ), then f (y ) = e− p(x)dx y (x) = f −1 I −1 e p(x)dx q (x)dx + c into u = f (y ) and using the fact that f is invertible, we obtain I (t)q (t)dt + c where I (x) = e p(t)dt and c is and arbitrary constant. dy 1 1 du dy +√ = sec2 y and the given equation tan y = √ . Let u = tan y so that dx dx dx 2 1+x 2 1+x du 1 1 becomes +√ u= √ which is first order linear. An integrating factor for this equation is dx 2 1+x 2 1+x√ √ √ √ √ √ e 1+x e 1+x d √1+x 1+x 1+x √ (e u) = √ =⇒ e u= =⇒ e 1+x u = e 1+x + c =⇒ u = I (x) = e =⇒ dx 2 1+x √ 2 1+x √ √ 1 + ce− 1+x . But u = tan y so tan y = 1 + ce− 1+x or y (x) = tan−1 (1 + ce− 1+x ). 65. sec2 y Solutions to Section 1.9 True-False Review: 1. FALSE. The requirement, as stated in Theorem 1.9.4, is that My = Nx , not Mx = Ny , as stated. 2. FALSE. A potential function φ(x, y ) is not an equation. The general solution to an exact differential equation takes the form φ(x, y, ) = c, where φ(x, y ) is a potential function. 3. FALSE. According to Definition 1.9.2, M (x)dx + N (y )dy = 0 is only exact if there exists a function φ(x, y ) such that φx = M and φy = N for all (x, y ) in a region R of the xy -plane. 4. TRUE. This is the content of part 1 of Theorem 1.9.11. 5. FALSE. If φ(x, y ) is a potential function for M (x, y )dx + N (x, y )dy = 0, then so is φ(x, y ) + c for any constant c. 6. TRUE. We have My = 2e2x − cos y and Nx = 2e2x − cos y, and so since My = Nx , this equation is exact. 7. TRUE. We have My = (x2 + y )2 (−2x) + 4xy (x2 + y ) = Nx , (x2 + y )4 and so this equation is exact. 8. FALSE. We have My = 2y and Nx = 2y 2 , 69 and since My = Nx , we conclude that this equation is not exact. 9. FALSE. We have My = ex sin y cos y + xex sin y cos y and Nx = cos y sin yex sin y , and since My = Nx , we conclude that this equation is not exact. Problems: 1. (y + 3x2 )dx + xdy = 0. M = y + 3x2 and N = x =⇒ My = 1 and Nx = 1 =⇒ My = Nx =⇒ the differential equation is exact. 2. [cos (xy ) − xy sin (xy )]dx − x2 sin (xy )dy = 0 =⇒ M = cos (xy ) − xy sin (xy ) and N = −x2 sin (xy ) =⇒ My = −2x sin (xy ) − x2 y cos (xy ) and Nx = −2x sin (xy ) − x2 y cos (xy ) =⇒ My = Nx =⇒ the differential equation is exact. 3. yexy dx + (2y − xe−xy )dy = 0. M = yexy and N = 2y − xe−xy =⇒ My = yxexy + exy and Nx = xye−xy − e−xy =⇒ My = Nx =⇒ the differential equation is not exact. 4. 2xydx + (x2 + 1)dy = 0. M = 2xy and N = x2 + 1 =⇒ My = 2x and Nx = 2x =⇒ My = Nx =⇒ ∂φ ∂φ = 2xy and (b) = the differential equation is exact so there exists a potential function φ such that (a) ∂x ∂x dh(x) dh(x) dh(x) 2xy + so from (a), 2xy = 2xy + =⇒ = 0 =⇒ h(x) is a constant. Since we need just one dx dx dx 2 potential function, let h(x) = 0. Thus, φ(x, y ) = (x + 1)y ; hence, (x2 + 1)y = c. 5. (y 2 + cos x)dx + (2xy + sin y )dy = 0. M = y 2 + cos x and N = 2xy + sin y =⇒ My = 2y and Nx = 2y =⇒ My = Nx =⇒ the differential equation is exact so there exists a potential function φ such that ∂φ ∂φ ∂φ dh(y ) (a) = y 2 + cos x and (b) = 2xy + sin y . From (a) φ(x, y ) = xy 2 + sin x + h(y ) =⇒ = 2xy + so ∂x ∂y ∂y dy dh(y ) dh = sin y =⇒ h(y ) = − cos y where the constant of integration has = 2xy + sin y =⇒ from (b) 2xy + dy dy been set to zero since we just need one potential function. φ(x, y ) = xy 2 +sin x − cos y =⇒ xy 2 +sin x − cos y = c. xy − 1 xy + 1 dx + dy = 0 then My = Nx = 1 =⇒ then the differential equation is exact so x y xy − 1 ∂φ xy + 1 ∂φ there exists a potential function φ such that (a) = and (b) = . From (a) φ(x, y ) = ∂x x ∂y y ∂φ dh(y ) dh(y ) xy + 1 dh(y ) xy − ln |x| + h(y ) =⇒ =x+ so from (b), x + = =⇒ = y −1 =⇒ h(y ) = ln |y | ∂y dy dy y dy where the constant of integration has been set to zero since we need just one potential function. φ(x, y ) = xy + ln |y/x| =⇒ xy + ln |x/y | = c. 6. Given 7. Given (4e2x + 2xy − y 2 )dx + (x − y )2 dy = 0 then My = Nx = 2y so the differential equation is exact ∂φ ∂φ and there exists a potential function φ such that (a) = 4e2x + 2xy − y 2 and (b) = (x − y )2 . ∂x ∂y y3 ∂φ dh(x) dh(x) From (b) φ(x, y ) = x2 y − xy 2 + + h(x) =⇒ = 2xy − y 2 + so from (a) 2xy − y 2 + = 3 ∂x dx dx dh(x) 4e2x + 2xy − y 2 =⇒ = 4e2x =⇒ h(x) = 2e2x where the constant of integration has been set to zero dx 70 since we need just one potential function. φ(x, y ) = x2 y − xy 2 + c1 =⇒ 6e2x + 3x2 y − 3xy 2 + y 3 = c. y3 y3 + 2e2x =⇒ x2 y − xy 2 + + 2e2x = 3 3 8. Given (y 2 − 2x)dx + 2xydy = 0 then My = Nx = 2xy so the differential equation is exact and there exists ∂φ ∂φ a potential function φ such that (a) = y 2 − 2x and (b) = 2xy . From (b) φ(x, y ) = xy 2 + h(x) =⇒ ∂x ∂y ∂φ dh(x) dh(x) dh(x) = y2 + so from (a) y 2 + = y 2 − 2x =⇒ = −2x =⇒ h(x) = −2x where the constant of ∂x dx dx dx integration has been set to zero since we just need one potential function. φ(x, y ) = xy 2 − x2 =⇒ xy 2 − x2 = c. x y 2 − x2 dy = 0 then My = Nx = so the differential equation 2 +y (x2 + y 2 )2 ∂φ ∂φ 1 y x is exact and there exists a potential function φ such that (a) = −2 and (b) =2 . ∂x x x + y2 ∂y x + y2 ∂φ y dh(x) y dh(x) From (b) φ(x, y ) = tan−1 (y/x) + h(x) =⇒ =− 2 + so from (a) − 2 + = ∂x x + y2 dx x + y2 dx 1 y dh − =⇒ = x−1 =⇒ h(x) = ln |x| where the constant of integration is set to zero since we only x x2 + y 2 dx need one potential function. φ(x, y ) = tan−1 (y/x) + ln |x| =⇒ tan−1 (y/x) + ln |x| = c. 9. Given y 1 − x x2 + y 2 dx + x2 x 10. Given [1+ln (xy )]dx + dy = 0 then My = Nx = y −1 so the differential equation is exact and there exists y ∂φ ∂φ dh(x) a potential function φ such that (a) = 1+ln (xy ) and (b) φ(x, y ) = x ln y + h(x) =⇒ = ln y + so ∂x ∂x dx dh dh(x) = 1 ln (xy ) =⇒ = 1+ln x =⇒ h(x) = c ln x where the constant of integration is set to from (a) ln y + dx dx zero since we only need one potential function. φ(x, y ) = x ln y +x ln x =⇒ x ln y +x ln x = c =⇒ x ln (xy ) = c. 11. Given [y cos (xy ) − sin x]dx + x cos (xy )dy = 0 then My = Nx = −xy sin (xy ) + cos (xy ) so the differential ∂φ = y cos (xy ) − sin x and (b) equation is exact so there exists a potential function φ such that (a) ∂x ∂φ ∂φ dh(x) = x cos (xy ). From (b) φ(x, y ) = sin (xy ) + h(x) =⇒ = y cos (xy ) + so from (a) y cos (xy ) + ∂x ∂x dx dh(x) dh = − sin x =⇒ h(x) = cos x where the constant of integration is set to zero = y cos (xy ) − sin x =⇒ dx dx since we only need one potential function. φ(x, y ) = sin (xy ) + cos x =⇒ sin (xy ) + cos x = c. 12. Given (2xy + cos y )dx + (x2 − x sin y − 2y )dy = 0 then My = Nx = 2x − sin y so the differential equation ∂φ ∂φ is exact so there is a potential function φ such that (a) = 2xy + cos y and (b) = x2 − x sin y − 2y . ∂x ∂y ∂φ dh(y ) dh(y ) From (a) φ(x, y ) = x2 y + x cos y + h(y ) =⇒ = x2 − x sin y + so from (b) x2 − x sin y + = ∂y dy dy dh x2 − x sin y − 2y =⇒ = −2y =⇒ h(y ) = −y 2 where the constant of integration has been set to zero since dy we only need one potential function. φ(x, y ) = x2 y + x cos y − y 2 =⇒ x2 y + x cos y − y 2 = c. 13. Given (3x2 ln x + x2 − y )dx − xdy = 0 then My = Nx = −1 so the differential equation is exact so ∂φ ∂φ there exists a potential function φ such that (a) = 3x2 ln x + x2 − y and (b) = x. From (b) ∂x ∂y 71 dh(x) dh(x) dh(x) ∂φ = −y + so from (a) −y + = 3x2 ln x + x2 − y =⇒ = ∂x dx dx dx 2 2 3 3x ln x + x =⇒ h(x) = x ln x where the constant of integration has been set to zero since we only need one potential function. φ(x, y ) = −xy + x3 ln x =⇒ −xy + x3 ln x = c. Now since y (1) = 5, c = −5; thus, x3 ln x + 5 x3 ln x − xy = −5 or y (x) = . x φ(x, y ) = −xy + h(x) =⇒ dx + 4xy = 3 sin x =⇒ (4xy − 3 sin x)dx + 2x2 dy = 0 then My = Nx = 4x so the differential dy ∂φ ∂φ equation is exact so there exists a potential function φ such that (a) = 4xy − 3 sin x and (b) = 2x2 . ∂x ∂y ∂φ dh(x) dh(x) dh(x) From (b) φ(x, y ) = 2x2 y + h(x) =⇒ = 4xy + so from (a) 4xy + = 4xy − 3 sin x =⇒ = ∂x dx dx dx −3 sin x =⇒ h(x) = 3 cos x where the constant of integration has been set to zero since we only need one potential function. φ(x, y ) = 2x2 y + 3 cos x =⇒ 2x2 y + 3 cos x = c. Now since y (2π ) = 0, c = 3; thus, 3 − 3 cos x . 2x2 y + 3 cos x = 3 or y (x) = 2x2 14. Given 2x2 15. Given (yexy + cos x)dx + xexy dy = 0 then My = Nx = xyexy + exy so the differential equation is ∂φ ∂φ exact so there exists a potential function φ such that (a) = yexy + cos x and (b) = xexy . From (b) ∂x ∂y ∂φ dh(x) dh(x) φ(x, y ) = exy +h(x) =⇒ = yexy + so from (a) yexy +cos x =⇒ = cos x =⇒ h(x) = sin x where ∂x dx dx the constant of integration is set to zero since we only need one potential function. φ(x, y ) = exy + sin x =⇒ ln (2 − sin x) . exy + sin x = c. Now since y (π/2) = 0, c = 2; thus, exy + sin x = 2 or y (x) = x 16. If φ(x, y ) is a potential function for M dx + N dy = 0 =⇒ d(φ(x, y )) = 0 so d(φ(x, y ) + c) = d(φ(x, y )) + d(c) = 0 + 0 = 0 =⇒ φ(x, y ) + c is also a potential function. 17. M = cos (xy )[tan (xy ) + xy ] and N = x2 cos (xy ) =⇒ My = 2x cos (xy ) − x2 y sin (xy ) = Nx =⇒ My = Nx =⇒ M dx = N dy = 0 is exact so I (x, y ) = cos (xy ) is an integrating factor for [tan (xy )+ xy ]dx + x2 dy = 0. 18. M = sec x[2x − (x2 + y 2 ) tan x] and N = 2y sec x =⇒ My = −2y sec x tan x and Nx = 2y sec x tan x =⇒ My = Nx =⇒ M dx + N dy = 0 is not exact so I (x) = sec x is not an integrating factor for [2x − (x2 + y 2 ) tan x]dx + 2ydy = 0. 19. M = e−x/y (x2 y −1 − 2x) and N = −e−x/y x3 y −2 =⇒ My = e−x/y (x3 y −3 − 3x2 y −2 ) = Nx =⇒ M dx + N dy = 0 is exact so I (x, y ) = y −2 e−x/y is an integrating factor for y [x2 − 2xy ]dx − x3 dy = 0. 20. Given (xy − 1)dx + x2 dy = 0 then M = xy − 1 and N = x2 . Thus My = x and Nx = 2x so My − N x = −x−1 = f (x) is a function of x alone so I (x) = e f (x)dx = x−1 is an integrating factor for the N given equation. Multiplying the given equation by I (x) results in the exact equation (y − x−1 )dx + xdy = 0. We find that φ(x, y ) = xy − ln |x| and hence, the general solution of our differential equation is xy − ln |x| = c. 21. Given ydx − (2x + y 4 )dy = 0 then M = y and N = −(2x + y 4 ). Thus My = 1 and Nx = −2 so My − N x = 3y −1 = g (y ) is a function of y alone so I (y ) = e− g(y)dy = 1/y 3 is an integrating factor M for the given differential equation. Multiplying the given equation by I (y ) results in the exact equation y −2 dx − (2xy −3 + y )dy = 0. We find that φ(x, y ) = xy −2 − y 2 /2 and hence, the general solution of our 72 differential equation is xy −2 − y 2 /2 = c1 =⇒ 2x − y 4 = cy 2 . 22. Given x2 ydx + y (x3 + e−3y sin y )dy = 0 then M = x2 y and N = y (x3 + e−3y sin y ). Thus My = x2 My − N x and Nx = 3x2 y so = y −1 − 3 = g (y ) is a function of y alone so I (y ) = e g(y)dy = e3y /y is an M integrating factor for the given equation. Multiplying the equation by I (y ) results in the exact equation x3 e3y x2 e3y dx + e3y (x3 + e−3y sin y )dy = 0. We find that φ(x, y ) = − cos y and hence the general solution 3 3 3y xe of our differential equation is − cos y = c. 3 My − N x 23. Given (y − x2 )dx + 2xdy = 0 then M = y − x2 and N = 2x. Thus My = 1 and Nx = 2 so = N 1 1 − = f (x) is a function of x alone so I (x) = e f (x)dx = √ is an integrating factor for the given equation. 2x x Multiplying the given equation by I (x) results in the exact equation (x−1/2 y − x3/2 )dx +2x1/2 dy = 0. We find 2x5/2 2x5/2 that φ(x, y ) = 2x1/2 y − and hence the general solution of our differential equation is 2x1/2 y − =c 5 5 5/2 c + 2x √. or y (x) = 10 x 24. Given xy [2 ln (xy ) + 1]dx + x2 dy = 0 then M = xy [2 ln (xy ) + 1] and N + x2 . Thus My = 3x + 1 MY − N x = y −1 = g (y ) is a function of y only so I (y ) = e g(y)dy = is an 2x ln (xy ) and Nx = 2x so M y integrating factor for the given equation. Multiplying the given equation by I (y ) results in the exact equation x[2 ln (xy ) + 1]dx + x2 y −1 dy = 0. We find that φ(x, y ) = x2 ln y + x2 ln x and hence the general solution of 2 our differential equation is x2 ln y + x2 ln x = c or y (x) = xec/x . dy 2xy 1 + = =⇒ (2xy + 2x3 y − 1)dx + (1 + x2 )2 dy = 0 then M = 2xy + 2x3 y − 1 and dx 1 + x2 (1 + x2 )2 My − N x 2x N = (1 + x2 )2 . Thus My = 2x + 2x3 and Nx = 4x(1 + x2 ) so =− = f (x) is a function of N 1 + x2 1 x alone so I (x) = e f (x)dx = is an integrating factor for the given equation. Multiplying the given 1 + x2 1 equation by I (x) yields the exact equation 2xy − dx + (1 + x2 )dy = 0. We find that φ(x, y ) = 1 + x2 (1 + x2 )y − tan−1 x and hence the general solution of our differential equation is (1 + x2 )y − tan−1 x = c or tan −1x + c y (x) = . 1 + x2 25. Given 26. Given (3xy − 2y −1 )dx + x(x + y −2 )dy = 0 then M = 3xy − 2y −1 and N = x(x + y −2 ). Thus My − N x 1 = = f (x) is a function of x alone so I (x) = e f (x)dx = x My = 3x + 2y −2 and Nx = 2x + y −2 so N x is an integrating factor for the given equation. Multiplying the given equation by I (x) results in the exact equation (3x2 y − 2xy −1 )dx + x2 (x + y −2 )dy = 0. We find that φ(x, y ) = x3 y − x2 y −1 and hence the general solution of our differential equation is x3 y − x2 y −1 = c. 27. Given (y −1 − x−1 )dx + (xy −2 − 2y −1 )dy = 0 =⇒ xr y s (y −1 − x−1 )dx + xr y s (xy −2 − 2y −1 )dy = 0 =⇒ (xr y s−1 − xr−1 y s )dx +(xr+1 y s−2 − 2xr y s−1 )dy = 0. Then M = xr y s−1 − xr−1 y s and N = xr+1 y s−2 − 2xr y s−1 so My = xr (s − 1)y s−2 − xr−1 sy s−1 and Nx = (r + 1)xr y s−2 − 2rxr−1 y s−1 . The equation is exact if and 73 only if My = Nx =⇒ xr y s−1 − xr−1 y s = (r + 1)xr y s−2 − 2rxr−1 y s−1 =⇒ s r+1 2r s−1 − = − =⇒ y2 xy y2 xy s − 2r s−r−2 = . From the last equation we require that s − r − 2 = 0 and s − 2r = 0. Solving this system y2 xy yields r = 2 and s = 4. 28. Given y (5xy 2 + 4)dx + x(xy 2 − 1)dy = 0 =⇒ xr y s y (5xy 2 + 4)dx + xr y s x(xy 2 − 1)dy = 0. Then M = xr y s+1 (5xy 2 + 4) and N = xr+1 y s (xy 2 − 1) so My = 5(s + 3)xr+1 y s+2 + 4(s + 1)xr y s and Nx = (r + 2)xr+1 y s−2 − (r + 1)xr y s . The equation is exact if and only if My = Nx =⇒ 5(s + 3)xr+1 y s+2 + 4(s + 1)xr y s = (r + 2)xr+1 y s+2 − (r + 1)xr y s =⇒ 5(s + 3)xy 2 + 4(s + 1) = (r + 2)xy 2 − (r + 1). From the last equation we require that 5(s + 3) = r + 2 and 4(s + 1) = −(r + 1). Solving this system yields r = 3 and s = −2. 29. Given 2y (y + 2x2 )dx + x(4y + 3x2 )dy = 0 =⇒ xr y s 2y (y + 2x2 )dx + xr y s x(4y + 3x2 )dy = 0. Then M = 2xr y s+2 + 4xr+2 y s+1 and N = 4xr+1 y s+1 + 3xr+3 y s so My = 2xr (s + 2)y s+1 + 4xr+2 (s + 1)y s and Nx = 4(r + 1)xr y s+1 + 3(r + 3)xr+2 y s . The equation is exact if and only if My = Nx =⇒ 2xr (s + 2)y s+1 + 4xr+2 (s + 1)y s = 4(r + 1)xr y s+1 + 3(r + 3)xr+2 y s =⇒ 2(s + 2)y + 4x2 (s + 1) = 4(r + 1)y + 3(r + 3)x2 . From this last equation we require that 2(s + 2) = 4(r + 1) and 4(s + 1) = 3(r + 3). Solving this system yields r = 1 and s = 2. My − N x = g (y ) is a function of y only. Then dividing the equation (1.9.21) by M , M it follows that I is an integrating factor for M (x, y )dx + N (x, y )dy = 0 if and only if it is a solution of N ∂I ∂I − = Ig (y ) (30.1). We must show that this differential equation has a solution I = I (y ). However, M ∂x ∂y dI if I = I (y ), then (30.1) reduces to = −Ig (y ), which is a separable equation with solution I (y ) = e− g(t)dt . dy 30. Suppose that dy + py = q can be written in the differential form as (py − q )dx + dy = 0 (31.1). This dx My − N x has M = py − q and N = 1 so that = p(x). Consequently, an integrating factor for (31.1) is N x I (x) = e p(t)dt . 31. (a) Note x x x φ(x, y ) = ye p(t)dt x − q (x)e p(t)dt p(t)dt x yields the exact equation e p(t)dt (py − q )dx + e p(t)dt dy = 0. x x ∂φ ∂φ Hence, there exists a potential function φ such that (i) = e p(t)dt (py − q ) and (ii) = e p(t)dt . From ∂x ∂y x x x x dh(x) dh(x) p(t)dt p(t)dt p(t)dt (i) p(x)ye + =e (py − q ) =⇒ = −q (x)e =⇒ h(x) = − q (x)e p(t)dt dx dx dx where the constant of integration has been set to zero since we just need one potential function. Consequently, (b) Multiplying (31.1) by I (x) = e dx =⇒ y (x) = I −1 x x Iq (t)dt + c , where I (x) = e Solutions to Section 1.10 True-False Review: 1. TRUE. This is well-illustrated by the calculations shown in Example 1.10.1. 2. TRUE. The equation y1 = y0 + f (x0 , y0 )(x1 − x0 ) p(t)dt . 74 dy is the tangent line to the curve dx = f (x, y ) at the point (x0 , y0 ). Once the point (x1 , y1 ) is determined, the procedure can be iterated over and over at the new points obtained to carry out Euler’s method. 3. FALSE. It is possible, depending on the circumstances, for the errors associated with Euler’s method to decrease from one step to the next. Problems: 1. Applying Euler’s method with y = 4y − 1, x0 = 0, y0 = 1, and h = 0.05 we have yn+1 = yn +0.05(4yn − 1). This generates the sequence of approximants given in the table below. n 1 2 3 4 5 6 7 8 9 10 xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.15 1.33 1.546 1.805 2.116 2.489 2.937 3.475 4.120 4.894 Consequently the Euler approximation to y (0.5) is y10 = 4.894. (Actual value: y (.05) = 5.792 rounded to 3 decimal places). xn yn 2xy , x0 = 0, y0 = 1, and h = 0.1 we have yn+1 = yn − 0.2 . 1 + x2 1 + x2 n This generates the sequence of approximants given in the table below. 2. Applying Euler’s method with y = − n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn 1 0.980 0.942 0.891 0.829 0.763 0.696 0.610 0.569 0.512 Consequently the Euler approximation to y (1) is y10 = 0.512. (Actual value: y (1) = 0.5). 2 3. Applying Euler’s method with y = x − y 2 , x0 = 0, y0 = 2, and h = 0.05 we have yn+1 = yn +0.05(xn − yn ). This generates the sequence of approximants given in the table below. 75 n 1 2 3 4 5 6 7 8 9 10 xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.80 1.641 1.511 1.404 1.316 1.242 1.180 1.127 1.084 1.048 Consequently the Euler approximation to y (0.5) is y10 = 1.048. (Actual value: y (.05) = 1.088 rounded to 3 decimal places). 4. Applying Euler’s method with y = −x2 y, x0 = 0, y0 = 1, and h = 0.2 we have yn+1 = yn − 0.2x2 yn . This n generates the sequence of approximants given in the table below. n 1 2 3 4 5 xn 0.2 0.4 0.6 0.8 1.0 yn 1 0.992 0.960 0.891 0.777 Consequently the Euler approximation to y (1) is y5 = 0.777. (Actual value: y (1) = 0.717 rounded to 3 decimal places). 2 5. Applying Euler’s method with y = 2xy 2 , x0 = 0, y0 = 1, and h = 0.1 we have yn+1 = yn + 0.1xn yn . This generates the sequence of approximants given in the table below. n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn 0.5 0.505 0.515 0.531 0.554 0.584 0.625 0.680 0.754 0.858 Consequently the Euler approximation to y (1) is y10 = 0.856. (Actual value: y (1) = 1). ∗ 6. Applying the modified Euler method with y = 4y − 1, x0 = 0, y0 = 1, and h = 0.05 we have yn+1 = yn + 0.05(4yn − 1) ∗ yn+1 = yn + 0.025(4yn − 1 + 4yn+1 − 1). This generates the sequence of approximants given in the table below. 76 n 1 2 3 4 5 6 7 8 9 10 xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.165 1.3663 1.6119 1.9115 2.2770 2.7230 3.2670 3.9308 4.7406 5.7285 Consequently the modified Euler approximation to y (0.5) is y10 = 5.7285. (Actual value: y (.05) = 5.7918 rounded to 4 decimal places). 2xy ∗ , x0 = 0, y0 = 1, and h = 0.1 we have yn+1 = 7. Applying the modified Euler method with y = − 1 + x2 xn yn yn − 0.2 1 + x2 n ∗ xn+1 yn+1 xn yn −2 . This generates the sequence of approximants given in the table yn+1 = yn + 0.05 − 1 + x2 1 + x2 +1 n n below. n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn 0.9900 0.9616 0.9177 0.8625 0.8007 0.7163 0.6721 0.6108 0.5536 0.5012 Consequently the modified Euler approximation to y (1) is y10 = 0.5012. (Actual value: y (1) = 0.5). ∗ 8. Applying the modified Euler method with y = x − y 2 , x0 = 0, y0 = 2, and h = 0.05 we have yn+1 = 2 yn − 0.05(xn − yn ) 2 ∗ yn+1 = yn + 0.025(xn − yn + xn+1 − (yn+1 )2 ). This generates the sequence of approximants given in the table below. 77 n 1 2 3 4 5 6 7 8 9 10 xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.8203 1.6725 1.5497 1.4468 1.3600 1.2866 1.2243 1.1715 1.1269 1.0895 Consequently the modified Euler approximation to y (0.5) is y10 = 1.0895. (Actual value: y (.05) = 1.0878 rounded to 4 decimal places). ∗ 9. Applying the modified Euler method with y = −x2 y, x0 = 0, y0 = 1, and h = 0.2 we have yn+1 = 2 yn − 0.2xn yn ∗ yn+1 = yn − 0.1[x2 yn + x2 +1 yn+1 ]. This generates the sequence of approximants given in the table below. n n n 1 2 3 4 5 xn 0.2 0.4 0.6 0.8 1.0 yn 0.9960 0.9762 0.9266 0.8382 0.7114 Consequently the modified Euler approximation to y (1) is y5 = 0.7114. (Actual value: y (1) = 0.7165 rounded to 4 decimal places). ∗ 10. Applying the modified Euler method with y = 2xy 2 , x0 = 0, y0 = 1, and h = 0.1 we have yn+1 = 2 yn + 0.1xn yn 2 ∗ yn+1 = yn +0.05[xn yn + xn+1 (yn+1 )2 ]. This generates the sequence of approximants given in the table below. n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn 0.5025 0.5102 0.5235 0.5434 0.5713 0.6095 0.6617 0.7342 0.8379 0.9941 Consequently the modified Euler approximation to y (1) is y10 = 0.9941. (Actual value: y (1) = 1). 11. We have y = 4y − 1, x0 = 0, y0 = 1, and h = 0.05. So, k1 = 0.05(4yn − 1), k2 = 0.05[4(yn + 1 k1 ) − 1], k3 = 2 1 0.05[4(yn + 2 k2 ) − 1], k4 = 0.05[4(yn + 1 k3 ) − 1], 2 78 yn+1 = yn + 1 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below 6 (computations rounded to five decimal places). n 1 2 3 4 5 6 7 8 9 10 xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.16605 1.36886 1.61658 1.91914 2.28868 2.74005 3.29135 3.96471 4.78714 5.79167 Consequently the Runge-Kutta approximation to y (0.5) is y10 = 5.79167. (Actual value: y (.05) = 5.79179 rounded to 5 decimal places). (xn + 0.05)(yn + k1 ) xy xn yn 2 ), k3 = , x0 = 0, y0 = 1, and h = 0.1. So, k1 = −0.2 , k2 = −0.2 [1 + (xn + 0.05)2 ] 1 + x2 1 + x2 n (xn + 0.05)(yn + k2 ) xn+1 (yn + k3 ) 2 , k4 = −0.2 , −0.2 [1 + (xn + 0.05)2 ] [1 + (xn+1 )2 ] 1 yn+1 = yn + 6 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below (computations rounded to seven decimal places). 12. We have y = −2 n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn 0.9900990 0.9615383 0.9174309 0.8620686 0.7999996 0.7352937 0.6711406 0.6097558 0.5524860 0.4999999 Consequently the Runge-Kutta approximation to y (1) is y10 = 0.4999999. (Actual value: y (.05) = 0.5). 2 13. We have y = x − y 2 , x0 = 0, y0 = 2, and h = 0.05. So, k1 = 0.05(xn − yn ), k2 = 0.05[xn + 0.025 − (yn + k1 2 ) ], k3 = 0.05[xn + 0.025 − (yn + k2 )2 ], k4 = 0.05[xn+1 − (yn + k3 )2 ]], 2 2 1 yn+1 = yn + 6 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below (computations rounded to six decimal places). 79 n 1 2 3 4 5 6 7 8 9 10 xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.1.81936 1.671135 1.548079 1.445025 1.358189 1.284738 1.222501 1.169789 1.125263 1.087845 Consequently the Runge-Kutta approximation to y (0.5) is y10 = 1.087845. (Actual value: y (0.5) = 1.087845 rounded to 6 decimal places). 14. We have y = −x2 y, x0 = 0, y0 = 1, and h = 0.2. So, k1 = −0.2x2 yn , k2 = −0.2(xn + 0.1)2 (yn + k1 ), k3 = n 2 −0.2(xn + 0.1)2 (yn + k2 ), k4 = −0.2(xn+1 )2 (yn + k3 ), 2 1 yn+1 = yn + 6 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below (computations rounded to six decimal places). n 1 2 3 4 5 xn 0.2 0.4 0.6 0.8 1.0 yn 0.997337 0.978892 0.930530 0.843102 0.716530 Consequently the Runge-Kutta approximation to y (1) is y10 = 0.716530. (Actual value: y (1) = 0.716531 rounded to 6 decimal places). 2 15. We have y = 2xy 2 , x0 = 0, y0 = 1, and h = 0.1. So, k1 = 0.2xn − yn , k2 = 0.2(xn + 0.05)(yn + k1 )2 , k3 = 2 k2 2 2 0.2(xn + 0.05)(yn + 2 ) , k4 = 0.2xn+1 (yn + k3 ) ]], yn+1 = yn + 1 (k1 + k2 + k3 + k4 ). This generates the sequence of approximants given in the table below 6 (computations rounded to six decimal places). n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn 0.502513 0.510204 0.523560 0.543478 0.571429 0.609756 0.662252 0.735295 0.840336 0.999996 Consequently the Runge-Kutta approximation to y (1) is y10 = 0.999996. (Actual value: y (1) = 1). 80 1 1 y = e−x/10 cos x, x0 = 0, y0 = 0, and h = 0.5 Hence, k1 = 0.5 − yn + exn /10 cos xn , k2 = 10 10 1 1 1 1 (−xn +0.25)/10 0.5 − yn + k1 + e cos (xn + 0.25) , k3 = 0.5 − yn + k2 + e(−xn +0.25)/10 cos (xn + 0.25) , 10 2 10 2 1 k4 = 0.5 − (yn + k3 ) + e−xn+1 /10 cos xn+1 , yn+1 = yn + 1 (k1 + 2k2 + 2k3 + k4 ). This generates the se6 10 quence of approximants plotted in the accompanying figure. We see that the solution appears to be oscillating with a diminishing amplitude. Indeed, the exact solution to the initial value problem is y (x) = e−x/10 sin x. The corresponding solution curve is also given in the figure. 16. We have y + y(x) 0.75 0.5 0.25 x 5 10 15 20 25 -0.25 -0.5 Figure 0.0.56: Figure for Exercise 16 Solutions to Section 1.11 Problems: d2 y 2 dy dy du d2 y = + 4x2 . Let u = so that = . Substituting these results into the first equation 2 dx x dx dx dx dx2 du 2 du 2 yields = u + 4x2 =⇒ − u = 4x2 . An appropriate integrating factor for this equation is I (x) = dx x dx x dx −2 dy x = x−2 =⇒ d(x−2 u) = 4 =⇒ x−2 u = 4 dx =⇒ x−2 u = 4x + c1 =⇒ u = 4x3 + c1 x2 =⇒ e = dx 3 2 3 4 4x + c1 x =⇒ y (x) = c1 x + x + c2 . 1. 2. d2 y 1 dy dy du d2 y = − 1 . Let u = so that = . Substituting these results into 2 dx (x − 1)(x − 2) dx dx dx dx2 81 1 du 1 1 du = (u − 1) =⇒ − u=− . An dx (x − 1)(x − 2) dx (x − 1)(x − 2) (x − 1)(x − 2) 1 − dx (x − 1)(x − 2) = x − 1 =⇒ d x − 1 u = appropriate integrating factor for this equation is I (x) = e x−2 dx x − 2 1 x−1 1 dy 1 −2 =⇒ u = (x − 2) dx =⇒ u = − + c1 =⇒ =− + c1 =⇒ y (x) = − ln |x − 1| + (x − 2)2 x−2 x−1 dx x−1 c1 x + c2 . the first equation yields d2 y 2 + dx2 y 2 dy dy du du . Let u = so that =u = dx dx dx dy du 2 du 2 first equation yields u + u2 = u =⇒ u = 0 or − u = 1. dy y dy y 2 dy y = y 2 =⇒ d (y 2 u) = y 2 =⇒ y 2 u last equation is I (y ) = e dy √ c1 y + 2 =⇒ ln |y 3 + c2 | = x + c3 =⇒ y (x) = 3 c4 ex + c5 . 3y 3. dy dx = d2 y . Substituting these results into the dx2 An appropriate integrating factor for the = y 2 dy =⇒ y 2 u = y3 dy + c1 =⇒ = 3 dx 2 dy du du d2 y so that =u = . Substituting these results into the first dx dx dy dx2 du du = u2 tan y . If u = 0 then = 0 =⇒ y equals a constant and this is a solution to the equation yields u dy dx du du dy equation. Now suppose that u = 0. Then = u tan y =⇒ = tan ydy =⇒ u = c1 sec y =⇒ = dy u dx c1 sec y =⇒ y (x) = sin−1 (c1 x + c2 ). 4. d2 y = dx2 dy dx tan y . Let u = 2 d2 y dy dy dy du d2 y + tan x = . Let u = so that = . Substituting these results into the first 2 dx dx dx dx dx dx2 1 du dz du + tan xu = u2 which is a Bernoulli equation. Letting z = u−1 gives 2 = =− . equation yields dx u dx dx dz Substituting these results into the last equation yields − tan xz = −1. Then an integrating factor for dx d this equation is I (x) = e− tan xdx = cos x =⇒ (z cos x) = − cos x =⇒ z cos x = − cos xdx =⇒ z = dx − sin x + c1 cos x dy cos x =⇒ u = =⇒ = =⇒ y (x) = c2 − ln |c1 − sin x|. cos x c1 − sin x dx c1 − sin x 5. 2 d2 x dx dx dx du d2 x = + 2 . Let u = so that = 2 . Substituting these results into the first equation 2 dt dt dt dt dt dt du du 2 2 yields = u + 2u =⇒ − 2u = u which is a Bernoulli equation. If u = 0 then x is a constant which dt dt dz 1 du satisfies the equation. Now suppose that u = 0. Let z = u−1 so that =− 2 . Substituting these dt u dt dz results into the last equation yields + 2z = −1. An integrating factor for this equation is I (x) = e2t =⇒ dt d 2t 1 2e2t 2e2t (e z ) = −e2t =⇒ z = ce−2t − =⇒ u = =⇒ x = dt =⇒ x(t) = c2 − ln |c1 − e2t |. dt 2 2c − e2t 2c − e2t 6. 7. d2 y dy dy du d2 y −2 = 6x4 . Let u = so that = . Substituting these results into the first equation dx2 dx dx dx dx2 82 dx −2 du 2 4 x = x−2 =⇒ yields − u = 6x . An appropriate integrating factor for this equation is I (x) = e dx x d −2 dy 1 (x u) = 6x2 =⇒ x−2 u = 6 x2 dx =⇒ u = 2x5 + cx2 =⇒ = 2x5 + cx2 =⇒ y (x) = x6 + c1 x3 + c2 . dx dx 3 d2 x dx dx du d2 x = 2 t+ . Let u = so that = 2 . Substituting these results into the first equation dt2 dt dt dt dt du 2 d yields − u = 2. An integrating factor for this equation is I (x) = t−2 =⇒ (t−2 u) = 2t−2 =⇒ u = dt t dt dx −2t + ct2 =⇒ = −2t + ct2 =⇒ x(t) = c1 t3 − t2 + c2 . dt 8. t 9. d2 y −α dx2 dy dx 2 −β dy dy du d2 y = 0. Let u = so that = . Substituting these results into the dx dx dx dx2 du − βu = αu2 which is a Bernoulli equation. If u = 0 then y is a constant and dx du dz satisfies the equation. Now suppose that u = 0. Let z = u−1 so that = −u−2 . Substituting dx dx dz these results into the last equation yields + βz = −α. The an integrating factor for this equation is dx α βeβx dy βeβx I (x) = eβ dx = eβx =⇒ eβx z = −α eβx dx =⇒ − + ce−βx =⇒ u = =⇒ = =⇒ β cβ − αeβx dx cβ − αeβx βx 1 βe y= dx =⇒ y (x) = − ln |c1 + c2 eβx |. cβ − αeβx α first equation yields 2 dy dy du d2 y d2 y − = 18x4 . Let u = so that = . Substituting these results into the first equation dx2 x dx dx dx dx2 du 2 d −2 yields − u = 18x4 which has I (x) = x−2 as an integrating factor so (x u) = 18x2 =⇒ u = dx x dx dy 6x5 + cx2 =⇒ = 6x5 + cx2 =⇒ y (x) = x6 + c1 x3 + c2 . dx 10. 2x dy dy du d2 y =− . Let u = so that = df racd2 ydx2 . If u = 0 then y is a constant and 2 2 dx dx 1+x dx dx satisfies the equation. Now suppose that u = 0. Substituting these results into the first equation yields du 2x c1 dy c1 =− u =⇒ ln |u| = − ln (1 + x2 ) + c =⇒ u = =⇒ = =⇒ y (x) = c1 tan−1 x + c2 . 2 2 dx 1+x 1+x dx 1 + x2 11. d2 y 1 + 2 dx y 2 3 dy dy du du d2 y =u = 2 . Substituting these results into . Let u = so that dx dx dx dy dx du 1 2 −y 3 the first equation yields u + u = ye u . If u = 0 then y is a constant and satisfies the equation. Now dx y du u suppose that u = 0. Substituting these results into the first equation yields + = ye−y u2 which is a dx y dv −1 −2 du Bernoulli equation. Let v = u so that = −u . Substituting these results into the last equation dy dy dv v d −1 (y v ) = −e−1 =⇒ yields − = −ye−y . Then I (y ) = y −1 is an integrating factor for the equation thus dy y dy ey dy ey v = y (e−1 + c) =⇒ u = =⇒ = =⇒ (ye−y + cy )dy = dx =⇒ e−y (y + 1) + c1 y 2 − x. y + cyey dx y + cyey 12. dy dx = ye−3 83 d2 y dy dy du d2 y − tan x = 1. Let u = so that u = . Substituting these results into the first equation 2 dx dx dx dy dx2 du yields − u tan x = 1. An appropriate integrating factor for this equation is I (x) = e− tan xdx = cos x =⇒ dx d dy (u cos x) = cos x =⇒ u cos x = sin x + c =⇒ u(x) = tan x + c sec x =⇒ = tan x + c sec x =⇒ y (x) = dx dx ln secx + c1 ln (sec x + tan x) + c2 . 13. d2 y =2 dx2 2 du du d2 y dy dy + y 2 . Let u = so that =u = . Substituting these results into the first dx dx dx dy dx2 2 du 1 dz du equation yields u − u2 = y , a Bernoulli equation. Let z = u2 so that u = . Substituting these dy y dy 2 dy dz 4 results into the last equation yields − z = 2y which has I (y ) = y −4 as an integrating factor. Therefore, dy y d −4 dy (y z ) = 2y −3 =⇒ z = c1 y 4 − y 2 =⇒ u2 = c1 y 4 − y 2 =⇒ u = ± c1 y 4 − y 2 =⇒ = ± c1 y 4 − y 2 =⇒ dy dx 1 = ±x + c2 . Using the facts that f (0) = 1 and y (0) = 0 we find that c1 = 1 and c2 = 0; thus cos−1 √ y c1 y (x) = sec x. 14. y dy du du d2 y d2 y = ω 2 y where ω > 0. Let u = so that =u = 2 . Substituting these results into the first 2 dx dx dx dy dx du 2 2 22 equation yields u = ω y =⇒ u = ω y + c2 . Using the given that y (0) = a and y (0) = 0 we find that dy dy 1 c2 = a2 ω 2 . Then = ±ω y 2 − a2 =⇒ cosh−1 (y/a) = ±x + c =⇒ y (x) = a cosh [ω (c ± x)] =⇒ y = dx ω ±aω sinh [ω (c ± x)] and since y (0) = 0, c = 0; hence, y (x) = a cosh (ωx). 15. dy du d2 y 16. Let u = so that u = . Substituting these results into the differential equation yields dx dx dx2 √ √ 1 1 du u = 1 + u2 . Separating the variables and integrating we obtain 1 + u2 = y + c. Imposing the initial dy a a √ 1 1 dy conditions y (0) = a, (0) = 0 gives c = 0. Hence, 1 + u2 = y so that 1 + u2 = 2 y 2 or equivalently, a a dx dy 1 1 2 /a2 − 1. Substituting u = u=± y and separating the variables gives = ± dx which 2 − a2 dx |a| y can be integrated to obtain cosh−1 (y/a) = ±x/a + c1 so that y = a cosh (±x/a + c1 ). Imposing the initial conditions y (0) = a gives c1 = 0 so that y (x) = a cosh (x/a). d2 y dy dy du d2 y + p(x) = q (x). Let u = so that = 2 . Substituting these results into the first equation dx2 dx dx dx dx du gives us the equivalent system: +p(x)u = q (x) which has a solution u = e− p(x)dx e− p(x)dx q (x)dx + c1 dx dy so = e− p(x)dx e− p(x)dx q (x)dx + c1 . Thus y = e− p(x)dx e− p(x)dx q (x)dx + c1 dx + c2 is dx a solution to the original equation. 17. 18. (a) u1 = y =⇒ u2 = du1 dy du2 d2 y du3 d3 y d3 y = =⇒ u3 = = =⇒ = ; thus =F 2 3 dx dx dx dx dx dx dx3 x, d2 y dx2 since 84 the latter equation is equivalent to du3 = F (x, u3 ). dx d3 y 1 d2 y du1 du2 = − 1 . Replace this equation by the equivalent first order system: = u2 , = u3 , dx3 x dx2 dx dx 1 du 3 dx du2 K du3 and = (u3 − 1) =⇒ = =⇒ u3 = Kx + 1 =⇒ = Kx + 1 =⇒ u2 = x2 + x + c2 =⇒ dx x u3 − 1 x dx 2 du1 K2 K3 12 12 3 = x + x + c2 =⇒ u1 = x + x + c2 x + c3 =⇒ y (x) = u1 = c1 x + x + c2 x + c3 . dx 2 6 2 2 (b) g dθ d2 θ + sin θ = 0, θ(0) = θ0 , and (0) = 0. 2 dt L dt 2 g dθ du d2 θ du dθ du dθ (a) + θ = 0. Let u = so that = = = u . Substituting these results 2 dt L dt dt dt2 dθ dt dθ du g g2 dθ 2 2 into the last equation yields u + θ = 0 =⇒ u = − θ + c1 , but (0) = 0 and θ(0) = θ0 so dθ L L dt g2 g2 g θ g 2 c2 = θ0 =⇒ u2 = (θ0 − θ2 ) =⇒ u = ± θ0 − θ2 =⇒ sin−1 =± t + c2 , but θ(0) = θ0 so 1 L L L θ0 L g π g π θ π g t . Yes, the predicted t =⇒ θ = θ0 cos t =⇒ θ = θ0 sin ± c2 = =⇒ sin−1 =± L L L 2 2 θ0 2 motion is reasonable. dθ du d2 θ du dθ du d2 θ g so that = 2= = u . Substituting these results into the last (b) 2 + sin θ = 0. Let u = dt L dt dt dt dθ dt dθ du g 2g dθ 2g 2 equation yields u + sin θ = 0 =⇒ u = cos θ + c. Since θ(0) = θ0 and (0) = 0, then c = − cos θ0 dθ L L dt L 2g 2g 2g 2g dθ 2g dθ [cos θ − cos θ0 ]1/2 . cos θ − cos θ0 =⇒ =± and so u2 = cos θ − cos θ0 =⇒ =± L L L dt L L dt L dθ (c) From part (b), = ±dt. When the pendulum goes from θ = θ0 to θ = 0 2g [cos θ − cos θ0 ]1/2 dθ is negative; hence, choose the negative sign. Thus, (which corresponds to one quarter of a period) dt L θ0 dθ L0 dθ . =⇒ T = T =− 1/2 2g 0 [cos θ − cos θ0 ]1/2 2g θ0 [cos θ − cos θ0 ] L θ0 dθ (d) T = =⇒ 2g 0 [cos θ − cos θ0 ]1/2 19. Given L 2g T= Let k = sin θ0 2 θ0 0 dθ 2 2 sin θ0 2 2 − 2 sin 1/2 θ 2 = 1 2 L 2g θ0 0 dθ 2 sin θ0 2 2 − sin θ 2 1/2 . so that T= 1 2 L 2g θ0 0 dθ k2 2 − sin θ 2 1/2 . Now let sin θ/2 = k sin u. When θ = 0, u = 0 and when θ = θ0 , u = π/2; moreover, dθ = (0.0.5) 2k cos (u)du =⇒ cos (θ/2) 85 2k dθ = 1 − sin2 (u)du =⇒ dθ = 2 k 2 − (k sin (u))2 du =⇒ dθ = L g π /2 0 du . Making this 1 − k 2 sin2 (u) 1 − sin2 (θ/2) 1 − k 2 sin2 (u) change of variables in equation (0.0.5) yields T= 2 k 2 − sin2 (θ/2)du where 1 − k 2 sin2 (u) k = sin θ0 /2. Solutions to Section 1.12 1. 2. 3. We first determine the slope of the given family at the point (x, y ). Differentiating y = cx3 (0.0.6) with respect to x yields From (0.0.6) we have c = y x3 dy = 3cx2 . dx which, when substituted into Equation (0.0.7) yields (0.0.7) dy 3y = . dx x Consequently, the differential equation for the orthogonal trajectories is dy x =− . dx 3y Separating the variables and integrating gives 32 1 y = − x2 + C, 2 2 which can be written in the equivalent form x2 + 3y 2 = k. 4. We first determine the slope of the given family at the point (x, y ). Differentiating y 2 = cx3 with respect to x yields 2y dy = 3cx2 dx (0.0.8) 86 so that From (0.0.8) we have c = 3cx2 dy = . dx 2y y2 x3 (0.0.9) which, when substituted into Equation (0.0.9) yields dy 3y = . dx 2x Consequently, the differential equation for the orthogonal trajectories is 2x dy =− . dx 3y Separating the variables and integrating gives 32 y = −x2 + C, 2 which can be written in the equivalent form 2x2 + 3y 2 = k. 5. We first determine the slope of the given family at the point (x, y ). Differentiating y = ln(cx) (0.0.10) with respect to x yields dy 1 =. dx x Consequently, the differential equation for the orthogonal trajectories is dy = −x. dx which can be integrated directly to obtain 1 y = − x2 + k. 2 6. We first determine the slope of the given family at the point (x, y ). Differentiating x4 + y 4 = c (0.0.11) with respect to x yields 4x3 + 4y 3 so that dy =0 dx dy x3 = − 3. dx y Consequently, the differential equation for the orthogonal trajectories is dy y3 = 3. dx x (0.0.12) 87 Separating the variables and integrating gives 1 1 − y −2 = − x−2 + C, 2 2 which can be written in the equivalent form y 2 − x2 = kx2 y 2 . 7. (a) We first determine the slope of the given family at the point (x, y ). Differentiating x2 + 3y 2 = 2cy (0.0.13) with respect to x yields 2x + 6y so that From (0.0.13) we have c = dy dy = 2c dx dx dy x = . dx c − 3y x2 +3y 2 2y (0.0.14) which, when substituted into Equation (0.0.14) yields dy = dx x x2 +3y 2 2y − 3y = 2xy , x2 − 3y 2 (0.0.15) as required. (b) It follows from Equation(0.0.15) that the differential equation for the orthogonal trajectories is 3y 2 − x2 dy =− . dx 2xy This differential equation is first-order homogeneous. Substituting y = xV into the preceding differential equation gives dV 3V 2 − 1 x +V = dx 2V which simplifies to dV V2−1 = . dx 2V Separating the variables and integrating we obtain ln(V 2 − 1) = ln x + C, or, upon exponentiation, V 2 − 1 = kx. Inserting V = y/x into the preceding equation yields y2 − 1 = kx, x2 that is, y 2 − x2 = kx3 . 88 y(x) x Figure 0.0.57: Figure for Exercise 8 y(x) x Figure 0.0.58: Figure for Exercise 9 8. See accompanying figure. 9. See accompanying figure. 10. (a) If v (t) = 25, then dv 1 = 0 = (25 − v ). dt 2 89 (b) The accompanying figure suggests that lim v (t) = 25. t→∞ v(t) 25 20 15 10 5 t 10 5 Figure 0.0.59: Figure for Exercise 10 11. (a) Separating the variables in Equation (1.12.6) yields mv dv =1 mg − kv 2 dy which can be integrated to obtain − m ln(mg − kv 2 ) = y + c. 2k Multiplying both sides of this equation by −1 and exponentiating gives 2k mg − kv 2 = c1 e− m y . The initial condition v (0) = 0 requires that c1 = mg , which, when inserted into the preceding equation yields 2k mg − kv 2 = mge− m y , or equivalently, v2 = as required. (b) See accompanying figure. 2k mg 1 − e− m y , k 90 v2(y) mg/k y Figure 0.0.60: Figure for Exercise 11 12. The given differential equation is separable. Separating the variables gives y ln x dy =2 , dx x which can be integrated directly to obtain 12 y = (ln x)2 + c, 2 or, equivalently, y 2 = 2(ln x)2 + c1 . 13. The given differential equation is first-order linear. We first divide by x to put the differential equation in standard form: dy 2 − y = 2x ln x. (0.0.16) dx x An integrating factor for this equation is I = e (−2/x)dx = x−2 . Multiplying Equation (0.0.16) by x−2 reduces it to d −2 (x y ) = 2x−1 ln x, dx which can be integrated to obtain x−2 y = (ln x)2 + c so that y (x) = x2 [(ln x)2 + c]. 91 14. We first re-write the given differential equation in the differential form 2xy dx + (x2 + 2y )dy = 0. (0.0.17) Then My = 2x = Nx so that the differential equation is exact. Consequently, there exists a potential function φ satisfying ∂φ = x2 + 2y. ∂y ∂φ = 2xy, ∂x Integrating these two equations in the usual manner yields φ(x, y ) = x2 y + y 2 . Therefore Equation (0.0.17) can be written in the equivalent form d(x2 y + y 2 ) = 0 with general solution x2 y + y 2 = c. 15. We first rewrite the given differential equation as dy y 2 + 3xy + x2 = , dx x2 which is first order homogeneous. Substituting y = xV into the preceding equation yields x so that x d + V = V 2 + 3V + 1 dx dV = V 2 + 2V + 1 = (V + 1)2 , dx or, in separable form, 1 dV 1 =. 2 dx (V + 1) x This equation can be integrated to obtain −(V + 1)−1 = ln x + c so that V +1= 1 . c1 − ln x Inserting V = y/x into the preceding equation yields y 1 +1= , x c1 − ln x so that y (x) = 1 − x. c1 − ln x 92 16. We first rewrite the given differential equation in the equivalent form dy + y · tan x = y 2 sin x, dx which is a Bernoulli equation. Dividing this equation by y 2 yields y −2 dy + y −1 tan x = sin x. dx Now make the change of variables u = y −1 in which case Equation (0.0.18) gives the linear differential equation − (0.0.18) dy = −y −2 dx . Substituting these results into du dx du + u · tan x = sin x dx or, in standard form, du − u · tan x = − sin x. dx (0.0.19) An integrating factor for this differential equation is I = e− tan x dx = cos x. Multiplying Equation (0.0.19) by cos x reduces it to d (u · cos x) = − sin x cos x dx which can be integrated directly to obtain u · cos x = so that 1 cos2 x + c, 2 cos2 x + c1 . cos x into the preceding equation and rearranging yields u= Inserting u = y −1 y (x) = 2 cos x . cos2 x + c1 17. The given differential equation is linear with integrating factor I=e 2e2x 1+e2x dx = eln(1+e 2x ) = 1 + e2x . Multiplying the given differential equation by 1 + e2x yields d e2x + 1 2e2x (1 + e2x )y = 2x = −1 + 2x dx e −1 e −1 which can be integrated directly to obtain (1 + e2x )y = −x + ln |e2x − 1| + c, so that y (x) = −x + ln |e2x − 1| + c . 1 + e2x 93 18. We first rewrite the given differential equation in the equivalent form x2 − y 2 , x y+ dy = dx which we recognize as being first order homogeneous. Inserting y = xV into the preceding equation yields dV |x| x +V =V + 1 − V 2, dx x that is, 1 1 dV √ =± . 2 dx x 1−V Integrating we obtain sin−1 V = ± ln |x| + c, so that V = sin(c ± ln |x|). Inserting V = y/x into the preceding equation yields y (x) = x sin(c ± ln |x|). 19. We first rewrite the given differential equation in the equivalent form (sin y + y cos x + 1)dx − (1 − x cos y − sin x)dy = 0. Then My = cos y + cos x = Nx so that the differential equation is exact. Consequently, there is a potential function satisfying ∂φ = sin y + y cos x + 1, ∂x ∂φ = −(1 − x cos y − sin x). ∂y Integrating these two equations in the usual manner yields φ(x, y ) = x − y + x sin y + y sin x, so that the differential equation can be written as d(x − y + x sin y + y sin x) = 0, and therefore has general solution x − y + x sin y + y sin x = c. 20. Writing the given differential equation as dy 1 25 −1 2 + y= y x ln x, dx x 2 we see that it is a Bernoulli equation with n = −1. We therefore divide the equation by y −1 to obtain y dy 1 25 2 + y2 = x ln x. dx x 2 94 We now make the change of variables u = y 2 , in which case, the preceding differential equation yields du dx dy = 2y dx . Inserting these results into 25 2 1 du 1 + u= x ln x, 2 dx x 2 or, in standard form, du 2 + u = 25x2 ln x. dx x (2/x)dx An integrating factor for this linear differential equation is I = e differential equation by x2 reduces it to = x2 . Multiplying the previous d2 (x u) = 25x4 ln x dx which can be integrated directly to obtain 1 15 x ln x − x5 5 25 x2 u = 25 +c so that u = x3 (5 ln x − 1) + cx−2 . Making the replacement u = y 2 in this equation gives y 2 = x3 (5 ln x − 1) + cx−2 . 21. The given differential equation can be written in the equivalent form dy ex−y = 2x+y = e−x e−2y , dx e which we recognize as being separable. Separating the variables gives e2y dy = e−x dx which can be integrated to obtain 1 2y e = −e−x + c 2 so that y (x) = 1 ln(c1 − 2e−x ). 2 22. The given differential equation is linear with integrating factor I = e given differential equation by sin x reduces it to d sin x (y sin x) = dx cos x which can be integrated directly to obtain y sin x = − ln(cos x) + c, so that y (x) = c − ln(cos x) . sin x cot x dx = sin x. Multiplying the 95 23. Writing the given differential equation as 1 2ex dy + y = 2y 2 e−x , dx 1 + ex 1 we see that it is a Bernoulli equation with n = 1/2. We therefore divide the equation by y 2 to obtain 1 y− 2 2ex 1 dy + y 2 = 2e−x . dx 1 + ex 1 We now make the change of variables u = y 2 , in which case, into the preceding differential equation yields 2 du dx = 1 − 1 dy 2 2y dx . Inserting these results du 2ex + u = 2e−x , dx 1 + ex or, in standard form, ex du + u = e−x . dx 1 + ex An integrating factor for this linear differential equation is I=e ex 1+ex dx = eln(1+e x ) = 1 + ex . Multiplying the previous differential equation by 1 + ex reduces it to d [(1 + ex )u] = e−x (1 + ex ) = e−x + 1 dx which can be integrated directly to obtain (1 + ex )u = −e−x + x + c so that u= x − e−x + c . 1 + ex 1 Making the replacement u = y 2 in this equation gives 1 y2 = x − e−x + c . 1 + ex 24. We first rewrite the given differential equation in the equivalent form dy y y = ln +1 . dx x x The function appearing on the right of this equation is homogeneous of degree zero, and therefore the differential equation itself is first order homogeneous. We therefore insert y = xV into the differential equation to obtain dV x + V = V (ln V + 1), x so that dV x = V ln V. dx 96 Separating the variables yields 1 dV 1 = V ln V dx x which can be integrated to obtain ln(ln V ) = ln x + c. Exponentiating both side of this equation gives ln V = c1 x, or equivalently, V = ec1 x . Inserting V = y/x in the preceding equation yields y = xec1 x . 25. For the given differential equation we have M (x, y ) = 1 + 2xey , N (x, y ) = −(ey + x), so that My − N x 1 + 2xey = = 1. M 1 + 2xey Consequently, an integrating factor for the given differential equation is I = e− dy = e−y . Multiplying the given differential equation by e−y yields the exact differential equation (2x + e−y )dx − (1 + xe−y )dy = 0. (0.0.20) Therefore, there exists a potential function φ satisfying ∂φ = 2x + e−y , ∂x ∂φ = −(1 + xe−y ). ∂y Integrating these two equations in the usual manner yields φ(x, y ) = x2 − y + xe−y . Therefore Equation (0.0.20) can be written in the equivalent form d(x2 − y + xe−y ) = 0 with general solution x2 − y + xe−y = c. 26. The given differential equation is first-order linear. However, it can also e written in the equivalent form dy = (1 − y ) sin x dx 97 which is separable. Separating the variables and integrating yields − ln |1 − y | = − cos x + c, so that 1 − y = c1 ecos x . Hence, y (x) = 1 − c1 ecos x . 27. For the given differential equation we have M (x, y ) = 3y 2 + x2 , N (x, y ) = −2xy, so that My − N x 4 =− . N x Consequently, an integrating factor for the given differential equation is I = e− 4 x dx = x−4 . Multiplying the given differential equation by x−4 yields the exact differential equation (3y 2 x−4 + x−2 )dx − 2yx−3 dy = 0. (0.0.21) Therefore, there exists a potential function φ satisfying ∂φ = 3y 2 x−4 + x−2 , ∂x ∂φ = −2yx−3 . ∂y Integrating these two equations in the usual manner yields φ(x, y ) = −y 2 x−3 − x−1 . Therefore Equation (0.0.21) can be written in the equivalent form d(−y 2 x−3 − x−1 ) = 0 with general solution −y 2 x−3 − x−1 = c, or equivalently, x2 + y 2 = c1 x3 . Notice that the given differential equation can be written in the equivalent form dy 3y 2 + x2 = , dx 2xy which is first-order homogeneous. Another equivalent way of writing the given differential equation is dy 3 1 − y = xy −1 , dx 2x 2 which is a Bernoulli equation. 98 28. The given differential equation can be written in the equivalent form 1 9 dy − y = − x2 y 3 dx 2x ln x 2 which is a Bernoulli equation with n = 3. We therefore divide the equation by y 3 to obtain y −3 1 9 dy − y −2 = − x2 . dx 2x ln x 2 We now make the change of variables u = y −2 , in which case, into the preceding differential equation yields − du dx dy = −2y −3 dx . Inserting these results 1 du 1 9 − u = − x2 , 2 dx 2x ln x 2 or, in standard form, du 1 + u = 9x2 . dx x ln x An integrating factor for this linear differential equation is I=e 1 x ln x dx = eln(ln x) = ln x. Multiplying the previous differential equation by ln x reduces it to d (ln x · u) = 9x2 ln x dx which can be integrated to obtain ln x · u = x3 (3 ln x − 1) + c so that u= x3 (3 ln x − 1) + c . ln x Making the replacement u = y 3 in this equation gives y3 = x3 (3 ln x − 1) + c . ln x 29. Separating the variables in the given differential equation yields 1 dy 2+x 1 = =1+ , y dx 1+x 1+x which can be integrated to obtain ln |y | = x + ln |1 + x| + c. Exponentiating both sides of this equation gives y (x) = c1 (1 + x)ex . 99 30. The given differential equation can be written in the equivalent form dy 2 +2 y=1 dx x − 1 (0.0.22) which is first-order linear. An integrating factor is I=e 2 dx x2 −1 1 1 x−1 = e ( x−1 − x+1 )dx = e[ln(x−1)−ln(x+1)] = . x+1 Multiplying (0.0.22) by (x − 1)/(x + 1) reduces it to the integrable form d dx x−1 ·y x+1 = x−1 2 =1− . x+1 x+1 Integrating both sides of this differential equation yields x−1 ·y x+1 = x − 2 ln(x + 1) + c so that y (x) = x+1 x−1 [x − 2 ln(x + 1) + c]. 31. The given differential equation can be written in the equivalent form [y sec2 (xy ) + 2x]dx + x sec2 (xy )dy = 0 Then My = sec2 (xy ) + 2xy sec2 (x) tan(xy ) = Nx so that the differential equation is exact. Consequently, there is a potential function satisfying ∂φ = y sec2 (xy ) + 2x, ∂x ∂φ = x sec2 (xy ). ∂y Integrating these two equations in the usual manner yields φ(x, y ) = x2 + tan(xy ), so that the differential equation can be written as d(x2 + tan(xy )) = 0, and therefore has general solution x2 + tan(xy ) = c, or equivalently, y (x) = tan−1 (c − x2 ) . x 100 32. CHANGE PROBLEM IN TEXT TO dy √ + 4xy = 4x y. dx The given differential equation is a Bernoulli equation with n = 1 . We therefore divide the equation 2 1 by y 2 to obtain 1 dy 1 − 4xy 2 = 4x. y− 2 dx 1 We now make the change of variables u = y 2 , in which case, into the preceding differential equation yields 2 du dx = 1 − 1 dy 2 2y dx . Inserting these results du + 4xu = 4x, dx or, in standard form, du + 2xu = 2x. dx An integrating factor for this linear differential equation is I=e 2x dx 2 = ex . 2 Multiplying the previous differential equation by ex reduces it to 2 2 d ex u = 2xex . dx which can be integrated directly to obtain 2 2 ex u = ex + c so that 2 u = 1 + cex . 1 Making the replacement u = y 2 in this equation gives 1 2 y 2 = 1 + cex . 33. CHANGE PROBLEM IN TEXT TO dy x2 y =2 + 2 dx x +y x then the answer is correct. The given differential equation is first-order homogeneous. Inserting y = xV into the given equation yields dV 1 x +V = + V, dx 1+V2 that is, (1 + V 2 ) dV 1 =. dx x 101 Integrating we obtain 1 V + V 3 = ln |x| + c. 3 Inserting V = y/x into the preceding equation yields y3 y + 3 = ln |x| + c. x 3x 34. For the given differential equation we have 1 = Nx y My = so that the differential equation is exact. Consequently, there is a potential function satisfying ∂φ x = + 2y. ∂y y ∂φ = ln(xy ) + 1, ∂x Integrating these two equations in the usual manner yields φ(x, y ) = x ln(xy ) + y 2 , so that the differential equation can be written as d[x ln(xy ) + y 2 ] = 0, and therefore has general solution x ln(xy ) + y 2 = c. 35. The given differential equation is a Bernoulli equation with n = −1. We therefore divide the equation by y −1 to obtain dy 1 25 ln x y + y2 = . dx x 2x3 We now make the change of variables u = y 2 , in which case, the preceding differential equation yields du dx 1 du 1 25 ln x + u= , 2 dx x 2x3 or, in standard form, du 2 + u = 25x−3 ln x. dx x An integrating factor for this linear differential equation is I=e 2 x dx = x2 . Multiplying the previous differential equation by x2 reduces it to d2 (x u) = 25x−1 ln x, dx dy = 2y dx . Inserting these results into 102 which can be integrated directly to obtain x2 u = 25 (ln x)2 + c 2 so that 25(ln x)2 + c . 2x2 Making the replacement u = y 2 in this equation gives u= y2 = 25(ln x)2 + c . 2x2 36. The problem as written is separable, but the integration does not work. CHANGE PROBLEM TO: (1 + y ) dy = xex−y . dx The given differential equation can be written in the equivalent form ey (1 + y ) dy = xex dx which is separable. Integrating both sides of this equation gives yey = ex (x − 1) + c. 37. The given differential equation can be written in the equivalent form cos x dy − y = − cos x dx sin x which is first order linear with integrating factor I = e− cos x sin x dx = e− ln(sin x) = Multiplying the preceding differential equation by d dx 1 sin x 1 ·y sin x 1 . sin x reduces it to =− cos x sin x which can be integrated directly to obtain 1 · y = − ln(sin x) + c sin x so that y (x) = sin x[c − ln(sin x)]. 38. The given differential equation is linear, and therefore can be solved using an appropriate integrating factor. However, if we rearrange the terms in the given differential equation then it can be written in the equivalent form 1 dy = x2 1 + y dx 103 which is separable. Integrating both sides of the preceding differential equation yields 13 x +c 3 ln(1 + y ) = so that 1 3 y (x) = c1 e 3 x − 1. Imposing the initial condition y (0) = 5 we find c1 = 6. Therefore the solution to the initial-value problem is 13 y (x) = 6e 3 x − 1. 39. The given differential equation can be written in the equivalent form e−6y dy = −e−4x dx which is separable. Integrating both sides of the preceding equation yields 1 1 − e−6y = e−4x + c 6 4 so that 1 3 y (x) = − ln c1 − e−4x . 6 2 Imposing the initial condition y (0) = 0 requires that 0 = ln c1 − Hence, c1 = 5 , and so 2 1 y (x) = − ln 6 3 2 . 5 − 3e−4x 2 . 40. For the given differential equation we have My = 4xy = Nx so that the differential equation is exact. Consequently, there is a potential function satisfying ∂φ = 3x2 + 2xy 2 , ∂x ∂φ = 2x2 y. ∂y Integrating these two equations in the usual manner yields φ(x, y ) = x2 y 2 + x3 , so that the differential equation can be written as d(x2 y 2 + x3 ) = 0, and therefore has general solution x2 y 2 + x3 = c. 104 Imposing the initial condition y (1) = 3 yields c = 10. Therefore, x2 y 2 + x3 = 10 so that 10 − x3 . x2 Note that the given differential equation can be written in the equivalent form y2 = 1 3 dy + y = − y −1 , dx x 2 which is a Bernoulli equation with n = −1. Consequently, the Bernoulli technique could also have been used to solve the differential equation. 41. The given differential equation is linear with integrating factor I = e− sin x dx = ecos x . Multiplying the given differential equation by ecos x reduces it to the integrable form d cos x (e · y ) = 1, dx which can be integrated directly to obtain ecos x · y = x + c.. Hence, y (x) = e− cos x (x + c). Imposing the given initial condition y (0) = 1 e requires that c = 1. Consequently, y (x) = e− cos x (x + 1). 42. (a) For the given differential equation we have My = my m−1 , Nx = −nxn−1 y 3 . We see that the only values for m and n for which My = Nx are m = n0. Consequently, these are the only values of m and n for which the differential equation is exact. (b) We rewrite the given differential equation in the equivalent form dy x5 + y m = , dx xn y 3 (0.0.23) from which we see that the differential equation is separable provided m = 0. In this case there are no restrictions on n. (c) From Equation (0.0.23) we see that the only values of m and n for which the differential equation is first-order homogeneous are m = 5 and n = 2. 105 (d) We now rewrite the given differential equation in the equivalent form dy − x−n y m−3 = x5−n y −3 . dx (0.0.24) Due to the y −3 term on the right-hand side of the preceding differential equation, it follows that there are no values of m and n for which the equation is linear. (e) From Equation (0.0.24) we see that the differential equation is a Bernoulli equation whenever m = 4. There are no constraints on n in this case. 43. In Newton’s Law of Cooling we have Tm = 180◦ F, T (0) = 80◦ F, T (3) = 100◦ F. We need to determine the time, t0 when T (t0 ) = 140◦ F. The temperature of the sandals at time t is governed by the differential equation dT = −k (T − 180). dt This separable differential equation is easily integrated to obtain T (t) = 180 + ce−kt . Since T (0) = 80 we have 80 = 180 + c =⇒ c = −100. Hence, T (t) = 180 − 100e−kt . Imposing the condition T (3) = 100 requires 100 = 180 − 100e−3k . Solving for k we find k = yields 1 3 ln 5 4 . Inserting this value for k into the preceding expression for T (t) t 5 T (t) = 180 − 100e− 3 ln( 4 ) . We need to find t0 such that t0 140 = 180 − 100e− 3 Solving for t0 we find t0 = 3 ln ln 5 2 5 4 ln( 5 ) 4 . ≈ 12.32 min. 44. In Newton’s Law of Cooling we have Tm = 70◦ F, T (0) = 150◦ F, T (10) = 125◦ F. We need to determine the time, t0 when T (t0 ) = 100◦ F. The temperature of the plate at time t is governed by the differential equation dT = −k (T − 70). dt 106 This separable differential equation is easily integrated to obtain T (t) = 70 + ce−kt . Since T (0) = 150 we have 150 = 70 + c =⇒ c = 80. Hence, T (t) = 70 + 80e−kt . Imposing the condition T (10) = 125 requires 125 = 70 + 80e−10k . Solving for k we find k = yields 1 10 ln 16 11 . Inserting this value for k into the preceding expression for T (t) t 16 T (t) = 70 + 80e− 10 ln( 11 ) . We need to find t0 such that t0 16 100 = 70 + 80e− 10 ln( 11 ) . Solving for t0 we find t0 = 10 ln ln 8 3 16 11 ≈ 26.18 min. 45. Let T (t) denote the temperature of the object at time t, and let Tm denote the temperature of the surrounding medium. Then we must solve the initial-value problem dT = k (T − Tm )2 , dt T (0) = T0 , where k is a constant. The differential equation can be written in separated form as dT 1 = k. 2 dt (T − Tm ) Integrating both sides of this differential equation yields − 1 = kt + c T − Tm so that T (t) = Tm − 1 . kt + c Imposing the initial condition T (0) = T0 we find that c= 1 Tm − T0 which, when substituted back into the preceding expression for T (t) yieldss T (t) = Tm − As t → ∞, T (t) approaches Tm . 1 kt + 1 Tm −T0 = Tm − T m − T0 . k (Tm − T0 )t + 1 107 46. We are given the differential equation dT = −k (T − 5 cos 2t) dt (0.0.25) dT (0) = 5. dt (0.0.26) together with the initial conditions T (0) = 0; (a) Setting t = 0 in (0.0.25) and using (0.0.26) yields 5 = −k (0 − 5) so that k = 1. (b) Substituting k = 1 into the differential equation (0.0.25) and rearranging terms yields dT + T = 5 cos t. dt An integrating factor for this linear differential equation is I = e preceding differential equation by et reduces it to dt = et . Multiplying the dt (e · T ) = 5et cos 2t dt which upon integration yields et · T = et (cos 2t + 2 sin 2t) + c, so that T (t) = ce−t + cos 2t + 2 sin 2t. Imposing the initial condition T (0) = 0 we find that c = −1. Hence, T (t) = cos 2t + 2 sin 2t − e−t . (c) For large values of t we have T (t) ≈ cos 2t + 2 sin 2t, which can be written in phase-amplitude form as √ T (t) ≈ 5 cos(2t − φ), where tan φ = 2. consequently, for large t, the temperature is approximately oscillatory with √ period π and amplitude 5. 47. If we let C (t) denote the number of sandhill cranes in the Platte River valley t days after April 1, then C (t) is governed by the differential equation dC = −kC dt (0.0.27) together with the auxiliary conditions C (0) = 500, 000; C (15) = 100, 000. (0.0.28) 108 Separating the variables in the differential equation (0.0.27) yields 1 dC = −k, C dt which can be integrated directly to obtain ln C = −kt + c. Exponentiation yields C (t) = c0 e−kt . The initial condition C (0) = 500, 000 requires c0 = 500, 000, so that C (t) = 500, 000e−kt . (0.0.29) Imposing the auxiliary condition C (15) = 100, 000 yields 100, 000 = 500, 000e−15k . Taking the natural logarithm of both sides of the preceding equation and simplifying we find that 1 k = 15 ln 5. Substituting this value for k into (0.0.29) gives t C (t) = 500, 000e− 15 ln 5 . (a) C (3) = 500, 000e−2 ln 5 = 500, 000 · 35 − 15 ln 5 (b) C (35) = 500, 000e 1 25 (0.0.30) = 20, 000 sandcranes. ≈ 11696 sandcranes. (c) We need to determine t0 such that t0 1000 = 500, 000e− 15 ln 5 that is, 1 . 500 Taking the natural logarithm of both sides of this equation and simplifying yields t0 e− 15 ln 5 = t0 = 15 · ln 500 ≈ 57.9 days after April 1. ln 5 48. Substituting P0 = 200, 000 into equation (1.5.3) in the text (page 45) yields P (t) = 200, 000C . 200, 000 + (C − 200, 000)e−rt (0.0.31) We are given P (3) = P (t1 ) = 230, 000, P (6) = P (t2 ) = 250, 000. Since t2 = 2t1 we can use the formulas (1.5.5), (1.5.6) on page 47 of the text to obtain r and C directly as follows: 1 25(23 − 20) 1 15 r = ln = ln ≈ 0.21. 3 20(25 − 23) 3 8 109 C= 230, 000[(23)(45) − (40)(25)] = 277586. (23)2 − (20)(25) Substituting these values for r and C into (0.0.31) yields P (t) = 55517200000 . 200, 000 + (77586)e−0.21t Therefore, P (10) = 55517200000 ≈ 264, 997, 200, 000 + (77586)e−2.1 P (20) = 55517200000 ≈ 275981. 200, 000 + (77586)e−4.2 and 49. The differential equation for determining q (t) is 3 dq 5 + q = cos 2t, dt 4 2 which has integrating factor I = e reduces it to the integrable form 5 4 dt 5 5 = e 4 t . Multiplying the preceding differential equation by e 4 t 5 d 35 e 4 t · q = e 4 t cos 2t. dt 2 Integrating and simplifying we find q (t) = 5 6 (5 cos 2t + 8 sin 2t) + ce− 4 t . 89 (0.0.32) The initial condition q (0) = 3 requires 3= so that c = 237 89 . 30 + c, 89 Making this replacement in (0.0.32) yields q (t) = 6 237 − 5 t (5 cos 2t + 8 sin 2t) + e 4. 89 89 The current in the circuit is i(t) = dq 12 1185 − 5 t = (8 cos 2t − 5 sin 2t) − e 4. dt 89 356 Answer in text has incorrect exponent. 50. The current in the circuit is governed by the differential equation di 100 + 10i = , dt 3 which has integrating factor I = e reduces it to the integrable form 10 dt = e10t . Multiplying the preceding differential equation by e10t d 10t 100 10t e ·i = e. dt 3 110 Integrating and simplifying we find 10 + ce−10t . 3 i(t) = (0.0.33) The initial condition i(0) = 3 requires 10 + c, 3 so that c = − 1 . Making this replacement in (0.0.33) yields 3 3= i(t) = 1 (10 − e−10t ). 3 51. We are given: r1 = 6 L/min, c1 = 3 g/L, r2 = 4 L/min, V (0) = 30 L, A(0) = 0 g, and we need to determine the amount of salt in the tank when V (t) = 60L. Consider a small time interval ∆t. Using the preceding information we have: ∆V = 6∆t − 4∆t = 2∆t, and A ∆t. V Dividing both of these equations by ∆t and letting ∆t → 0 yields ∆A ≈ 18∆t − 4 dV = 2. dt (0.0.34) dA A + 4 = 18. dt V Integrating (0.0.34) and imposing the initial condition V (0) = 30 yields V (t) = 2(t + 15). (0.0.35) (0.0.36) We now insert this expression for V (t) into (0.0.35) to obtain dA 2 + A = 18. dt t + 15 2 An integrating factor for this differential equation is I = e t+15 dt = (t +15)2 . Multiplying the preceding differential equation by (t + 15)2 reduces it to the integrable form d (t + 15)2 A = 18(t + 15)2 . dt Integrating and simplifying we find A(t) = 6(t + 15)3 + c . (t + 15)2 Imposing the initial condition A(0) = 0 requires 0= 6(15)3 + c , (15)2 111 so that c = −20250. Consequently, A(t) = 6(t + 15)3 − 20250 . (t + 15)2 We need to determine the time when the solution overflows. Since the tank can hold 60 L of solution, from (0.0.36) overflow will occur when 60 = 2(t + 15) =⇒ t = 15. The amount of chemical in the tank at this time is A(15) = 6(30)3 − 20250 ≈ 157.5 g. (30)2 52. Applying Euler’s method with y = x2 + 2y 2 , x0 = 0, y0 = −3, and h = 0.1 we have yn+1 = yn + 2 0.1(x2 + 2yn ). This generates the sequence of approximants given in the table below. n n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn −1.2 −0.911 −0.74102 −0.62219 −0.52877 −0.44785 −0.371736 −0.29510 −0.21368 −0.12355 Consequently the Euler approximation to y (1) is y10 = −0.12355. 53. Applying Euler’s method with y = 3x + 2, x0 = 1, y0 = 2, and h = 0.05 we have y yn+1 = yn + 0.05 3xn +2 . yn This generates the sequence of approximants given in the table below. n 1 2 3 4 5 6 7 8 9 10 xn 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 yn 2.1750 2.34741 2.51770 2.68622 2.85323 3.01894 3.18353 3.34714 3.50988 3.67185 112 Consequently, the Euler approximation to y (1.5) is y10 = 3.67185. 54. Applying the modified Euler method with y = x2 + 2y 2 , x0 = 0, y0 = −3, and h = 0.1 generates the sequence of approximants given in the table below. n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn −1.9555 −1.42906 −1.11499 −0.90466 −0.74976 −0.62555 −0.51778 −0.41723 −0.31719 −0.21196 Consequently, the modified Euler approximation to y (1) is y10 = −0.21196. Comparing this to the corresponding Euler approximation from Problem 52 we have |yME − yE | = |0.21196 − 0.12355| = 0.8841. 3x + 2, x0 = 1, y0 = 2, and h = 0.05 generates the y sequence of approximants given in the table below. 55. Applying the modified Euler method with y = n 1 2 3 4 5 6 7 8 9 10 xn 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 yn 2.17371 2.34510 2.51457 2.68241 2.84886 3.01411 3.17831 3.34159 3.50404 3.66576 Consequently, the modified Euler approximation to y (1.5) is y10 = 3.66576. Comparing this to the corresponding Euler approximation from Problem 53 we have |yME − yE | = |3.66576 − 3.67185| = 0.00609. 56. Applying the Runge-Kutta method with y = x2 + 2y 2 , x0 = 0, y0 = −3, and h = 0.1 generates the sequence of approximants given in the table below. 113 n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn −1.87392 −1.36127 −1.06476 −0.86734 −0.72143 −0.60353 −0.50028 −0.40303 −0.30541 −0.20195 Consequently the Runge-Kutta approximation to y (1) is y10 = −0.20195. Comparing this to the corresponding Euler approximation from Problem 52 we have |yRK − yE | = |0.20195 − 0.12355| = 0.07840. 3x + 2, x0 = 1, y0 = 2, and h = 0.05 generates the y sequence of approximants given in the table below. 57. Applying the Runge-Kutta method with y = n 1 2 3 4 5 6 7 8 9 10 xn 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 yn 2.17369 2.34506 2.51452 2.68235 2.84880 3.01404 3.17823 3.34151 3.50396 3.66568 Consequently the Runge-Kutta approximation to y (1.5) is y10 = 3.66568. Comparing this to the corresponding Euler approximation from Problem 53 we have |yRK − yE | = |3.66568 − 3.67185| = 0.00617. Last digit in answer the text needs changing. Solutions to Section 2.1 True-False Review: 1. TRUE. A diagonal matrix has no entries below the main diagonal, so it is upper triangular. Likewise, it has no entries above the main diagonal, so it is also lower triangular. 2. FALSE. An m × n matrix has m row vectors and n column vectors. 3. TRUE. Since A is symmetric, A = AT . Thus, (AT )T = A = AT , so AT is symmetric. 114 4. FALSE. The trace of a matrix is the sum of the entries along the main diagonal. 5. TRUE. If A is skew-symmetric, then AT = −A. But A and AT contain the same entries along the main diagonal, so for AT = −A, both A and −A must have the same main diagonal. This is only possible if all entries along the main diagonal are 0. 6. TRUE. If A is both symmetric and skew-symmetric, then A = AT = −A, and A = −A is only possible if all entries of A are zero. 7. TRUE. Both matrix functions are defined for values of t such that t > 0. 8. FALSE. The (3, 2)-entry contains a function that is not defined for values of t with t ≤ 3. So for example, this matrix functions is not defined for t = 2. 9. TRUE. Each numerical entry of the matrix function is a constant function, which has domain R. 10. FALSE. For instance, the matrix function A(t) = [t] and B (t) = [t2 ] satisfy A(0) = B (0), but A and B are not the same matrix function. Problems: 1. a31 = 0, a24 = −1, a14 = 2, a32 = 2, a21 = 7, a34 = 4. 1 −1 2. 3. 4. 5. 6. 5 3 ; 2 × 2 matrix. 2 1 −1 ; 2 × 3 matrix. 0 4 −2 −1 1 ; 4 × 1 matrix. 1 −5 1 −3 −2 3 6 0 ; 4 × 3 matrix. 2 7 4 −4 −1 5 0 −1 2 1 0 3 ; 3 × 3 matrix. −2 −3 0 7. tr(A) = 1 + 3 = 4. 8. tr(A) = 1 + 2 + (−3) = 0. 9. tr(A) = 2 + 2 + (−5) = −1. 1 −1 , . 3 5 Row vectors: [1 − 1], [3 5]. 1 3 −4 11. Column vectors: −1 , −2 , 5 . 2 6 7 Row vectors: [1 3 − 4], [−1 − 2 5], [2 6 7]. 10. Column vectors: 115 12. Column vectors: 2 5 , 10 −1 , 6 3 . Row vectors: [2 10 6], [5 − 1 3]. 2 1 2 4 . Column vectors: 3 , 4 . 1 5 1 2 501 −1 7 0 2 . Row vectors: [2 5 0 1], [−1 7 0 2], [4 − 6 0 3]. 14. B = 4 −6 0 3 1 13. A = 3 5 15. A = [a1 , a2 , . . . , ap ] has p columns and each column dimensions q × p. 20 0 0 . 16. One example: 0 3 0 0 −1 2312 0 5 6 2 17. One example: 0 0 3 5 . 0001 1 3 −1 2 −3 0 4 −3 . 18. One example: 1 −4 0 1 −2 3 −1 0 300 19. One example: 0 2 0 . 005 00 20. The only possibility here is the zero matrix: 0 0 00 √ 1 √ t+2 0 3−t 21. One example: . 0 0 0 2 t −t 0 0 0 . 22. One example: 0 0 0 0 23. One example: 24. One example: t2 + 1 1 1 t2 +1 0 1 1 q -vector has q rows, so the resulting matrix has 0 0 . 0 1. . 25. One example: Let A and B be 1 × 1 matrix functions given by A(t) = [t] and B (t) = [t2 ]. 26. Let A be a symmetric upper-triangular matrix. Then all elements below the main diagonal are zeros. Consequently, since A is symmetric, all elements above the main diagonal must also be zero. Hence, the 116 only nonzero entries can occur along the main diagonal. That is, A is a diagonal matrix. 27. Since A is skew-symmetric, a11 a22 = a33 =0. Further, a12 = −a21 = −1, a13 = −a31 = −3, and = 0 −1 −3 0 −1 . a32 = −a23 = 1. Consequently, A = 1 3 1 0 Solutions to Section 2.2 True-False Review: 1. FALSE. The correct statement is (AB )C = A(BC ), the associative law. A counterexample to the particular statement given in this review item can be found in Problem 7. 2. TRUE. Multiplying from left to right, we note that AB is an m × p matrix, and right multiplying AB by the p × q matrix C , we see that ABC is an m × q matrix. 3. TRUE. We have (A + B )T = AT + B T = A + B , so A + B is 010 00 4. FALSE. For example, let A = −1 0 0 , B = 0 0 000 −3 0 00 0 but AB = 0 0 −3 is not symmetric. 00 0 symmetric. 3 0 . Then A and B are skew-symmetric, 0 5. FALSE. The correct equation is (A + B )2 = A2 + AB + BA + B 2 . The statement is false since AB + BA 10 01 11 does not necessarily equal 2AB . For instance, if A = and B = , then (A + B )2 = 00 00 00 12 and A2 + 2AB + B 2 = = (A + B )2 . 00 6. FALSE. For example, let A = 0 0 1 0 and B = 1 0 0 0 . Then AB = 0 even though A = 0 and B = 0. 00 00 and let B = . Then A is not upper triangular, despite 10 00 the fact that AB is the zero matrix, hence automatically upper triangular. 7. FALSE. For example, let A = 8. FALSE. For instance, the matrix A = 1 0 0 0 is neither the zero matrix nor the identity matrix, and 2 yet A = A. 9. TRUE. The derivative of each entry of the matrix is zero, since in each entry, we take the derivative of a constant, thus obtaining zero for each entry of the derivative of the matrix. 10. FALSE. The correct statement is given in Problem 41. The problem with the statement as given is that the second term should be dA B , not B dA . dt dt 11. FALSE. For instance, the matrix function A = form cet 0 0 cet . 2et 0 0 3et satisfies A = dA dt , but A does not have the 117 12. TRUE. This follows by exactly the same proof as given in the text for matrices of numbers (see part 3 of Theorem 2.2.21). Problems: 1. 2A = 2 6 4 −2 10 4 −6 3 −9 −3 −12 −15 , −3B = A − 2B = = 1 3 , 2 −1 5 2 − 4 −2 2 8 6 10 −3 4 −7 1 −3 −8 , 3A + 4B = = 3 9 11 13 6 −3 15 6 2 31 9 26 + 8 −4 12 4 16 20 . 2. Solving for D, we have 2A + B − 3C + 2D = A + 4C 2D = −A − B + 7C 1 D = (−A − B + 7C ). 2 When appropriate substitutions are made for A,B , and C , we obtain: −5 −2.5 2.5 6.5 9 . D = 1.5 −2.5 2.5 −0.5 3. 5 10 −3 27 22 3 AB = 9 BC = 8 , −6 , DC = [10], 2 −2 3 2 −3 . CD = −2 4 −4 6 DB = [6 14 − 4], CA and AD cannot be computed. 4. AB = 2−i 1+i −i 2 + 4i = (2 − i)i + (1 + i)0 −i(i) + (2 + 4i)0 = 1 + 2 i 2 − 2i 1 1 + 17i i 1 − 3i 0 4+i (2 − i)(1 − 3i) + (1 + i)(4 + i) −i(1 − 3i) + (2 + 4i)(4 + i) . 118 5. AB = 3 + 2i 2 − 4i 5 + i −1 + 3i −1 + i 3 + 2i 4 − 3i 1 + i = (3 + 2i)(−1 + i) + (2 − 4i)(4 − 3i) (5 + i)(−1 + i) + (−1 + 3i)(4 − 3i) = −9 − 21i 11 + 10i −1 + 19i 9 + 15i 6. 3 − 2i i −i 1 AB = (3 + 2i)(3 + 2i) + (2 − 4i)(1 + i) (5 + i)(3 + 2i) + (−1 + 3i)(1 + i) . −1 + i 2 − i 0 1 + 5i 0 3 − 2i = (3 − 2i)(−1 + i) + i(1 + 5i) −i(−1 + i) + 1(1 + 5i) = (3 − 2i)(2 − i) + i · 0 −i(2 − i) + 1 · 0 −6 + 6i 4 − 7i 2 + 3i 2 + 6i −1 − 2i 3 − 2i 7. (3 − 2i)0 + i(3 − 2i) −i · 0 + 1(3 − 2i) 1 −1 2 3 −2 346 ABC = 7 7 = CAB = = 3 2 1 5 4 −3 C −1 6 −3 2 1 −4 −3 2 1 −4 9 35 1 −1 2 3 −2 346 −7 9 2 3 9 −13 −14 −21 8. (2A − 3B )C = = 2 = −12 −22 14 −126 B 3 2 1 5 4 −3 = −1 6 1 −2 3 1 5 −10 −9 −7 . −3 −1 5 3 −1 −7 43 −21 −131 2 3 C 25 −20 = . . 9. Ac = 10. 11. 1 −5 3 4 6 −2 =6 1 −5 + (−2) 3 4 = 0 −38 . 3 −1 4 2 3 −1 4 −13 1 5 3 = 2 2 + 3 1 + (−4) 5 = −13 . Ac = 2 7 −6 3 −4 7 −6 3 −16 −1 2 7 Ac = 4 5 −4 5 −1 −1 2 −7 = 5 4 + (−1) 7 = 13 . 5 −4 29 119 12. The dimensions of B should be n × r in order that ABC is defined. The elements of the ith row of A are ai1 , ai2 , . . . , ain and the elements of the j th column of BC are r r r b1m cmj , b2m cmj , . . . , m=1 m=1 bnm cmj , m=1 so the element in the ith row and j th column of ABC = A(BC ) is r r ai1 m=1 m=1 n = r b2m cmj + · · · + ain b1m cmj + ai2 r aik n r k=1 bkm cmj m=1 = m=1 k=1 bnm cmj m=1 aik bkm cmj . −1 −4 8 7 . 13. (a): A2 = AA = 1 −1 2 3 1 −1 2 3 A3 = A2 A = −1 −4 8 7 A4 = A3 A = −9 −11 22 13 = 1 −1 2 3 −9 −11 22 13 = 1 −1 2 3 . −31 −24 48 17 = . (b): 0 10 0 10 −2 0 1 0 1 −2 0 1 = 4 −3 0 . A2 = AA = −2 4 −1 0 4 −1 0 2 4 −1 −2 0 1 0 10 4 −3 0 0 −2 0 1 = 6 4 −3 . A3 = A2 A = 4 −3 2 4 −1 4 −1 0 −12 3 4 4 −3 0 0 10 6 4 −3 6 4 −3 −2 0 1 = −20 9 4 . A4 = A3 A = −12 3 4 4 −1 0 10 −16 3 14. (a): (A + B )2 = (A + B )(A + B ) = A(A + B ) + B (A + B ) = A2 + AB + BA + B 2 . (b): (A − B )2 = (A + (−1)B )2 = A2 + A(−1)B + (−1)BA + [(−1)B ]2 2 by part (a) 2 = A − AB − BA + B . 15. A2 − 2A − 8I2 = 14 −10 −2 6 −2 = 14 −10 −2 6 + 3 −1 −5 −1 −6 2 10 2 + −8 10 01 −8 0 0 −8 = 02 . 120 16. 100 A2 = 0 1 0 − 001 1xz Substituting A = 0 1 y for A, we have 001 1xz 1 0 1 y 0 001 0 0 −1 0 1 0 0 −1 = 0 0 0 0 0 xz 1 1 y = 0 01 0 1 1 0 0 1 . 1 0 1 , 1 1 1 0 that is, 1 0 0 2x 2z + xy 1 = 0 1 2y 0 1 0 1 1 0 0 1 . 1 Since corresponding elements of equal matrices are equal, we obtain the following implications: 1 Thus, A = 0 0 2y = 1 =⇒ y = 1/2, 2x = 1 =⇒ x = 1/2, 2z + xy = 0 =⇒ 2z + (1/2)(1/2) = 0 =⇒ z = −1/8. 1/2 −1/8 1 1/2 . 0 1 x1 x1 −2 y −2 y x1 x2 − x − 2 x+y−1 , or equivalently, −2 y −2x − 2y + 2 y 2 − y − 2 trices are equal, it follows that 17. In order that A2 = A, we require = x1 −2 y , that is, x2 − 2 −2x − 2y y 2 − y − 2 = 0 =⇒ y = −1 or y = 2. Two cases arise from x + y − 1 = 0: (a): If x = −1, then y = 2. (b): If x = 2, then y = −1. Thus, 18. 1 0 −1 −2 1 2 0 −i i 0 σ1 σ2 = 0 1 σ2 σ3 = 0 −i i 0 1 0 0 −1 σ3 σ1 = 1 0 0 −1 0 1 1 0 2 1 −2 −1 or A = = = = = = 02 . Since corresponding elements of equal ma- x2 − x − 2 = 0 =⇒ x = −1 or x = 2, and A= x+y −2 + y 2 . i 0 0 −i =i 1 0 0 −1 0i i0 =i 0 1 =i 0 −i i 0 0 −1 1 0 1 0 = iσ3 . = iσ1 . = iσ2 . 121 19. [A, B ] = AB − BA = 3 4 = −1 −1 10 4 − = 20. 1 −1 2 1 −6 2 1 6 1 2 1 −1 2 1 31 42 − 5 −2 8 −2 = 02 . [A1 , A2 ] = A1 A2 − A2 A1 = 1 0 0 1 = 0 0 1 0 01 00 0 0 − 01 00 − 1 0 1 0 0 1 = 02 , thus A1 and A2 commute. [A1 , A3 ] = A1 A3 − A3 A1 = 1 0 0 1 = 0 1 0 0 0 1 0 0 0 1 − 0 1 − 0 0 0 0 1 0 0 1 = 02 , thus A1 and A3 commute. [A2 , A3 ] = A2 A3 − A3 A2 = 21. 1 0 = Then [A3 , A2 ] = −[A2 , A3 ] = 0 0 1 0 0 0 −1 0 0 1 0 1 0 0 0 0 − 0 1 − 0 1 = 0 0 1 0 0 −1 0 0 1 0 = 02 . = 02 . Thus, A2 and A3 do not commute. [A1 , A2 ] = A1 A2 − A2 A1 1 4 1 = 4 1 = 4 = 0 −1 1 0 0i i0 i 0 0 −i 1 4 − 2i 0 0 −2i = − 1 4 0 −1 1 0 0i i0 −i 0 0i 1 2 i 0 0 −i = A3 . [A2 , A3 ] = A2 A3 − A3 A2 1 4 1 = 4 1 = 4 = 0 −1 1 0 0i i0 0 2i 2i 0 i 0 0 −i − 1 4 = − 1 4 i 0 0 −1 0 −i −i 0 1 2 0i i0 = A1 . 0 −1 1 0 122 [A3 , A1 ] = A3 A1 − A1 A3 1 4 1 = 4 1 = 4 i 0 0 −i 0i i0 0 −1 −1 0 = − 0 −2 2 0 = − 1 4 1 2 0 1 1 4 0i i0 i 0 0 −i 1 0 0 −1 1 0 = A2 . 22. [A, [B, C ]] + [B, [C, A]] + [C, [A, B ]] = [A, BC − CB ] + [B, CA − AC ] + [C, AB − BA] = A(BC − CB ) − (BC − CB )A + B (CA − AC ) − (CA − AC )B + C (AB − BA) − (AB − BA)C = ABC − ACB − BCA + CBA + BCA − BAC − CAB + ACB + CAB CBA − ABC + BAC = 0. 23. Proof that A(BC ) = (AB )C : Let A = [aij ] be of size m × n, B = [bjk ] be of size n × p, and C = [ckl ] be of size p × q . Consider the (i, j )-element of (AB )C : p n [(AB )C ]ij = k=1 p n aih bhk ckj = h=1 aih bhk ckj h=1 = [A(BC )]ij . k=1 Proof that A(B + C ) = AB + AC : We have n [A(B + C )]ij = aik (bkj + ckj ) k=1 n = (aij bkj + aik ckj ) k=1 n n = aik bkj + k=1 aik ckj k=1 = [AB + AC ]ij . 24. Proof that (AT )T = A: Let A = [aij ]. Then AT = [aji ], so (AT )T = [aji ]T = aij = A, as needed. Proof that (A + C )T = AT + C T : Let A = [aij ] and C = [cij ]. Then [(A + C )T ]ij = [A + C ]ji = [A]ji + [C ]ji = aji + cji = [AT ]ij + [C T ]ij = [AT + C T ]ij . Hence, (A + C )T = AT + C T . 25. We have m (IA)ij = δik akj = δii aij = aij , k=1 for 1 ≤ i ≤ m and 1 ≤ j ≤ p. Thus, Im Am×p = Am×p . 26. Let A = [aij ] and B = [bij ] be n × n matrices. Then n n tr(AB ) = n aki bik k=1 i=1 n = n bik aki k=1 i=1 n i=1 k=1 = bik aki = tr(BA). 123 27. 1 2 3 −1 0 4 AT = 1 2 −1 4 −3 0 1 −1 1 0 2 AAT = 2 3 4 −1 1 −1 1 2 0 2 AB = 3 4 −1 3 4 −1 0 −1 1 2 1 211 BT = , , 1 2 3 4 19 −8 −2 −1 0 4 = −8 17 −3 4 , 1 2 −1 0 −2 4 26 4 −3 0 4 01 10 4 −3 −1 2 = −4 1 , 0 1 1 −5 10 0 21 1 2 3 0 −1 1 2 −1 0 4 = 10 −4 −5 . 1 2111 2 −1 4 1 10 4 −3 0 B T AT = 28. (a): We have z 2z , z −x −y y S = [s1 , s2 , s3 ] = 0 x −y so 2 AS = 2 1 2 5 2 1 −x −y 2 0 y 2 x −y z −x −y y 2z = 0 z x −y 7z 14z = [s1 , s2 , 7s3 ]. 7z (b): −x 0 S T AS = S T (AS ) = −y y z 2z x −x −y −y 0 y z x −y 7z 2x2 = 0 14z 7z 0 but S T AS = diag(1, 1, 7), so we have the following √ 2 2 √ 3 3y 2 = 1 =⇒ y = ± 3 √ 6 2 6z = 1 =⇒ z = ± . 6 2 2x = 1 =⇒ x = ± 29. 2 0 (a): 0 0 0 2 0 0 0 0 2 0 0 0 . 0 2 0 3y 2 0 0 0 , 42z 2 124 7 (b): 0 0 0 7 0 0 0 . 7 30. Suppose A is an n × n scalar matrix with trace k . If A = aIn , then tr(A) = na = k , so we conclude that k a = k/n. So A = n In , a uniquely determined matrix. 31. We have ST = and TT = T 1 (A + AT ) 2 1 (A − AT ) 2 = T = 1 1 (A + AT )T = (AT + A) = S 2 2 1T 1 (A − A) = − (A − AT ) = −T. 2 2 Thus, S is symmetric and T is skew-symmetric. 32. 1 1 3 S= 2 7 1 1 T = 3 2 7 −5 3 1 2 4 + −5 −2 6 3 −5 3 1 2 4 − −5 −2 6 3 3 7 2 −2 10 1 −1 5 1 −2 4 2 = −1 2 −2 = 2 1 . 2 4 6 10 2 12 5 16 3 7 0 −8 −4 0 −4 −2 1 2 −2 = 8 0 6 = 4 0 3 . 2 4 6 4 −6 0 2 −3 0 33. If A is an n × n symmetric matrix, then AT = A, so it follows that T= 1 1 (A − AT ) = (A − A) = 0n . 2 2 If A is an n × n skew-symmetric matrix, then AT = −A and it follows that S= 1 1 (A + AT ) = (A + (−A)) = 0n . 2 2 34. Let A be any n × n matrix. Then A= 1 1 1 1 (2A) = (A + AT + A − AT ) = (A + AT ) + (A − AT ), 2 2 2 2 a sum of a symmetric and skew-symmetric matrix, respectively, by Problem 31. 35. If A = [aij ] and D = diag(d1 , d2 , . . . , dn ), then we must show that the (i, j )-entry of DA is di aij . In index notation, we have n di δik akj = di δii aij = di aij . (DA)ij = k=1 Hence, DA is the matrix obtained by multiplying the ith row vector of A by di , where 1 ≤ i ≤ n. 36. (a): We have (AAT )T = (AT )T AT = AAT , so that AAT is symmetric. (b): We have (ABC )T = [(AB )C ]T = C T (AB )T = C T (B T AT ) = C T B T AT , as needed. 37. A (t) = −2e−2t cos t . 125 38. A (t) = 1 − sin t 39. A (t) = et 2et cos t 4 2e2t 8e2t . 2t 10t . − sin t 0 cos t 1 . 3 0 cos t 40. A (t) = sin t 0 41. We show that the (i, j )-entry of both sides of the equation agree. First, recall that the (i, j )-entry of n d AB is k=1 aik bkj , and therefore, the (i, j )-entry of dt (AB ) is (by the product rule) n n n aik bkj + aik bkj + aik bkj = k=1 k=1 aik bkj . k=1 The former term is precise the (i, j )-entry of the matrix dA B , while the latter term is precise the (i, j )-entry dt d of the matrix A dB . Thus, the (i, j )-entry of dt (AB ) is precisely the sum of the (i, j )-entry of dA B and the dt dt (i, j )-entry of A dB . Thus, the equation we are proving follows immediately. dt 42. We have π /2 cos t sin t 0 sin t − cos t dt = π /2 0 sin(π/2) − cos(π/2) = − sin 0 − cos 0 = 1 0 0 −1 1 1 . e − 1 1 − 1/e 2e − 2 5 − 5/e . − = 43. We have 1 et 2et 0 e−t 5e−t dt = 44. We have 1 0 −e−t −5e−t et 2et 1 0 = e −1/e 2e −5/e − − 1 cos 2t 2 t tet − et 1 0 3 − 5t 32 tan t t + cos t 2 1 e2 − cos 2 2 2 2 − 0 = −14/3 0 0 tan 1 3 + cos 1 2 e2t sin 2t t2 − 5 dt = tet 2 sec t 3t − sin t 1 −1 2 −5 = 1 2t 2e 3 e2 −1 −1 2 2 −1 = −14/3 1 tan 1 1−cos 2 2 1 2 . 1 + cos 1 45. We have 1 0 et 2et e2t 4e2t t2 5t2 dt = = = 46. 2t 3t2 dt = t2 t3 . t3 3 53 3t 1 0 e e2 /2 1/3 2e 2e2 5/3 − et 2et 1 2t 2e 2t 2e 1 2 1/2 0 2 0 = e−1 2e − 2 e2 −1 2 2 2e − 2 1/3 5/3 . 126 47. 48. 49. sin t − cos t 0 cos t sin t 3t −e−t . −5e−t 1 2t e2t sin 2t 2e t2 − 5 dt = t3 − 5t tet 3 sec2 t 3t − sin t tan t et 2et e−t 5e−t sin t 0 − cos t t2 /2 . 3t2 /2 t − cos t 0 t dt = − sin t 1 0 dt = et 2et − 1 cos 2t 2 tet − et . 32 2 t + cos t Solutions to Section 2.3 True-False Review: 1. FALSE. The last column of the augmented matrix corresponds to the constants on the right-hand side of the linear system, so if the augmented matrix has n columns, there are only n − 1 unknowns under consideration in the system. 2. FALSE. Three distinct planes can intersect in a line (e.g. Figure 2.3.1, lower right picture). For instance, the xy -plane, the xz -plane, and the plane y = z intersect in the x-axis. 3. FALSE. The right-hand side vector must have m components, not n components. 4. TRUE. If a linear system has two distinct solutions x1 and x2 , then any point on the line containing x1 and x2 is also a solution, giving us infinitely many solutions, not exactly two solutions. 5. TRUE. The augmented matrix for a linear system has one additional column (containing the constants on the right-hand side of the equation) beyond the matrix of coefficients. 6. FALSE. For instance, if A = to AT x = 0 take the form 0 t 0 0 1 0 , then solutions to Ax = 0 take the form . The solution sets are not the same. Problems: 1. 2 · 1 − 3(−1) + 4 · 2 = 13, 1 + (−1) − 2 = −2, 5 · 1 + 4(−1) + 2 = 3. 2. 2 + (−3) − 2 · 1 = −3, 3 · 2 − (−3) − 7 · 1 = 2, 2 + (−3) + 1 = 0, 2 · 2 + 2(−3) − 4 · 1 = −6. 3. (1 − t) + (2 + 3t) + (3 − 2t) = 6, (1 − t) − (2 + 3t) − 2(3 − 2t) = −7, t 0 , while solutions 127 5(1 − t) + (2 + 3t) − (3 − 2t) = 4. 4. s + (s − 2t) − (2s + 3t) + 5t = 0, 2(s − 2t) − (2s + 3t) + 7t = 0, 4s + 2(s − 2t) − 3(2s + 3t) + 13t = 0. 5. The lines 2x + 3y = 1 and 2x + 3y = 2 are system has no solution. 1 2 −3 1 1 6. A = 2 4 −5 , b = 2 , A# = 2 7 7 2 −1 3 parallel in the xy -plane, both with slope −2/3; thus, the 2 −3 1 4 −5 2 . 2 −1 3 11 1 −1 3 11 ,b = , A# = 2 4 −3 7 2 24 1 2 −1 0 1 2 −1 8. A = 2 3 −2 , b = 0 , A# = 2 3 −2 5 6 −5 0 5 6 −5 7. A = 1 −1 3 −3 72 0 0 . 0 . 9. It is acceptable to use any variable names. We will use x1 , x2 , x3 , x4 : x1 − x2 x1 + x2 3x1 + x2 +2x3 + 3x4 −2x3 + 6x4 +4x3 + 2x4 = 1, = −1, = 2. 10. It is acceptable to use any variable names. We will use x1 , x2 , x3 : 2x1 + x2 4x1 − x2 7x1 + 6x2 +3x3 = 3, +2x3 = 1, +3x3 = −5. 11. Given Ax = 0 and Ay = 0, and an arbitrary constant c, (a): we have Az = A(x + y) = Ax + Ay = 0 + 0 = 0 and Aw = A(cx) = c(Ax) = c0 = 0. (b): No, because A(x + y) = Ax + Ay = b + b = 2b = b, and A(cx) = c(Ax) = cb = b in general. 12. x1 x2 = −4 3 6 −4 x1 x2 + 4t t2 . 128 13. x1 x2 = t2 − sin t −t 1 14. x1 x2 = 0 − sin t e2t 0 x1 0 15. x2 = −et x3 −t − sin t 0 t2 x1 x2 . x1 + x2 1 x1 t2 x2 0 x3 0 . 1 t + t3 . 1 16. We have 4e4t −2(4e4t ) x (t) = and e4t −2e4t 2 −1 −2 3 Ax + b = 17. We have x (t) = + 0 0 = 4(−2e−2t ) + 2 cos t 3(−2e−2t ) + sin t = 4e4t −8e4t 2e4t + (−1)(−2e4t ) + 0 −2e4t + 3(−2e4t ) + 0 = = 4e4t −8e4t . −8e−2t + 2 cos t −6e−2t + sin t and Ax + b = = 1 −4 −3 2 4e−2t + 2 sin t 3e−2t − cos t + −2(cos t + sin t) 7 sin t + 2 cos t 4e−2t + 2 sin t − 4(3e−2t − cos t) − 2(cos t + sin t) −3(4e−2t + 2 sin t) + 2(3e−2t − cos t) + 7 sin t + 2 cos t = −8e−2t + 2 cos t −6e−2t + sin t . Solutions to Section 2.4 True-False Review: 1. TRUE. The precise row-echelon form obtained for a matrix depends on the particular elementary row operations (and their order). However, Theorem 2.4.15 states that there is a unique reduced row-echelon form for a matrix. 2. FALSE. Upper triangular matrices could have pivot entries that are not 1. For instance, the following 20 matrix is upper triangular, but not in row echelon form: . 00 3. TRUE. The pivots in a row-echelon form of an n × n matrix must move down and to the right as we look from one row to the next beneath it. Thus, the pivots must occur on or to the right of the main diagonal of the matrix, and thus all entries below the main diagonal of the matrix are zero. 4. FALSE. This would not be true, for example, if A was a zero matrix with 5 rows and B was a nonzero matrix with 4 rows. 5. FALSE. If A is a nonzero matrix and B = −A, then A + B = 0, so rank(A + B ) = 0, but rank(A), rank(B ) ≥ 1 so rank(A)+ rank(B ) ≥ 2. 6. FALSE. For example, if A = B = 1 + 1 = 2. 0 0 1 0 , then AB = 0, so rank(AB ) = 0, but rank(A)+ rank(B ) = 129 7. TRUE. A matrix of rank zero cannot have any pivots, hence no nonzero rows. It must be the zero matrix. 8. TRUE. The matrices A and 2A have the same reduced row-echelon form, since we can move between the two matrices by multiplying the rows of one of them by 2 or 1/2, a matter of carrying out elementary row operations. If the two matrices have the same reduced row-echelon form, then they have the same rank. 9. TRUE. The matrices A and 2A have the same reduced row-echelon form, since we can move between the two matrices by multiplying the rows of one of them by 2 or 1/2, a matter of carrying out elementary row operations. Problems: 1. Row-echelon form. 2. Neither. 3. Reduced row-echelon form. 4. Neither. 5. Reduced row-echelon form. 6. Row-echelon form. 7. Reduced row-echelon form. 8. Reduced row-echelon form. 9. 2 1 1 −3 1 ∼ 1 −3 2 1 2 ∼ 2 −4 −4 8 1 ∼ 1 −2 −4 8 1. M1 ( 1 ) 2 11. 2 14 3 −2 6 1 2 −3 4 ∼ 2 −3 4 3 −2 6 2 14 1 1 5 1 ∼ 0 0 −5 1. P13 12. 2. A21 (−1) 0 0 0 1 1 3 2 ∼ 1 −3 0 1 3 ∼ 2. A12 (−2) 1. P12 10. 1 −3 0 7 , Rank (A) = 2. 3. M2 ( 1 ) 7 1 −2 0 0 , Rank (A) = 1. 2. A12 (4) 1 12 1 12 1 12 3 4 ∼ 2 −3 4 ∼ 0 −5 0 ∼ 0 −1 0 2 14 0 −1 0 0 −5 0 2 112 6 0 ∼ 0 1 0 , Rank (A) = 2. 0 000 2 3. A12 (−2), A13 (−3) 3 0 1 4 ∼ 0 5 0 1 0 0 3 0 2 1 ∼ 0 4 0 4. P23 1 0 0 5. M2 (−1) 3 1 , Rank (A) = 2. 0 6. A32 (5) 130 1. A12 (−1), A13 (−3) 2. A23 (−4) 13. 2 −1 3 2 1 3 1 3 1 3 13 1 2 3 4 5 3 2 ∼ 2 −1 ∼ 2 −1 ∼ 0 −7 ∼ 0 −1 ∼ 0 1 , Rank (A) = 2. 2 5 2 5 2 5 0 −1 0 −7 00 1. P12 2. A21 (−1) 3. A12 (−2), A13 (−2) 4. P23 5. M2 (−1), A23 (7). 14. 2 −1 3 3 1 −2 1 1 2 3 1 −2 ∼ 2 −1 3 ∼ 2 2 −2 1 2 −2 1 0 1 2 −5 12 5 6 1 2 ∼ 0 1 ∼ 0 00 0 −5 13 2. A21 (−1), A23 (−1) 1. P12 15. 2 −1 1 −2 1 −5 3 1 0 2 3 1 2 3. A12 (−2) 4. P23 1 2 −5 1 2 −5 4 0 −5 13 ∼ 0 −1 −2 0 −1 −2 0 −5 13 −5 2 , Rank (A) = 3. 1 5. M2 (−1) 1 −2 4 1 −2 1 3 1 −2 1 3 1 2 3 3 ∼ 2 −1 3 4 ∼ 0 3 1 −2 ∼ 0 1 5 1 −5 0 5 0 00 0 0 0 1. P12 16. 2 −5 3 −1 3 ∼ −1 −2 −5 12 7 2 ∼ 0 1 23 00 2. A12 (−2), A13 (−1) 2. A12 (−3), A13 (−2), A14 (−2) 1 1 3 0 3 7. M3 (1/23). − 2 , Rank (A) = 2. 3 0 3. M2 (1/3) −2 −1 3 1 −1 10 1 −1 10 −2 3 1 1 3 −2 3 120 1 01 ∼ ∼ −1 1 0 2 −2 −1 3 0 0 −3 3 −1 22 2 −1 22 0 1 02 1 −1 1 0 10 1 40 , Rank (A) = 4. ∼ 0 0 1 −1 0 00 1 1. P13 6. A23 (5) 1 −1 10 30 1 0 1 ∼ 0 0 −3 3 0 0 01 3. A24 (−1) 4. M3 (1/3) 17. 4 74 7 1 21 2 3 53 513 53 5 2 −2 2 −2 ∼ 2 −2 2 −2 5 −2 5 −2 5 −2 5 −2 1 2 2 0 −1 ∼ 0 −6 0 −12 1 2 1 21 2 0 −1 3 0 10 1 ∼ 0 −6 0 −6 0 −6 0 −12 0 −12 0 −12 131 1 40 ∼ 0 0 2 1 0 0 1 0 0 0 2 1 , Rank (A) = 2. 0 0 1. A21 (−1) 2. A12 (−3), A13 (−2), A14 (−5) 3. M2 (−1) 4. A23 (6), A24 (12) 18. 2 1 2 1 0 3 3 2 1 4 1 5 2 1 1 3 ∼ 2 7 2 0 1 3 2 3 1 1 4 5 3 1 2 2 ∼ 0 7 0 0 21 3 1 3 1 −1 2 −4 ∼ 0 3 −3 3 1 0 0 2 1 3 1 −1 2 −4 0 0 −3 1 10 21 3 4 ∼ 0 1 −1 2 −4 , Rank (A) = 3. 00 0 1 −1 3 2. A12 (−2), A13 (−2), 1. P12 19. 3 2 1 −1 1 ∼ 1 −1 3 2 2 ∼ 1. P12 20. 3 2 1 1. P13 21. 1 −1 0 5 3 ∼ 2. A12 (−3) 3. A23 (−3) 1 −1 0 1 4 ∼ 3. M2 ( 1 ) 5 1 0 4. M3 (− 1 ) 3 0 1 = I2 , Rank (A) = 2. 4. A21 (1) 7 10 12 1 1 2 1 121 1 1 2 3 4 3 −1 ∼ 2 3 −1 ∼ 0 −1 −3 ∼ 0 1 3 ∼ 0 2 1 3 7 10 0 1 7 017 0 1 0 −5 100 5 6 3 ∼ 0 1 0 = I3 , Rank (A) = 3. ∼ 0 1 00 1 001 2. A12 (−2), A13 (−3) 3. M2 (−1) 4. A21 (−2), A23 (−1) 1 5. M3 ( 4 ) 0 −5 1 3 0 4 6. A31 (5), A32 (−3) 3 −3 6 1 −1 2 1 2 −2 4 ∼ 0 0 0 , Rank (A) = 1. 6 −6 12 0 00 1. M1 ( 1 ), A12 (−2), A13 (−6) 3 22. 3 5 −12 1 2 −5 1 2 −5 10 1 1 2 3 2 3 −7 ∼ 0 −1 3 ∼ 0 1 −3 ∼ 0 1 −3 , Rank (A) = 2. −2 −1 1 0 3 −9 0 3 −9 00 0 132 1. A21 (−1), A12 (−2), A13 (2) 23. 1 3 2 4 −1 −1 2 1 −1 −1 2 −2 0 710 1 31 ∼ −1 2 4 0 1 40 −2 38 0 2 70 100 5 1 450 40 1 0 ∼ 0 0 1 −1 ∼ 0 000 1 0 1. A12 (−3), A13 (−2), A14 (−4) 1 2 3 0 0 0 0 1 0 0 0 0 1 0 1 20 ∼ 0 0 2 3 1 3 130 ∼ 1 −1 0 1 −2 0 0 1 0 0 0 1 0 0 0 5 0 4 1 −1 0 −1 0 0 = I4 , Rank (A) = 4. 0 1 3. A31 (−2), A32 (−3), A34 (−1) 5. A41 (−5), A42 (−4), A43 (1) 3 1 −2 1 3 1 −2 1 3 1 −2 0 1 1 2 3 7 ∼ 0 0 −1 −2 ∼ 0 0 1 2 ∼ 0 0 1 2 , Rank (A) = 2. 10 0 0 −1 −2 0 0 −1 −2 0 000 1. A12 (−3), A13 (−4) 25. 3. A21 (−2), A23 (−3) 2. A21 (1), A23 (−1), A24 (−2) 4. M4 (−1) 24. 1 −2 3 −6 4 −8 2. M2 (−1) 1 3 2 2 1 0 1 1 2 ∼ 1 0 4 ∼ 0 0 1. A12 (−3), A13 (−2) 2. M2 (−1) 01 2 1 0 2 0 0 −6 −2 ∼ 0 0 0 −4 −1 0 1 0 1/3 01 5 0 1 1/3 ∼ 0 0 00 1 00 2. M2 (− 1 ) 6 3. A21 (−1), A23 (1) 1 2 1 01 3 0 1 1/3 ∼ 0 0 0 −4 −1 00 00 1 0 , Rank (A) = 3. 01 3. A21 (−2), A23 (4) 4. M3 (3) 0 1 0 1/3 1/3 1/3 5. A32 (− 1 ), A31 (− 1 ) 3 3 Solutions to Section 2.5 True-False Review: 1. FALSE. This process is known as Gaussian elimination. Gauss-Jordan elimination is the process by which a matrix is brought to reduced row echelon form via elementary row operations. 2. TRUE. A homogeneous linear system always has the trivial solution x = 0, hence it is consistent. 3. TRUE. The columns of the row-echelon form that contain leading 1s correspond to leading variables, while columns of the row-echelon form that do not contain leading 1s correspond to free variables. 4. TRUE. If the last column of the row-reduced augmented matrix for the system does not contain a pivot, then the system can be solved by back-substitution. On the other hand, if this column does contain 133 a pivot, then that row of the row-reduced matrix containing the pivot in the last column corresponds to the impossible equation 0 = 1. 5. FALSE. The linear system x = 0, y = 0, z = 0 has a solution in (0, 0, 0) even though none of the variables here is free. 6. FALSE. The columns containing the leading 1s correspond to the leading variables, not the free variables. Problems: For the problems of this section, A will denote the coefficient matrix of the given system, and A# will denote the augmented matrix of the given system. 1. Converting the given system of equations to an augmented obtain the following equivalent matrices: 12 1 2 1 1 1211 2 1 3 5 1 3 ∼ 0 −1 −2 0 ∼ 0 1 02 0 2 5 −1 2671 1. A12 (−3), A13 (−2) matrix and using Gaussian elimination we 1 1 1 3 2 0 ∼ 0 0 5 −1 2. M2 (−1) 2 1 0 1 1 2 0 . 1 −1 3. A23 (−2) The last augmented matrix results in the system: x1 + 2x2 + x3 = 1, x2 + 2x3 = 0, x3 = −1. By back substitution we obtain the solution (−2, 2, −1). 2. Converting the given system of equations to an augmented matrix and using Gaussian elimination, we obtain the following equivalent matrices: 1 −2 −5 −3 1 −2 −5 −3 3 −1 0 1 1 2 2 4 ∼ 0 4 ∼ 2 1 5 5 15 10 1 5 7 −5 −8 −3 0 9 27 18 7 −5 −8 −3 1 −2 −5 −3 1011 3 4 1 3 2 ∼ 0 1 3 2 . ∼ 0 0 9 27 18 0000 1. A21 (−1) 2. A12 (−2), A13 (−7) 3. M2 ( 1 ) 5 4. A21 (2), A23 (−9) The last augmented matrix results in the system: x1 + x3 = 1, x2 + 3x3 = 2. Let the free variable x3 = t, a real number. By back substitution we find that the system has the solution set {(1 − t, 2 − 3t, t) : for all real numbers t}. 3. Converting the given system of equations to an augmented matrix and using Gaussian elimination we obtain the following equivalent matrices: 134 3 5 −1 14 12 1 1 1 2 1 3 ∼ 3 5 −1 25 6 25 62 121 4 ∼ 0 1 4 000 1. P12 1 3 2 4 ∼ 0 2 0 3 5 5 ∼ −9 2. A12 (−3), A13 (−2) 1 2 1 3 3 −1 −4 −5 ∼ 0 0 1 4 −4 1213 0 1 4 5 . 0001 3. M2 (−1) 4. A23 (−1) 2 1 1 1 3 4 5 4 −4 1 5. M4 (− 9 ) This system of equations is inconsistent since 2 = rank(A) < rank(A# ) = 3. 4. Converting the given system of equations to an augmented matrix and using Gaussian elimination we obtain the following equivalent matrices: 6 −3 3 12 1 −1 1 −1 −1 2 2 2 2 1 1 ∼ 0 2 −1 ∼ 2 −1 1 4 0 1 4 −4 2 −2 −8 0 0 −4 2 −2 −8 1. M1 ( 1 ) 6 1 2 0 0 2 0 . 0 2. A12 (−2), A13 (4) Since x2 and x3 are free variables, let x2 = s and x3 = t. The single equation obtained from the augmented matrix is given by x1 − 1 x2 + 1 x3 = 2. Thus, the solution set of our system is given by 2 2 {(2 + s t − , s, t) : s, t any real numbers }. 22 5. Converting the given system of equations to an augmented matrix and using Gaussian elimination we obtain the following equivalent matrices: 2 −1 3 14 3 1 −2 −1 1 2 −5 −15 3 1 −2 −1 1 2 −1 3 14 2 2 −1 3 −14 ∼ ∼ 7 2 −3 3 7 2 −3 3 7 2 −3 3 5 −1 −2 5 5 −1 −2 5 5 −1 −2 5 1 2 4 0 −12 ∼ 0 −5 0 −11 12 70 1 ∼ 00 00 1. P12 5. A42 (−1) −5 −15 1 32 108 5 0 ∼ 13 44 0 23 80 0 −5 −15 1 −9 −28 8 0 ∼ −32 −96 0 −76 −228 0 2. A21 (−1) 6. M2 (−1) 2 −5 −15 1 −1 9 28 6 0 ∼ −5 13 44 0 −11 23 80 0 2 −5 −15 1 1 −9 −28 9 0 ∼ 0 32 96 0 0 −76 −228 0 3. A12 (−2), A13 (−7), A14 (−5) 7. A23 (5), A24 (11) 1 2 −5 −15 3 0 −5 13 44 ∼ 0 −12 32 108 0 −11 23 80 2 −5 −15 1 −9 −28 −5 13 44 −11 23 80 2 −5 −15 1 −9 −28 . 0 1 3 0 0 0 4. P23 8. M3 (−1) 1 9. M3 ( 32 ), A34 (76). 135 The last augmented matrix results in the system of equations: x1 − 2x2 − 5x3 = −15, x2 − 9x3 = −28, x3 = 3. Thus, using back substitution, the solution set for our system is given by {(2, −1, 3)}. 6. Converting the given system of equations to an augmented obtain the following equivalent matrices: 1 1 1 1 −3 −3 2 −1 −4 5 3 8 2 0 −1 813 2 −5 2 −5 ∼ ∼ 5 1 6 −6 20 0 6 −6 20 5 0 −3 2 −1 −4 −5 1 1 −3 −3 1 1 −3 −3 1 1 −3 −3 −4 −17 5 0 1 −4 −17 40 1 ∼ ∼ 0 0 1 4 13 52 0 0 0 0 −10 −40 0 0 −10 −40 1. P14 2. A12 (−3), A13 (−5), A14 (−2) 3. M2 (−1) matrix and using Gaussian elimination we −3 −3 4 17 9 35 2 11 1 60 ∼ 0 0 1 1 −3 −3 30 1 −4 −17 ∼ 0 1 9 35 0 −3 2 11 1 −3 −3 1 −4 −17 . 0 1 4 0 0 0 4. A23 (−1), A24 (3) 1 5. M3 ( 13 ) 6. A34 (10) The last augmented matrix results in the system of equations: x1 + x2 − 3x3 = − 3, x2 − 4x3 = −17, x3 = 4. By back substitution, we obtain the solution set {(10, −1, 4)}. 7. Converting the given system of equations to an augmented matrix obtain the following equivalent matrices: 1 2 −1 1 1 2 −1 1 1 1 2 4 −2 2 2 ∼ 0 0 00 5 10 −5 5 5 00 00 and using Gaussian elimination we 1 0 . 0 1. A12 (−2), A13 (−5) The last augmented matrix results in the equation x1 + 2x3 − x3 + x4 = 1. Now x2 , x3 , and x4 are free variables, so we let x2 = r, x3 = s, and x4 = t. It follows that x1 = 1 − 2r + s − t. Consequently, the solution set of the system is given by {(1 − 2r + s − t, r, s, t) : r, s, t and real numbers }. 8. Converting the given system of equations to an augmented obtain the following equivalent matrices: 1 2 −1 11 1 2 −1 1 1 2 −3 1 −1 2 1 0 −7 3 −3 0 ∼ 1 −5 2 −2 1 0 −7 3 −3 0 4 1 −1 13 0 −7 3 −3 −1 matrix and using Gaussian elimination we 1 2 −1 1 1 3 20 0 1 −3 7 7 ∼ 0 −7 3 −3 0 0 −7 3 −3 −1 136 1 30 ∼ 0 0 2 −1 1 −3 7 0 0 0 0 1 3 7 0 0 1 1 2 −1 0 4 0 1 −3 7 ∼ 0 0 0 0 −1 00 0 1. A12 (−2), A13 (−1), A14 (−4) 2. M2 (− 1 ) 7 1 1 050 ∼ −1 0 0 0 1 3 7 0 0 2 −1 1 −3 7 0 0 0 0 3. A23 (7), A24 (7) 1 3 7 0 0 4. P34 1 0 . 1 0 5. M3 (−1) The given system of equations is inconsistent since 2 = rank(A) < rank(A# ) = 3. 9. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 1 0 2 1 12 1 1 −2 3 2 1 1 −2 3 2 1 1 4 −3 2 ∼ 0 0 1 4 −3 2 ∼ 0 0 0 0 0 −3 −12 9 −6 4 −1 −10 50 1. A13 (−2) 2 0 0 1 1 0 1 −2 3 4 −3 2 . 0 00 2. A23 (3) The last augmented matrix indicates that the first two equations of the initial system completely determine its solution. We see that x4 and x5 are free variables, so let x4 = s and x5 = t. Then x3 = 2 − 4x4 + 3x5 = 2 − 4s +3t. Moreover, x2 is a free variable, say x2 = r, so then x1 = 3 − 2r − (2 − 4s +3t) − s +2t = 1 − 2r +3s − t. Hence, the solution set for the system is {(1 − 2r + 3s − t, r, 2 − 4s + 3t, s, t) : r, s, t any real numbers }. 10. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 1 4 1 1 4 1 2 −1 −2 2 4 4 1 2 4 3 −2 −1 ∼ 4 3 −2 −1 ∼ 0 −13 −6 −17 1 4 1 4 2 −1 −1 2 0 −9 −3 −6 1 4 1 4 1 4 1 4 1 4 1 4 5 6 12 4 8 ∼ 0 12 4 8 ∼ 0 −1 −2 ∼ 0 0 −13 −6 −17 0 −1 −2 −9 0 12 4 1 0 −7 −32 1 0 −7 −32 1 8 9 10 2 9 ∼ 0 1 2 9 ∼ 0 ∼ 0 1 0 0 −20 −100 00 1 5 0 1. P13 6. P23 2. A12 (−4), A13 (−2) 7. M2 (−1) 3. P23 8. A21 (−4), A23 (−12) 4. M2 (− 4 ) 3 9. 1 4 1 3 ∼ 0 −9 −3 0 −13 −6 4 141 7 −9 ∼ 0 1 2 8 0 12 4 00 3 1 0 −1 . 01 5 4 −6 −17 4 9 8 5. A23 (1) 1 M3 (− 20 ) 10. A31 (7), A32 (−2) The last augmented matrix results in the solution (3, −1, 5). 11. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 31 52 1 1 −1 1 1 1 −1 1 1 2 1 1 −1 1 ∼ 3 1 5 2 ∼ 0 −2 8 −1 21 23 21 23 0 −1 4 1 137 1 1 −1 1 −4 ∼ 0 0 −1 4 3 1 1 −1 1 1 ∼ 0 1 −4 1/2 . 2 00 0 3/2 1 1 4 We can stop here, since we see from this last augmented matrix that the system is inconsistent. In particular, 2 = rank(A) < rank(A# ) = 3. 1. P12 2. A12 (−3), A13 (−2) 3. M2 (− 1 ) 2 4. A23 (1) 12. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 1 0 −2 −3 1 0 −2 −3 1 0 −2 −3 1 0 −2 −3 3 2 1 3 −2 0 1 −1 0 ∼ 0 1 −1 2 0 ∼ 0 4 −9 ∼ 0 −2 00 0 0 0 −4 4 0 0 −4 4 0 1 −4 2 −3 . 2. M2 (− 1 ) 2 1. A12 (−3), A13 (−1) 3. A23 (4) The last augmented matrix results in the following system of equations: x1 − 2x3 = −3 and x2 − x3 = 0. Since x3 is free, let x3 = t. Thus, from the system we obtain the solutions {(2t − 3, t, t) : t any real number }. 13. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 1 −2 3 1 1 −2 3 1 2 −1 3 −1 3 6 6 1 2 3 2 1 −5 −6 ∼ 3 2 1 −5 −6 ∼ 0 8 −8 −8 −24 1 −2 3 1 6 2 −1 3 −1 3 0 3 −3 −3 −9 10 1 −1 0 1 −2 3 1 6 4 3 1 −1 −1 −3 ∼ 0 1 −1 −1 −3 . ∼ 0 0 3 −3 −3 −9 00 0 0 0 1. P13 2. A12 (−3), A13 (−2) 3. M2 ( 1 ) 8 4. A21 (2), A23 (−3) The last augmented matrix results in the following system of equations: x1 + x3 − x4 = 0 and x2 − x3 − x4 = −3. Since x3 and x4 are free variables, we can let x3 = s and x4 = t, where s and t are real numbers. It follows that the solution set of the system is given by {(t − s, s + t − 3, s, t) : s, t any real numbers }. 14. Converting the given system of equations to an augmented matrix and using we obtain the following equivalent matrices: 1 1 1 −1 4 1 1 1 −1 4 11 1 −1 −1 −1 2 1 0 −2 −2 0 −2 2 0 1 ∼ ∼ 1 1 −1 1 −2 0 0 −2 2 −6 0 0 1 −1 1 1 −8 0 −2 0 2 −12 01 Gauss-Jordan elimination 1 −1 4 1 0 1 1 −1 3 0 −1 6 138 10 0 −1 1 0 30 1 ∼ 00 1 −1 0 0 −1 −1 3 3 1 0 0 −1 140 1 0 1 −2 5 ∼ ∼ 0 0 1 −1 3 3 5 0 0 0 −2 8 1. A12 (−1), A13 (−1), A14 (−1) 4. A32 (−1), A34 (1) 5. 1 0 0 0 3 0 0 −1 1 0 0 0 −1 10 1 −2 6 0 1 0 0 2 ∼ 0 1 −1 3 0 0 1 0 −1 00 1 −4 0 0 0 1 −4 2. M2 (− 1 ), M3 (− 1 ), M4 (− 1 ) 2 2 2 M4 (− 1 ) 2 . 3. A24 (−1) 6. A41 (1), A42 (−1), A43 (1) It follows from the last augmented matrix that the solution to the system is given by (−1, 2, −1, −4). 15. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 1 −3 −2 −1 −2 2 1 −3 −2 −1 −2 2 2 −1 3 1 −1 11 1 −3 −2 −1 −2 5 7 3 3 7 3 1 −1 11 0 2 2 −1 2 1 ∼ 0 10 ∼ 3 3 4 2 7 −8 1 −2 −1 1 −2 1 −2 −1 1 −2 1 5 3 3 5 −5 2 1 2 3 −3 0 2 1 2 3 −3 1 0 12 7 6 12 −8 5 −3 −3 1 2 2 5 −3 −3 1 2 2 4 31 11 4 31 11 10 −1 −1 10 1 −3 −2 −1 −2 2 5 5 5 5 5 5 5 5 7 3 3 7 7 3 3 7 3 3 7 7 0 1 0 1 0 1 5 5 5 5 5 5 5 5 5 5 4 5 5 5 3 1 11 2 ∼ 0 0 ∼ 0 0 −10 −4 1 −22 4 2 7 −8 1 − 10 ∼ 0 10 5 5 0 −4 0 2 −12 0 0 5 3 3 5 −5 0 0 −4 0 2 −12 6 24 49 24 0 12 7 6 12 −8 − 124 0 0 − 5 −5 − 124 0 0 − 49 − 6 5 5 5 5 5 5 1 34 2 6 1 1 34 2 1 0 0 − 25 1 0 0 0 10 1 0 0 − 25 50 25 5 50 25 37 42 1 7 1 37 42 0 1 0 0 0 1 0 0 1 0 − 25 −8 − 25 25 50 10 5 25 50 8 7 6 2 1 11 1 2 1 11 − 10 3 − 10 ∼ 0 0 1 5 ∼ 0 0 1 5 5 ∼ 0 0 1 0 −2 5 8 16 8 0 0 0 1 0 0 0 0 0 0 1 −2 1 1 −2 −5 5 5 11 191 68 68 191 0 0 0 0 11 000 − 81 000 − 81 5 10 25 25 50 25 25 50 1 6 1 0 0 0 10 1 10000 5 7 0 1 0 0 − 8 0 1 0 0 0 −3 10 5 10 9 4 3 ∼ 0 0 1 0 0 ∼ 0 0 1 0 −1 2 0 0 0 1 1 −2 0 0 0 1 0 −4 00001 2 0000 1 2 1. P12 8. 2. A12 (−2), A13 (−3), A14 (−1), A15 (−5) 2 A41 ( 25 ), 3. M2 ( 1 ) 5 1 5. M3 (− 10 ) 6. A31 (− 11 ), A32 (− 7 ), 5 5 1 A42 (− 25 ), A43 (− 2 ), A45 (− 68 ) 9. M5 ( 10 ) 5 25 11 4. A21 (3), A23 (−10), A24 (−5), A25 (−12) A34 (4), A35 ( 49 ) 7. M4 ( 5 ) 5 8 1 7 10. A51 (− 10 ), A52 (− 10 ), A53 ( 1 ), A54 (−1) 2 It follows from the last augmented matrix that the solution to the system is given by (1, −3, 4, −4, 2). 16. The equation Ax = b reads 1 −3 1 x1 8 5 −4 1 x2 = 15 . −4 2 4 −3 x3 Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 1 −3 1 8 1 −3 1 8 1 −3 1 8 1 2 5 −4 1 15 ∼ 0 11 −4 −25 ∼ 0 1 1 −5 2 4 −3 −4 0 10 −5 −20 0 10 −5 −20 139 1 1 0 4 −7 10 4 −7 4 5 1 −5 ∼ 0 1 1 −5 ∼ 0 ∼ 0 1 0 0 0 1 −2 0 0 −15 30 3 1. A12 (−5), A13 (−2) 2. A32 (−1) 0 1 0 1 0 0 −3 . 1 −2 1 4. M3 (− 15 ) 3. A21 (3), A23 (−10) 5. A31 (−4), A32 (−1) Thus, from the last augmented matrix, we see that x1 = 1, x2 = −3, and x3 = −2. 17. The equation Ax = b reads 1 0 5 x1 0 3 −2 11 x2 = 2 . 2 −2 6 x3 2 Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 1 0 5 0 1 0 50 1 0 50 2 1 3 −2 11 2 ∼ 0 −2 −4 2 ∼ 0 1 2 −1 0 −2 −4 2 0 −2 −4 2 2 −2 6 2 105 0 3 ∼ 0 1 2 −1 . 000 0 1. A12 (−3), A13 (−2) 2. M2 (−1/2) 3. A23 (2) Hence, we have x1 + 5x3 = 0 and x2 + 2x3 = −1. Since x3 is a free variable, we can let x3 = t, where t is any real number. It follows that the solution set for the given system is given by {(−5t, −2t − 1, t) : t ∈ R}. 18. The equation Ax = b reads 0 0 0 1 −1 x1 −2 5 1 x2 = 8 . 2 1 x3 5 Converting the given system of equations to an augmented matrix using Gauss-Jordan elimination we obtain the following equivalent matrices: 0 1 −1 −2 0 1 −1 −2 0 1 −1 −2 0101 1 2 3 0 5 1 8 ∼ 0 0 6 18 ∼ 0 0 1 3 ∼ 0 0 1 3 . 02 1 5 00 3 9 00 3 9 0000 1. A12 (−5), A13 (−2) 2. M2 (1/6) 3. A21 (1), A23 (−3) Consequently, from the last augmented matrix it follows that the solution set for the matrix equation is given by {(t, 1, 3) : t ∈ R}. 19. The equation Ax = b reads 1 −1 0 −1 x1 2 2 13 7 x2 = 2 . 3 −2 1 0 x3 4 140 Converting the given system of equations to an augmented matrix and using Gauss-Jordan obtain the following equivalent matrices: 1 −1 0 −1 2 1 −1 0 −1 2 1 −1 0 −1 2 10 1 2 3 2 13 7 2 ∼ 0 33 9 −2 ∼ 0 11 3 −2 ∼ 0 1 3 −2 1 04 0 11 3 −2 0 33 9 −2 00 1. A12 (−2), A13 (−3) 2. P23 elimination we 12 0 1 3 −2 . 00 4 3. A21 (1), A23 (−3) From the last row of the last augmented matrix, it is clear that the given system is inconsistent. 20. The equation Ax = b reads 1 3 2 −2 1 0 −1 x1 1 −2 3 x2 3 1 1 x3 3 5 −2 x4 Converting the given system of equations to obtain the following equivalent matrices: 1 1 0 11 0 1 2 3 1 −2 3 8 1 0 −2 −2 ∼ 23 1 1 1 2 3 0 0 5 5 −2 3 5 −2 −9 2 8 = 3 . −9 an augmented matrix and using Gauss-Jordan elimination we 1 0 −1 1 3 1 1 01 2 1 2 1 0 −1 1 1 0 −1 3 0 1 0 220 . ∼ ∼ 00 2 0 0 0 −1 0 −2 −2 0 00 00 0 0 5 5 0 −5 0 −5 1. A12 (−3), A13 (−2), A14 (2) 2. P23 3. A21 (−1), A23 (2), A24 (−5) From the last augmented matrix, we obtain the system of equations: x1 − x3 + x4 = 3, x2 + x3 = −1. Since both x3 and x4 are free variables, we may let x3 = r and x4 = t, where r and t are real numbers. The solution set for the system is given by {(3 + r − t, −r − 1, r, t) : r, t ∈ R}. 21. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 12 −1 3 1 2 −1 3 12 −1 3 1 2 2 5 ∼ 0 1 . 1 7 ∼ 0 1 3 1 3 1 2 2 2 1 1 −k −k 0 −1 1 − k −3 − k 0 0 4 − k −2 − k 1. A12 (−2), A13 (−1) 2. A23 (1) (a): If k = 2, then the last row of the last augmented matrix reveals an inconsistency; hence the system has no solutions in this case. (b): If k = −2, then the last row of the last augmented matrix consists entirely of zeros, and hence we have only two pivots (first two columns) and a free variable x3 ; hence the system has infinitely many solutions. (c): If k = ±2, then the last augmented matrix above contains a pivot for each variable x1 , x2 , and x3 , and can be solved for a unique solution by back-substitution. 141 22. Converting the given system of equations to an augmented matrix and using Gauss-Jordan elimination we obtain the following equivalent matrices: 2 1 −1 10 1 1 1 −1 0 1 1 1 −1 0 1 1 1 −1 0 1 2 1 −1 1 0 2 0 −1 −3 3 0 ∼ ∼ 4 2 −1 1 0 4 2 −1 1 0 0 −2 −5 5 0 3 −1 1 k0 3 −1 1 k0 0 −4 −2 k + 3 0 1 1 1 −1 0 1 3 −3 3 ∼ 0 −2 −5 5 0 −4 −2 k + 3 1. P12 0 1 040 ∼ 0 0 0 0 1 1 0 0 2. A12 (−2), A13 (−4), A14 (−3) 1 −1 3 −3 1 −1 10 k − 9 3. M2 (−1) 0 1 050 ∼ 0 0 0 0 1 1 0 0 1 −1 3 −3 1 −1 0 k+1 4. A23 (2), A24 (4) 0 0 . 0 0 5. A34 (−10) (a): Note that the trivial solution (0, 0, 0, 0) exists under all circumstances, so there are no values of k for which there is no solution. (b): From the last row of the last augmented matrix, we see that if k = −1, then the variable x4 corresponds to an unpivoted column, and hence it is a free variable. In this case, therefore, we have infinitely solutions. (c): Provided that k = −1, then each variable in the system corresponds to a pivoted column of the last augmented matrix above. Therefore, we can solve the system by back-substitution. The conclusion from this is that there is a unique solution, (0, 0, 0, 0). 23. Converting the given system of equations to an augmented matrix and using we obtain the following equivalent matrices: 1 1 −2 4 1 1 −2 4 1 1 −2 4 3 2 1 3 5 −4 16 ∼ 0 2 1 2 ∼ 2 4 ∼ 0 1 0 1 4−a b−8 0 1 4−a b−8 2 3 −a b 1. A12 (−3), A13 (−2) 2. M2 ( 1 ) 2 Gauss-Jordan elimination 1 0 0 0 −3 2 . 1 1 2 0 3 − a b − 10 3. A21 (−1), A23 (−1) (a): From the last row of the last augmented matrix above, we see that there is no solution if a = 3 and b = 10. (b): From the last row of the augmented matrix above, we see that there are infinitely many solutions if a = 3 and b = 10, because in that case, there is no pivot in the column of the last augmented matrix corresponding to the third variable x3 . (c): From the last row of the augmented matrix above, we see that if a = 3, then regardless of the value of b, there is a pivot corresponding to each variable x1 , x2 , and x3 . Therefore, we can uniquely solve the corresponding system by back-substitution. 24. Converting the given system of equations we obtain the following equivalent matrices: 1 −a 2 1 −3 a + b to an augmented matrix and using Gauss-Jordan elimination 3 1 −a 3 1 6 ∼ 0 1 + 2a 0 . 1 0 b − 2a 10 142 From the middle row, we see that if a = − 1 , then we must have x2 = 0, but this leads to an inconsistency in 2 solving for x1 (the first equation would require x1 = 3 while the last equation would require x1 = − 1 . Now 3 1 −1/2 3 1 suppose that a = − 2 . Then the augmented matrix on the right reduces to . If b = −1, 0 b + 1 10 then once more we have an inconsistency in the last row. However, if b = −1, then the row-echelon form obtained has full rank, and there is a unique solution. Therefore, we draw the following conclusions: (a): There is no solution to the system if a = − 1 or if a = − 1 and b = −1. 2 2 (b): Under no circumstances are there an infinite number of solutions to the linear system. 1 (c): There is a unique solution if a = − 2 and b = −1. 25. The corresponding augmented matrix for this linear system can be reduced to row-echelon form via 11 1 y1 11 1 y1 1 1 1 y1 2 1 . 2 3 1 y2 ∼ 0 1 −1 y2 − 2y1 ∼ 0 1 −1 y2 − 2y1 00 0 y 1 − 2y 2 + y 3 0 2 −2 y3 − 3y1 3 5 1 y3 1. A12 (−2), A13 (−3) 2. A23 (−2) For consistency, we must have rank(A) = rank(A# ), which requires (y1 , y2 , y3 ) to satisfy y1 − 2y2 + y3 = 0. If this holds, then the system has an infinite number of solutions, because the column of the augmented matrix corresponding to y3 will be unpivoted, indicating that y3 is a free variable in the solution set. 26. Converting the given system of equations to an augmented matrix and using Gaussian elimination we obtain the following row-equivalent matrices. Since a11 = 0: a11 a21 a12 a22 b1 b2 1 ∼ 1 0 a12 a11 a22 a11 −a21 a12 a11 1. M1 (1/a11 ), A12 (−a21 ) b1 a11 a11 b2 −a21 b1 a11 2 ∼ 1 0 a12 a11 ∆ a11 b1 a11 ∆2 a11 . 2. Definition of ∆ and ∆2 (a): If ∆ = 0, then rank(A) = rank(A# ) = 2, so the system has a unique solution (of course, we are assuming ∆2 a11 = 0 here). Using the last augmented matrix above, a∆ x2 = a11 , so that x2 = ∆2 . Using this, we can ∆ 11 solve x1 + a12 a11 x2 = b1 a11 for x1 to obtain x1 = ∆1 ∆, where we have used the fact that ∆1 = a22 b1 − a12 b2 . a12 b1 1 a11 a11 , so it follows that 00 ∆2 the system has (i) no solution if ∆2 = 0, since rank(A) < rank(A# ) = 2, and (ii) an infinite number of solutions if ∆2 = 0, since rank(A# ) < 2. (b): If ∆ = 0 and a11 = 0, then the augmented matrix of the system is (c): An infinite number of solutions would be represented as one line. No solution would be two parallel lines. A unique solution would be the intersection of two distinct lines at one point. 27. We first use the partial pivoting algorithm 1211 3 1 3 5 1 3 ∼ 1 2 2671 to reduce the augmented matrix of the system: 513 3 5 1 3 2 2 1 1 ∼ 0 1/3 2/3 0 671 0 8/3 19/3 −1 143 35 ∼ 0 8/3 0 1/3 3 3 3 3 5 1 1 4 19/3 −1 ∼ 0 8/3 19/3 −1 . 0 0 −1/8 1/8 2/3 0 2. A12 (−1/3), A13 (−2/3) 1. P12 3. P23 4. A23 (−1/8) Using back substitution to solve the equivalent system yields the unique solution (−2, 2, −1). 28. We first use the partial pivoting algorithm to reduce the augmented matrix of the system: 7 2 −3 3 2 −1 3 14 7 2 −3 3 3 1/7 −5/7 −16/7 1 −2 −1 2 0 1 −2 −1 1 3 ∼ ∼ 7 92/7 3 2 −1 3 14 0 −11/7 27/7 2 −3 0 −17/7 1/7 20/7 5 −1 −2 5 5 −1 −2 5 7 2 −3 3 7 2 −3 3 1/7 20/7 1/7 20/7 4 0 −17/7 3 0 −17/7 ∼ ∼ 192/17 92/7 0 0 64/17 0 −11/7 27/7 0 0 −12/17 −36/17 0 1/7 −5/7 −16/7 7 2 −3 3 1/7 20/7 5 0 −17/7 . ∼ 0 0 64/17 192/17 0 0 0 0 1. P13 2. A12 (−3/7), A13 (−2/7), A14 (−5/7) 4. A23 (−11/17), A24 (1/17) 3. P24 5. A34 (3/16) Using back substitution to solve the equivalent system yields the unique solution (2, −1, 3). 29. We first use the partial pivoting algorithm to 2 −1 −4 5 5 6 3 2 −5 813 2 ∼ 5 6 −6 20 2 −1 1 1 −3 −3 1 1 reduce the augmented matrix of the −6 −20 5 6 −6 −5 8 2 0 −8/5 −7/5 ∼ −4 5 0 −17/5 −8/5 −3 −3 0 −1/5 −9/5 5 6 −6 20 5 3 0 −17/5 −8/5 −3 4 0 ∼ 0 −8/5 −7/5 −4 ∼ 0 0 −1/5 −9/5 −7 0 5 6 −6 20 0 −17/5 −8/5 6 −3 5 ∼ ∼ 0 0 −29/17 −116/17 0 0 −11/17 −44/17 1. P13 6 −6 20 −17/5 −8/5 −3 0 −11/17 −44/17 0 −29/17 −116/17 5 6 −6 20 0 −17/5 −8/5 −3 . 0 0 −29/17 −116/17 0 0 0 0 2. A12 (−3/5), A13 (−2/5), A14 (−1/5) 4. A23 (−8/17), A24 (−1/17) system: 20 −4 −3 −7 5. P34 3. P23 6. A34 (−11/29) 144 Using back substitution to solve the equivalent system yields the unique solution (10, −1, 4). 30. We first use the partial pivoting algorithm to reduce the augmented matrix of the system: 4 3 −2 −1 2 −1 −1 4 3 −2 −1 2 1 2 4 0 5/2 2 ∼ 0 −5/2 3 −2 −1 ∼ 2 −1 −1 4 0 13/4 3/2 17/4 4 1 4 1 1 4 1 −1 4 3 −2 −1 4 3 −2 4 3 17/4 . 3/2 ∼ 0 13/4 3/2 17/4 ∼ 0 13/4 0 0 15/13 75/13 0 −5/2 0 5/2 1. P12 2. A12 (−1/2), A13 (−1/4) 3. P23 4. A23 (10/13) Using back substitution to solve the equivalent system yields the unique solution (3, −1, 5). 31. (a): Let A = # a11 a21 a31 ... an1 0 a22 a32 ... an2 0 0 a33 ... an3 ... ... ... ... ... 0 0 0 ... ann b1 b2 b3 ... bn represent the corresponding augmented matrix of the given system. Since a11 x1 = b1 , we can solve for x1 easily: b1 x1 = , (a11 = 0). a11 Now since a21 x1 + a22 x2 = b2 , by using the expression for x1 we just obtained, we can solve for x2 : x2 = a11 b2 − a21 b1 . a11 a22 In a similar manner, we can solve for x3 , x4 , . . . , xn . (b): We solve instantly for x1 from the first equation: x1 = 2. Substituting this into the middle equation, we obtain 2 · 2 − 3 · x2 = 1, from which it quickly follows that x2 = 1. Substituting for x1 and x2 in the bottom equation yields 3 · 2 + 1 − x3 = 8, from which it quickly follows that x3 = −1. Consequently, the solution of the given system is (2, 1, −1). 32. This system of equations is not linear in x1 , x2 , and x3 ; however, the system is linear in x3 , x2 , and x3 , 1 2 so we can first solve for x3 , x2 , and x3 . Converting the given system of equations to an augmented matrix 1 2 and using Gauss-Jordan elimination we obtain the following equivalent matrices: 4 2 3 12 1 −1 12 1 −1 1 2 1 2 1 −1 1 2 ∼ 4 2 3 12 ∼ 0 6 −1 4 3 1 −1 2 3 1 −1 2 0 4 −4 −4 1 −1 1 2 1 −1 1 2 10 0 1 3 4 5 4 −4 −4 ∼ 0 1 −1 −1 ∼ 0 1 −1 −1 ∼ 0 0 6 −1 4 0 6 −1 4 00 5 10 10 0 1 1001 6 7 ∼ 0 1 −1 −1 ∼ 0 1 0 1 . 00 1 2 0012 145 2. A12 (−4), A13 (−3) 1. P12 5. A21 (1), A23 (−6) 3. P23 6. M2 (1/5) 4. M2 (1/4) 7. A32 (1) Thus, taking only real solutions, we have x3 = 1, x2 = 1, and x3 = 2. Therefore, x1 = 1, x2 = ±1, and 1 2 x3 = 2, leading to the two solutions (1, 1, 2) and (1, −1, 2) to the original system of equations. There is no contradiction of Theorem 2.5.9 here since, as mentioned above, this system is not linear in x1 , x2 , and x3 . 33. Reduce the augmented matrix of the system: 10 30 1 1 −2 0 1 1 −2 0 3 2 −1 0 3 2 1 2 1 −5 0 ∼ 0 1 −5 0 5 0 ∼ 0 1 1 0 ∼ 0 −1 0 0 −34 0 0 −9 11 0 0 −9 11 0 5 −4 10 10 10 30 5 4 ∼ 0 1 −5 0 ∼ 0 1 00 00 10 1. A21 (−1), A12 (−2), A13 (−5) 4. M3 (−1/34) 00 0 0 . 10 2. M2 (−1) 3. A21 (−1), A23 (9) 5. A31 (−3), A32 (5) Therefore, the unique solution to this system is x1 = x2 = x3 = 0: (0, 0, 0). 34. Reduce the augmented matrix of the system: 1 −1 −1 0 2 1 −1 0 3 −1 20 2 0 1 3 −1 1 −1 −1 0 ∼ 2 1 −1 0 5 2 −2 0 5 2 −2 0 1 −1 −1 1 −4 40 ∼ 0 2 5 0 7 3 1. P13 1 0 050 ∼ 0 0 0 0 0 −5 1 −4 0 13 0 31 1 −1 −1 20 2 5 ∼ 0 3 1 0 7 3 1 0 −5 0 0 0 6 0 1 −4 0 ∼ 10 0 0 0 0 0 0 31 0 2. A12 (−3), A13 (−2), A14 (−5) 5. A21 (1), A23 (−2), A24 (−7) 0 03 ∼ 0 0 6. M3 (1/13) 1 −1 −1 0 3 1 0 2 5 0 7 3 10 70 1 ∼ 0 0 00 3. P23 0 0 1 0 4. A32 (−1) 7. A31 (5), A32 (4), A34 (−31) Therefore, the unique solution to this system is x1 = x2 = x3 = 0: (0, 0, 0). 35. Reduce the augmented matrix 2 −1 −1 5 −1 2 1 1 4 of the system: 0 1 1 40 1 1 40 1 2 0 ∼ 5 −1 2 0 ∼ 0 −6 −18 0 0 2 −1 −1 0 0 −3 −9 0 1 1 40 10 4 1 3 0 ∼ 0 1 ∼ 0 0 −3 −9 0 00 3 1. P13 2. A12 (−5), A13 (−2) 3. M2 (−1/6) 10 3 0 . 00 4. A21 (−1), A23 (3) 0 0 0 0 0 0 . 0 0 146 It follows that x1 + x3 = 0 and x2 + 3x3 = 0. Setting x3 = t, where t is a free variable, we get x2 = −3t and x1 = −t. Thus we have that the solution set of the system is {(−t, −3t, t) : t ∈ R}. 36. Reduce the augmented matrix of the system: 0 0 0 1 + 2i 1 − i 1 i 1+i −i 1 1−i −1 1 2 i 1+i −i 0 ∼ 1 + 2i 1 − i 1 0 ∼ 1 + 2i 1 − i 1 0 2i 1 1 + 3i 0 2i 1 1 + 3i 0 2i 1 1 + 3i 0 1 1−i −1 0 4 ∼ 0 −2 − 2i 1 + 2i 0 ∼ 0 −1 − 2i 1 + 5i 0 1 1−i −1 0 6 0 1 3i ∼ 0 0 0 −5 + 8i 0 3 2. M1 (−i) 1. P12 6. P23 1 1−i −1 0 −2 − 2i 1 + 2i 0 1 3i 1 1 − i −1 7 ∼ 0 1 3i 0 0 1 3. A12 (−1 − 2i), A13 (−2i) 7. 1 M3 ( −5+8i ) 0 1 1−i −1 0 5 0 ∼ 0 0 −5 + 8i 0 0 0 1 3i 0 1000 0 8 0 ∼ 0 1 0 0 . 0 0010 4. A23 (−1) 5. A32 (2 + 2i) 8. A21 (−1 + i), A31 (1), A32 (−3i) Therefore, the unique solution to this system is x1 = x2 = x3 = 0: (0, 0, 0). 37. Reduce the augmented matrix of the system: 2 3 210 1 3 1 6 −1 2 0 ∼ 6 −1 12 640 12 6 2 1 3 3 1 ∼ 0 0 −2 1. M1 (1/3) 1 3 0 0 1 3 2 4 2 1 0 3 2 0 ∼ 0 −5 0 −2 0 10 0 4 0 ∼ 0 1 00 0 2. A12 (−6), A13 (−12) 1 3 0 0 3. M2 (−1/5) 1 3 0 0 0 0 0 0 0 . 0 4. A21 (−2/3), A23 (2) From the last augmented matrix, we have x1 + 1 x3 = 0 and x2 = 0. Since x3 is a free variable, we let x3 = t, 3 where t is a real number. It follows that the solution set for the given system is given by {(t, 0, −3t) : t ∈ R}. 38. Reduce the augmented matrix 2 1 −8 3 −2 −5 5 −6 −3 3 −5 1 of the system: 0 3 −2 −5 012 1 −8 ∼ 0 5 −6 −3 0 3 −5 1 1 −3 3 7 −14 30 ∼ 0 9 −18 0 4 −8 1. P12 2. A21 (−1) 0 1 −3 3 022 1 −8 ∼ 0 5 −6 −3 0 3 −5 1 0 1 −3 3 040 1 −2 ∼ 0 0 9 −18 0 0 4 −8 3. A12 (−2), A13 (−5), A14 (−3) 0 0 0 0 0 1 0 −3 0 0 5 0 1 −2 0 ∼ 0 0 0 00 0 00 00 4. M2 (1/7) . 5. A21 (3), A23 (−9), A24 (−4) 147 From the last augmented matrix we have: x1 − 3x3 = 0 and x2 − 2x3 = 0. Since x3 is a free variable, we let x3 = t, where t is a real number. It follows that x2 = 2t and x1 = 3t. Thus, the solution set for the given system is given by {(3t, 2t, t) : t ∈ R}. 39. Reduce the augmented matrix of the system: 1 1+i 1−i 0 1 1 i 1 i 0 ∼ 0 1 − 2i −1 + i 1 − 3i 0 0 1 1+i 1−i 3 −2−i 1 ∼ 0 5 0 0 0 1. A12 (−i), A13 (−1 + 2i) 1+i 2−i −4 + 2i 0 4 0 ∼ 0 2. A23 (2) 1−i 0 1 1+i 1−i 0 2 −1 0 ∼ 0 2 − i −1 0 2 0 0 0 0 0 1 0 6−2i 0 5 0 1 −2−i 0 . 5 0 00 0 1 3. M2 ( 2−i ) 4. A21 (−1 − i) From the last augmented matrix we see that x3 is a free variable. We set x3 = 5s, where s ∈ C. Then x1 = 2(i − 3)s and x2 = (2 + i)s. Thus, the solution set of the system is {(2(i − 3)s, (2 + i)s, s) : s ∈ C}. 40. Reduce the augmented matrix of the system: 1 −1 10 1 −1 10 0 3 20 3 2 010 ∼ 3 3 −4 0 0 −1 0 0 0 6 −6 0 5 1 −1 0 1 0 5/3 1 0 5/3 0 2/3 0 4 0 1 2/3 30 1 ∼ 0 0 −6 0 ∼ 0 0 1 0 0 −10 0 0 −10 0 1. A13 (−3), A14 (−5) 4. M3 (−1/6) 2. M2 (1/3) 1 20 ∼ 0 0 0 05 ∼ 0 0 −1 10 1 2/3 0 3 −4 0 6 −6 0 1000 0 1 0 0 . 0 0 1 0 0000 3. A21 (1), A23 (−3), A24 (−6) 5. A31 (−5/3), A32 (−2/3), A34 (10) Therefore, the unique solution to this system is x1 = x2 = x3 = 0: (0, 0, 0). 41. Reduce the augmented matrix of the system: 2 −4 60 1 −2 3 3 −6 9 0 1 3 −6 9 1 −2 3 0 ∼ 2 −4 6 5 −10 15 0 5 −10 15 1. M1 (1/2) 0 1 −2 3 0 020 0 0 0 ∼ . 0 0 0 0 0 0 0 000 2. A12 (−3), A13 (−2), A14 (−5) From the last matrix we have that x1 − 2x3 + 3x3 = 0. Since x2 and x3 are free variables, let x2 = s and let x3 = t, where s and t are real numbers. The solution set of the given system is therefore {(2s − 3t, s, t) : s, t ∈ R}. 42. Reduce the augmented matrix of the system: 4 −2 −1 −1 0 1 −3 1 −4 0 1 −3 1 −4 0 1 2 3 1 −2 3 0 ∼ 3 1 −2 3 0 ∼ 0 10 −5 15 0 5 −1 −2 10 5 −1 −2 10 0 14 −7 21 0 148 −4 0 3/2 0 . 0 0 1 −3 1 1 −3 1 −4 0 1 −3 1 −4 0 4 5 1 −1/2 2 −1 3 0 ∼ 0 2 −1 3 0 ∼ 0 ∼ 0 0 0 0 0 0 0 00 0 2 −1 30 3 1. A21 (−1) 2. A12 (−3), A13 (−5) 4. A23 (−1) 3. M2 (1/5), M3 (1/7) 5. M2 (1/2) From the last augmented matrix above we have that x2 − 1 x3 + 3 x4 = 0 and x1 − 3x2 + x3 − 4x4 = 0. Since x3 2 2 and x4 are free variables, we can set x3 = 2s and x4 = 2t, where s and t are real numbers. Then x2 = s − 3t and x1 = s − t. It follows that the solution set of the given system is {(s − t, s − 3t, 2s, 2t) : s, t ∈ R}. 43. Reduce the augmented matrix of the system: 1 1 2 1 −1 10 1 1 1 1 −1 0 1 2 ∼ 3 −1 1 −2 0 3 −1 4 2 4 2 −1 10 10 1 1 1 −1 0 1 3 −3 0 4 0 1 30 ∼ ∼ 0 −4 −2 1 0 0 0 00 0 −2 −5 50 100 0 1 0 −2 20 0 3 −3 0 7 0 1 0 60 1 ∼ ∼ 0 0 1 −1 0 0 0 1 −1 0 0 0 −1 0 0 10 −11 0 1 1 1 −1 1 −1 0 3 −1 1 0 2 0 −1 −3 ∼ 1 1 −2 0 0 −4 −2 0 −2 −5 5 −1 10 1 0 −2 2 −2 20 3 −3 3 −3 0 5 0 1 ∼ 3 10 −11 0 0 0 −3 0 0 10 −11 −3 30 10 100 00 0 0 090 1 080 1 0 ∼ ∼ 0 0 0 1 −1 0 0 0 00 000 10 0 1. P12 2. A12 (−2), A13 (−3), A14 (−4) 5. P34 6. M3 (−1/3) 3. M2 (−1) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 . 0 0 4. A21 (−1), A23 (4), A24 (2) 7. A31 (2), A32 (−3), A34 (−10) 8. M4 (−1) 9. A43 (1) From the last augmented matrix, it follows that the solution set to the system is given by {(0, 0, 0, 0)}. 44. The equation Ax = 0 is 2 −1 3 4 x1 x2 0 0 = . Reduce the augmented matrix of the system: 2 −1 0 3 40 1 ∼ 1 −1 2 3 4 1. M1 (1/2) 0 0 2 ∼ 1 −1 2 0 11 2 2. A12 (−3) 0 0 1 −1 2 0 1 3 ∼ 3. M2 (2/11) 0 0 4 ∼ 100 010 4. A21 (1/2) From the last augmented matrix, we see that x1 = x2 = 0. Hence, the solution set is {(0, 0)}. 45. The equation Ax = 0 is 1 − i 2i 1 + i −2 x1 x2 = 0 0 1 −1 + i 0 1+i −2 0 ∼ . Reduce the augmented matrix of the system: 1 − i 2i 0 1 + i −2 0 1 ∼ 2 1 −1 + i 0 0 0 0 . . 149 1. M1 ( 1+i ) 2 2. A12 (−1 − i) It follows that x1 + (−1 + i)x2 = 0. Since x2 is a free variable, we can let x2 = t, where t is a real number. The solution set to the system is then given by {(t(1 − i), t) : t ∈ R}. 46. The equation Ax = 0 is 1 + i 1 − 2i −1 + i 2 + i x1 x2 = 0 0 . Reduce the augmented matrix of the system: 1 + i 1 − 2i 0 −1 + i 2 + i 0 1 ∼ 1 − 1+3i 2 −1 + i 2 + i 1. M1 ( 1−i ) 2 0 0 1 − 1+3i 2 0 0 2 ∼ 0 0 . 2. A12 (1 − i) It follows that x1 − 1+3i x2 = 0. Since x2 is a free variable, we can let x2 = r, where r is any complex 2 number. Thus, the solution set to the given system is {( 1+3i r, r) : r ∈ C}. 2 47. The equation Ax = 0 is 1 23 x1 0 2 −1 0 x2 = 0 . 1 11 x3 0 Reduce the augmented 1 23 2 −1 0 1 11 matrix of 0 1 0 ∼ 0 10 4 ∼ 0 1 00 1. A12 (−2), A13 (−1) the system: 1 1 2 30 2 0 −5 −6 0 ∼ 0 0 0 −1 −2 0 1 0 −1 −1 0 5 2 2 0 ∼ 0 1 00 1 40 2. P23 3. M2 (−1) 2 3 −1 −2 −5 −6 0 6 0 ∼ 0 1 2 30 0 3 1 2 0 0 ∼ 0 0 −5 −6 0 0 1000 0 1 0 0 . 0010 4. A21 (−2), A23 (5) 5. M3 (1/4) 6. A31 (1), A32 (−2) From the last augmented matrix, we see that the only solution to the given system is x1 = x2 = x3 = 0: {(0, 0, 0)}. 48. The equation Ax = 0 is x 11 1 −1 1 x −1 0 −1 2 2 x3 13 2 2 x4 Reduce the augmented 11 1 −1 −1 0 −1 2 13 2 2 matrix of the 0 1 1 0 ∼ 0 0 0 0 0 = . 0 0 system: 1 1 2 1 −1 0 10 2 0 1 0 ∼ 0 1 1 30 00 1. A12 (1), A13 (−1) 1 −2 0 10 3 0 1 0 ∼ 0 1 1 10 00 2. A21 (−1), A23 (−2) 3. A31 (−1) 0 −3 0 0 1 0 . 1 10 150 From the last augmented matrix, we see that x4 is a free variable. We set x4 = t, where t is a real number. The last row of the reduced row echelon form above corresponds to the equation x3 + x4 = 0. Therefore, x3 = −t. The second row corresponds to the equation x2 + x4 = 0, so we likewise find that x2 = −t. Finally, from the first equation we have x1 − 3x4 = 0, so that x1 = 3t. Consequently, the solution set of the original system is given by {(3t, −t, −t, t) : t ∈ R}. 49. The equation Ax = 0 is 2 − 3i 1 + i i−1 x1 0 3 + 2i −1 + i −1 − i x2 = 0 . 5−i 2i −2 x3 0 Reduce the augmented matrix of this system: −5−i −1+5i 2 − 3i 1 + i i−1 0 1 1 0 13 13 2 1 3 + 2i −1 + i −1 − i 0 ∼ 3 + 2i −1 + i −1 − i 0 ∼ 0 5−i 2i −2 0 0 0 5−i 2i −2 1. M1 ( 2+3i ) 13 −1+5i 13 −5−i 13 0 0 0 0 0 0 . 0 2. A12 (−3 − 2i), A13 (−5 + i) 5− From the last augmented matrix, we see that x1 + −1+5i x2 + −13 i x3 = 0. Since x2 and x3 are free variables, 13 we can let x2 = 13r and x3 = 13s, where r and s are complex numbers. It follows that the solution set of the system is {(r(1 − 5i) + s(5 + i), r, s) : r, s ∈ C}. 50. The equation Ax = 0 is 1 30 x1 0 −2 −3 0 x2 = 0 . 1 40 x3 0 Reduce the augmented matrix of the system: 1 1300 1 300 2 1 −2 −3 0 0 ∼ 0 3 0 0 ∼ 0 0 0100 1 400 1. A12 (2), A13 (−1) 2. P23 3 1 3 1 00 3 0 0 ∼ 0 0 00 0 1 0 00 0 0 . 00 3. A21 (−3), A23 (−3) From the last augmented matrix we see that the solution set of the system is {(0, 0, t) : t ∈ R}. 51. The equation Ax = 0 is 1 0 3 3 −1 7 2 1 8 1 1 5 −1 1 −1 Reduce the augmented matrix of the system: 1 0 3 1 0 30 3 −1 7 0 0 −1 −2 1 2 1 8 0 ∼ 0 1 2 1 1 5 0 0 1 2 −1 1 −1 0 0 1 2 x1 0 x2 = 0 . x3 0 0 0 0 0 0 2 ∼ 1 0 0 0 0 0 1 1 1 1 3 2 2 2 2 0 0 0 0 0 3 ∼ 1 0 0 0 0 0 1 0 0 0 3 2 0 0 0 0 0 0 0 0 . 151 1. A12 (−3), A13 (−2), A14 (−1), A15 (1) 2. M2 (−1) 3. A23 (−1), A24 (−1), A25 (−1) From the last augmented matrix, we obtain the equations x1 + 3x3 = 0 and x2 + 2x3 = 0. Since x3 is a free variable, we let x3 = t, where t is a real number. The solution set for the given system is then given by {(−3t, −2t, t) : t ∈ R}. 52. The equation Ax = 0 is x 1 −1 0 1 1 3 −2 0 5 x2 x3 −1 201 x4 Reduce the augmented matrix of 1 −1 0 3 −2 0 −1 20 0 = 0 . 0 the system: 1 −1 0 1 0 10 10 1 2 5 0 ∼ 0 1 0 2 0 ∼ 0 1 10 0 1020 00 1. A12 (−3), A13 (1) 030 0 2 0 . 000 2. A21 (1), A23 (−1) From the last augmented matrix we obtain the equations x1 + 3x4 = 0 and x2 + 2x4 = 0. Because x3 and x4 are free, we let x3 = t and x4 = s, where s and t are real numbers. It follows that the solution set of the system is {(−3s, −2s, t, s) : s, t ∈ R}. 53. The equation Ax = 0 is 1 3 −2 x1 0 −3 0 0 x2 = 0 . 0 −9 0 x3 0 60 0 x4 Reduce the augmented matrix of the system: 1 0 −3 0 0 1 0 −3 0 0 1 3 0 −9 0 0 ∼ 0 0 0 0 0 . 00 000 −2 0 600 1. A12 (−3), A13 (2) From the last augmented matrix we obtain x1 − 3x3 = 0. Therefore, x2 , x3 , and x4 are free variables, so we let x2 = r, x3 = s, and x4 = t, where r, s, t are real numbers. The solution set of the given system is therefore {(3s, r, s, t) : r, s, t ∈ R}. 54. The equation Ax = 0 is 2+i i 3 − 2i x1 0 i 1 − i 4 + 3i x2 = 0 . 3 − i 1 + i 1 + 5i x3 0 Reduce the augmented matrix 2+i i 3 − 2i i 1 − i 4 + 3i 3 − i 1 + i 1 + 5i of the system: 0 i 1 − i 4 + 3i 0 1 −1 − i 3 − 3i 0 1 2 0 ∼ 2+i i 3 − 2i 0 ∼ 2 + i i 3 − 2i 0 0 3 − i 1 + i 1 + 5i 0 3 − i 1 + i 1 + 5i 0 152 1 −1 − i 3 − 4i 1 −1 − i 3 − 4i 0 4 5+31i 1 ∼ 0 1 + 4i −7 + 3i 0 ∼ 0 17 0 5 + 3i −4 + 20i 0 0 5 + 3i −4 + 20i 1 0 25−32i 0 100 17 6 7 ∼ 0 1 5+31i 0 ∼ 0 1 0 17 001 00 1 0 3 1. P12 2. M1 (−i) 3. A12 (−2 − i), A13 (−3 + i) 6. M3 (−i/10) 7. 4. M2 ( 1−4i ) 17 A31 ( −25+32i ), 17 25−32i 17 5+31i 17 0 10 5 0 ∼ 0 1 0 00 0 0 . 0 10i 0 0 0 5. A21 (1 + i), A23 (−5 − 3i) − A32 ( −51731i ) From the last augmented matrix above, we see that the only solution to this system is the trivial solution. Solutions to Section 2.6 True-False Review: 1. FALSE. An invertible matrix is also known as a nonsingular matrix. 1 2 2. FALSE. For instance, the matrix 1 2 does not contain a row of zeros, but fails to be invertible. 3. TRUE. If A is invertible, then the unique solution to Ax = b is x = A−1 b. 10 100 4. FALSE. For instance, if A = and B = 0 0 , then AB = I2 , but A is not even a square 001 01 matrix, hence certainly not invertible. 5. FALSE. For instance, if A = In and B = −In , then A and B are both invertible, but A + B = 0n is not invertible. 6. TRUE. We have (AB )B −1 A−1 = In and B −1 A−1 (AB ) = In , and therefore, AB is invertible, with inverse B −1 A−1 . 7. TRUE. From A2 = A, we subtract to obtain A(A − I ) = 0. Left multiplying both sides of this equation by A−1 (since A is invertible, A−1 exists), we have A − I = A−1 0 = 0. Therefore, A = I , the identity matrix. 8. TRUE. From AB = AC , we left-multiply both sides by A−1 (since A is invertible, A−1 exists) to obtain A−1 AB = A−1 AC . Since A−1 A = I , we obtain IB = IC , or B = C . 9. TRUE. Any 5 × 5 invertible matrix must have rank 5, not rank 4 (Theorem 2.6.5). 10. TRUE. Any 6 × 6 matrix of rank 6 is invertible (Theorem 2.6.5). Problems: 1. We have AA−1 = 2 −1 3 −1 −1 −3 1 2 = (2)(−1) + (−1)(−3) (3)(−1) + (−1)(−3) (2)(1) + (−1)(2) (3)(1) + (−1)(2) 1 0 = 0 1 = I2 . 2. We have AA−1 = 49 37 7 −9 −3 4 = (4)(7) + (9)(−3) (3)(7) + (7)(−3) (4)(−9) + (9)(4) (3)(−9) + (7)(4) = 1 0 0 1 = I2 . 153 3. We have AA−1 351 8 −29 3 19 −2 = 1 2 1 −5 267 2 −8 1 (3)(8) + (5)(−5) + (1)(2) (3)(−29) + (5)(19) + (1)(−8) (3)(3) + (5)(−2) + (1)(1) = (1)(8) + (2)(−5) + (1)(2) (1)(−29) + (2)(19) + (1)(−8) (1)(3) + (2)(−2) + (1)(1) (2)(8) + (6)(−5) + (7)(2) (2)(−29) + (6)(19) + (7)(−8) (2)(3) + (6)(−2) + (7)(1) 100 = 0 1 0 = I3 . 001 4. We have [A|I2 ] = 121 130 0 1 12 10 0 1 −1 1 1 ∼ 1 0 2 ∼ 0 3 −2 1 −1 1 = [I2 |A−1 ]. Therefore, 3 −2 −1 1 A−1 = 1. A12 (−1) . 2. A21 (−2) 5. We have [A|I2 ] = 1 1+i 1 1−i 1 0 0 1 3 ∼ 1 0 1 1+i 0 −1 −1 + i 1 1 ∼ 1 0 −1 1 + i 0 1 1 − i −1 2 ∼ 1 0 1 0 1+i 1 1 − i −1 = [I2 |A−1 ]. Thus, −1 1 + i 1 − i −1 A−1 = 1. A12 (−1 + i) 2. M2 (−1) . 3. A21 (−1 − i) 6. We have [A|I2 ] = 1 −i 1 i−1 20 3 ∼ 0 1 1 ∼ −i 1 0 1−i 1−i 1 1 0 1 0 1+i 1 01 −1+i 2 1+i 2 2 ∼ 1 −i 1 0 11 = [I2 |A−1 ]. Thus, A−1 = 1. A12 (1 − i) 1+i 1 −1+i 2 1+i 2 2. M2 (1/(1 − i)) . 3. A21 (i) 0 1+i 2 154 7. Note that AB = 02 for all 2 × 2 matrices B . Therefore, A is not invertible. 8. We have 1 −1 2 1 11 [A|I3 ] = 2 4 −3 10 104 3 ∼ 0 1 2 001 1 0 0 0 1 −1 1 0 ∼ 0 3 1 0 1 −3 0 1 1 4 −4 0 1 ∼ 0 10 1 −3 0 0 1 0 00 10 1 −1 2 2 1 0 ∼ 0 1 2 −4 0 01 0 3 7 −2 1 0 0 −43 −4 13 1 0 −24 −2 7 = [I3 |A−1 ]. 01 10 1 −3 1 2 7 −2 2 −4 0 1 0 Thus, A−1 1. A12 (−2), A13 (−4) −43 −4 13 7 . = −24 −2 10 1 −3 3. A21 (1), A23 (−3) 2. P23 4. A31 (−4), A32 (−2) 9. We have 35110 [A|I3 ] = 1 2 1 0 1 26700 121 0 10 3 4 3 0 ∼ ∼ 0 1 2 −1 025 0 −2 1 1 2 10 10 10 2 0 0 ∼ 0 −1 −2 1 −3 0 0 2 5 0 −2 1 01 100 8 −29 3 1 0 −3 2 −5 0 5 19 −2 = [I3 |A−1 ]. 01 2 −1 3 0 ∼ 0 1 0 −5 001 2 −8 1 00 1 2 −8 1 1 0 1 0 ∼ 3 2 1 2 5 6 10 11 70 Thus, A−1 1. P12 2. A12 (−3), A13 (−2) 8 −29 3 19 −2 . = −5 2 −8 1 3. M2 (−1) 4. A21 (−2), A23 (−2) 5. A31 (3), A32 (−2) 10. This matrix is not invertible, because the column of zeros guarantees that the rank of the matrix is less than three. 11. We have 42 [A|I3 ] = 2 1 32 1 1 11 3 ∼ 0 −1 −29 0 −2 −57 0 32 4 1 0 ∼ 2 1 −7 1 4 2 −13 0 −1 1 1 1 11 0 4 0 3 −2 ∼ 0 1 29 0 1 4 −4 0 −2 −57 1 1 0 18 −34 −1 6 ∼ 0 1 0 −29 55 00 1 1 −2 −13 1 −7 0 40 0 1 0 0 0 1 1 1 2 0 ∼ 2 0 4 −1 1 1 5 −3 2 ∼ 0 4 −4 0 0 1 0 2 = [I3 |A−1 ]. 0 1 11 0 −1 1 1 −7 0 1 0 2 −13 1 00 0 −18 0 2 −1 1 29 0 −3 2 0 1 1 −2 0 155 Thus, 18 −34 −1 55 2 . = −29 1 −2 0 A−1 2. A21 (−1) 1. P13 3. A12 (−2), A13 (−4) 5. A21 (−1), A23 (2) 4. M2 (−1) 6. A31 (18), A32 (−29) 12. We have 1 2 −3 [A|I3 ] = 2 6 −2 −1 1 4 1 3 ∼ 0 0 1 2 −3 0 1 4 0 ∼ 0 2 03 1 1 0 −7 3 −1 0 4 1 0 ∼ 1 2 −1 2 0 −5 4 −3 1 2 11 1 0 0 − 13 5 10 5 1 3 − 10 ∼ 0 1 0 5 4 3 0 0 1 −5 10 1 0 0 0 1 0 1 1 2 −3 100 2 −2 1 0 ∼ 0 1 2 −1 101 03 1 1 1 0 −7 3 −1 0 1 0 01 2 −1 2 4 3 1 00 1 − 5 10 − 5 7 −5 2 = [I3 |A−1 ]. 5 1 −5 0 1 2 0 0 0 1 Thus, A−1 = 2. M2 ( 1 ) 2 1. A12 (−2), A13 (1) 13. We have 1 i 2 [A|I3 ] = 1 + i −1 2i 2 2i 5 10 3 ∼ 0 1 00 − 13 5 3 5 −4 5 11 10 1 − 10 3 10 2 5 −1 5 . 4. M3 (− 1 ) 5 3. A21 (−2), A23 (−3) 0 1 i 2 1 0 ∼ 0 −i −2 0 0 1 1 0 −i 1 0 1 4 −2i 1 − i i 0 ∼ 0 1 −2 0 1 0 1 0 0 −7 5 0 1 0 5. A31 (7), A32 (−2) 1 00 1i 2 1 00 2 −1 − i 1 0 ∼ 0 1 −2i 1 − i i 0 00 1 −2 0 1 −2 01 00 −i 10 1 0 1 − 5i i 2i = [I3 |A−1 ]. 01 −2 01 Thus, A−1 −i 10 = 1 − 5i i 2i . −2 01 1. A12 (−1 − i), A13 (−2) 2. M2 (i) 3. A21 (−i) 4. A32 (2i) 14. We have 2 131 [A|I3 ] = 1 −1 2 0 3 340 0 1 0 0 1 −1 2 0 1 0 ∼ 2 131 1 3 340 1 0 0 0 1 −1 20 10 2 0 ∼ 0 3 −1 1 −2 0 1 0 6 −2 0 −3 1 156 0 10 1 −1 2 1 −2 0 3 −1 ∼ 0 0 0 0 −2 11 3 Since 2 = rank(A) < rank(A# ) = 3, we know that A−1 does not exist (we have obtained a row of zeros in the block matrix on the left. 2. A12 (−2), A13 (−3) 1. P12 3. A23 (−2) 15. We have 1 −1 2 31 2 0 3 −4 0 [A|I4 ] = 3 −1 7 80 1 03 50 0 1 0 0 0 0 1 0 1 −1 2 3 1 0 2 −1 −10 −2 010 ∼ 2 1 −1 −3 0 0 0 1 1 2 −1 1 1 1 −1 2 3 1000 1 1 2 −1 0 0 1 3 0 20 ∼ ∼ 0 2 1 −1 −3 0 1 0 0 0 0 2 −1 −10 −2 1 0 0 1 10 3 5 00 0 1 0 1 1 2 −1 0 0 150 4 ∼ ∼ 0 0 1 5 1 0 −1 2 0 0 0 0 −3 −14 01 0 −2 27 10 −27 1000 7 3 −8 60 1 0 0 ∼ 0 0 1 0 −14 −5 14 0001 3 1 −3 0 3 5 0 1 1 2 −1 0 −1 −5 −1 0 −3 −14 0 0 0 0 1 0 0 −10 −3 0 1 0 −3 −2 0 01 5 10 00 1 31 35 11 = [I4 |A−1 ]. −18 4 000 1 0 0 0 1 0 001 0 1 0 1 1 −2 0 −2 3 −5 1 −1 −1 2 −3 4 Thus, A−1 27 10 −27 35 7 3 −8 11 . = −14 −5 14 −18 3 1 −3 4 1. A12 (−2), A13 (−3), A14 (−1) 4. M3 (−1) 2. P13 5. A31 (−3), A32 (−1), A34 (3) 3. A21 (1), A23 (−2), A24 (−2) 6. A41 (10), A42 (3), A43 (5) 16. We have 0 −2 −1 −3 2 0 2 1 [A|I4 ] = 1 −2 0 2 3 −1 −2 0 1 −2 0 2 4 2 −3 20 ∼ 0 −2 −1 −3 0 5 −2 −6 0 0 1 0 0 1 −2 0 2 012 0 2 1 ∼ 0 0 −2 −1 −3 1 3 −1 −2 0 0 0 1 0 0 1 0 0 1 −2 0 20 0 10 1 1 −2 0 3 0 −3 0 1 2 4 ∼ 0 0 0 0 −2 −1 −3 1 0 −3 1 0 5 −2 −6 0 0 0 −1 0 2 0 0 −3 1 1 0 0 0 0 1 0 0 0 0 1 0 1 4 0 0 1 0 0 0 1 0 0 0 1 157 1 10 1 2 3 1 0 1 −4 4 2 ∼ 9 0 0 0 −2 9 0 0 −2 −9 4 1 101 2 1 −3 60 1 2 4 ∼ 1 001 2 0 0 0 −9 2 1 0 2 1 0 4 1 1 2 0 −5 4 1 2 1 4 5 18 1 2 0 0 0 1 2 0 100 0 9 1 0 1 0 −1 0 8 9 ∼ 1 5 001 0 18 2 000 1 −2 −1 9 9 −1 9 −5 9 1 9 2 9 1 1 0 0 00 10 1 2 2 1 1 −1 0 5 0 1 −3 0 −1 0 2 2 4 4 2 ∼ 0 0 −9 −9 0 −5 −1 1 −1 0 2 4 4 2 1 −1 1 00 0 −9 1 −1 0 2 2 2 2 1 2 0 0 100 0 0 9 −9 9 1 −1 0 7 0 1 0 −1 0 1 − 5 2 9 9 9 1 5 1 2 ∼ 1 −9 001 0 18 −2 9 2 9 9 −1 0 0 0 0 − 9 1 1 −1 0 2 2 2 2 2 1000 0 −1 9 9 9 9 1 1 1 0 1 0 0 −2 0 −3 9 9 ∼ 9 9 = [I |A−1 ]. 4 2 1 1 0 0 1 0 0 −2 −9 9 3 9 2 1 2 0 0 0 1 −9 −9 0 0 9 0 Thus, 0 −2 A−1 = 9 1 9 −2 9 5. P34 6. M3 (− 2 ) 9 2 −1 9 9 1 0 −1 3 9 . 1 0 −2 3 9 2 −1 0 9 9 3. M2 ( 1 ) 4 2. A12 (−2), A14 (−3) 1. P13 2 9 7. A31 (−1), 4. A21 (2), A23 (2), A24 (−5) 1 A32 (− 2 ) 8. M4 (− 2 ) 9 1 9. A42 (1), A43 (− 2 ) 17. To determine the second column of A−1 without determining the whole inverse, we solve the vector 2 −1 4 x 0 1 2 y = 1 . linear system 5 1 −1 3 z 0 1 2 18. We have A = 3 5 ,b= 1 3 , and the Gauss-Jordan method yields A−1 = −5 3 2 −1 . Therefore, we have x = A−1 b = So we have x1 = 4 and 1 19. We have A = 0 2 Therefore, we have −5 3 2 −1 1 3 = 4 −1 . x2 = −1. 1 −2 −2 7 5 −3 1 1 , b = 3 , and the Gauss-Jordan method yields A−1 = −2 −1 1 . 4 −3 1 2 2 −1 7 5 −3 −2 −2 1 3 = 2 . x = A−1 b = −2 −1 2 2 −1 1 1 Hence, we have x1 = −2, x2 = 2, and x3 = 1. 20. We have A = 1 −2i 2−i 4i ,b= 2 −i , and the Gauss-Jordan method yields A−1 = Therefore, we have x = A−1 b = 1 2 + 8i 4i 2i −2 + i 1 2 −i = 1 2 + 8i 2 + 8i −4 + i . 1 2+8i 4i 2i −2 + i 1 . 158 Hence, we have x1 = 1 3 21. We have A = 2 4 Therefore, we have and x2 = −4+ii . 2+8 45 1 −79 27 46 10 1 , b = 1 , and the Gauss-Jordan method yields A−1 = 12 −4 −7 . 18 1 38 −13 −22 −79 27 46 1 −6 x = A−1 b = 12 −4 −7 1 = 1 . 38 −13 −22 1 3 Hence, we have x1 = −6, x2 = 1, and x3 = 3. 1 1 2 12 2 −1 , b = 24 , and the Gauss-Jordan method yields A−1 = 22. We have A = 1 2 −1 1 −36 Therefore, we have −1 3 5 12 −10 1 3 3 −3 24 = 18 . x = A−1 b = 12 5 −3 −1 −36 2 −1 3 5 1 3 3 −3 . 12 5 −3 −1 Hence, x1 = −10, x2 = 18, and x3 = 2. 23. We have AAT = 01 −1 0 0 −1 1 0 = (0)(0) + (1)(1) (−1)(0) + (0)(1) (0)(−1) + (1)(0) (−1)(−1) + (0)(0) = 1 0 0 1 = I2 , = 1 0 0 1 = I2 , so AT = A−1 . 24. We have √ AAT = = √ 3/2 √ /2 1 3/2 √1/2 − −1/2 3/2 1/2 3/2 √ √ √ √ ( 3/2)( 3/2) + √ /2)(1/2) (1 ( 3/2)(−1/2) + √ /2)( √ /2) (1 3 √ (−1/2)( 3/2) + ( 3/2)(1/2) (−1/2)(−1/2) + ( 3/2)( 3/2) so AT = A−1 . 25. We have AAT = = cos α − sin α sin α cos α cos α sin α − sin α cos α cos2 α + sin2 α (cos α)(− sin α) + (sin α)(cos α) (− sin α)(cos α) + (cos α)(sin α) (− sin α)2 + cos2 α = 1 0 0 1 so AT = A−1 . 26. We have AAT = 1 1 + 2x2 −2x 1 − 2x2 2x 1 2x 2x2 = 1 1 + 4x2 + 4x4 2x2 −2x 1 1 + 4x2 + 4x4 0 0 1 1 + 2x2 1 2x −2x 1 − 2x2 2x2 −2x 0 1 + 4x2 + 4x4 0 2x2 2x 1 0 = I3 , 0 2 4 1 + 4x + 4x = I2 , 159 so AT = A−1 . 27. For part 2, we have (B −1 A−1 )(AB ) = B −1 (A−1 A)B = B −1 In B = B −1 B = In , and for part 3, we have T (A−1 )T AT = (AA−1 )T = In = In . 28. We prove this by induction on k , with k = 1 trivial and k = 2 proven in part 2 of Theorem 2.6.9. Assuming the statement is true for a product involving k − 1 matrices, we may proceed as follows: (A1 A2 · · · Ak )−1 = ((A1 A2 · · · Ak−1 )Ak )−1 = A−1 (A1 A2 · · · Ak−1 )−1 k 1 1 = A−1 (A−−1 · · · A−1 A−1 ) = A−1 A−−1 · · · A−1 A−1 . 2 1 2 1 k k k k In the second equality, we have applied part 2 of Theorem 2.6.9 to the two matrices A1 A2 · · · Ak−1 and Ak , and in the third equality, we have assumed that the desired property is true for products of k − 1 matrices. 29. Since A is symmetric, we know that AT = A. We wish to show that (A−1 )T = A−1 . We have (A−1 )T = (AT )−1 = A−1 , which shows that A−1 is symmetric. The first equality follows from part 3 of Theorem 2.6.9, and the second equality results from the assumption that A is symmetric. 30. Since A is skew-symmetric, we know that AT = −A. We wish to show that (A−1 )T = −A−1 . We have (A−1 )T = (AT )−1 = (−A)−1 = −(A−1 ), which shows that A−1 is skew-symmetric. The first equality follows from part 3 of Theorem 2.6.9, and the second equality results from the assumption that A−1 is skew-symmetric. 31. We have (In − A)(In + A + A2 + A3 ) = In (In + A + A2 + A3 ) − A(In + A + A2 + A3 ) = In + A + A2 + A3 − A − A2 − A3 − A4 = In − A4 = In , where the last equality uses the assumption that A4 = 0. This calculation shows that In − A and In + A + A2 + A3 are inverses of one another. 32. We have B = BIn = B (AC ) = (BA)C = In C = C. 33. YES. Since BA = In , we know that A−1 = B (see Theorem 2.6.11). Likewise, since CA = In , A−1 = C . Since the inverse of A is unique, it must follow that B = C . 34. We can simply compute 1 ∆ a22 −a21 −a12 a11 a11 a21 a12 a22 1 ∆ 1 = ∆ = a22 a11 − a12 a21 −a21 a11 + a11 a21 a11 a22 − a12 a21 0 a22 a12 − a12 a22 −a21 a12 + a11 a22 0 a11 a22 − a12 a21 = 1 0 0 1 = I2 . 160 Therefore, a11 a21 a12 a22 −1 = 1 ∆ −a12 a11 a22 −a21 . 35. Assume that A is an invertible matrix and that Axi = bi for i = 1, 2, . . . , p (where each bi is given). Use elementary row operations on the augmented matrix of the system to obtain the equivalence [A|b1 b2 b3 . . . bp ] ∼ [In |c1 c2 c3 . . . cp ]. The solutions to the system can be read from the last matrix: xi = ci for each i = 1, 2, . . . , p. 36. We have 1 −1 2 −1 1 1 10 2 ∼ 0 1 00 1 −1 1 1 1 −1 2 1 −1 2 1 4 1 2 3 ∼ 0 1 2 −1 4 −1 6 −1 52 0 2 5 −2 6 0 100 3 0 3 1 0 9 −5 3 2 −1 4 −1 ∼ 0 1 0 −1 8 −5 . 1 0 −2 2 001 0 −2 2 Hence, x1 = (0, −1, 0), x2 = (9, 8, −2), 1. A12 (−2), A13 (−1) x3 = (−5, −5, 2). 2. A21 (1), A23 (−2) 3. A31 (−3), A32 (−2) 37. (a): Let ei denote the ith column vector of the identity matrix Im , and consider the m linear systems of equations Axi = ei for i = 1, 2, . . . , m. Since rank(A) = m and each ei is a column m-vector, it follows that rank(A# ) = m = rank(A) and so each of the systems Axi = ei above has a solution (Note that if m < n, then there will be an infinite number of solutions). If we let B = [x1 , x2 , . . . , xm ], then AB = A [x1 , x2 , . . . , xm ] = [Ax1 , Ax2 , . . . , Axm ] = [e1 , e2 , . . . , em ] = In . ad (b): A right inverse for A in this case is a 3 × 2 matrix b e such that cf a + 3b + c d + 3e + f 2a + 7b + 4c 2d + 7e + 4f = 1 0 0 1 . Thus, we must have a + 3b + c = 1, d + 3e + f = 0, 2a + 7b + 4c = 0, 2d + 7e + 4f = 1. 161 The first and third equation comprise a linear system with augmented matrix 1 2 311 740 for a, b, and 131 1 . Setting c = t, we have b = −2 − 2t 0 1 2 −2 and a = 7 + 5t. Next, the second and fourth equation above comprise a linear system with augmented matrix 1310 1310 for d, e, and f . The row-echelon form of this augmented matrix is . Setting 2741 0121 f = s, we have e = 1 − 2s and d = −3 + 5s. Thus, right inverses of A are precisely the matrices of the form 7 + 5t −3 + 5s −2 − 2t 1 − 2s . t s c. The row-echelon form of this augmented matrix is Solutions to Section 2.7 True-False Review: 1. TRUE. Since every elementary matrix corresponds to a (reversible) elementary row operation, the reverse elementary row operation will correspond to an elementary matrix that is the inverse of the original elementary matrix. 2 0 2. FALSE. For instance, the matrices product, 2 0 0 2 0 1 and 10 02 are both elementary matrices, but their , is not. 3. FALSE. Every invertible matrix can be expressed as a product of elementary matrices. Since every elementary matrix is invertible and products of invertible matrices are invertible, any product of elementary matrices must be an invertible matrix. 4. TRUE. Performing an elementary row operation on a matrix does not alter its rank, and the matrix EA is obtained from A by performing the elementary row operation associated with the elementary matrix E . Therefore, A and EA have the same rank. 2 5. FALSE. If Pij is a permutation matrix, then Pij = In , since permuting the ith and j th rows of In twice − 2 yields In . Alternatively, we can observe that Pij = In from the fact that Pij 1 = Pij . 6. FALSE. For example, consider the elementary matrices E1 = have E1 E2 = 11 07 and E2 E1 = 1 0 7 7 10 07 1 Then we have E1 E2 = 0 0 1 0 1 1 . Then we . 7. FALSE. For example, and E2 = 1 0 consider the elementary matrices E1 = 0 36 130 1 2 and E2 E1 = 0 1 2 . 001 01 1 30 1 0 and E2 = 0 01 0 0 1 0 0 2 . 1 8. FALSE. The only matrices we perform an LU factorization for are invertible matrices for which the reduction to upper triangular form can be accomplished without permuting rows. 9. FALSE. The matrix U need not be a unit upper triangular matrix. 162 10. FALSE. As can be seen in Example 2.7.8, a 4 × 4 matrix with LU factorization will have 6 multipliers, not 10 multipliers. Problems: 1. 0 Permutation Matrices: P12 = 1 0 k0 Scaling Matrices: M1 (k ) = 0 1 00 0 0 , 1 1 0 0 0 0 , 1 1 1 0 , P23 = 0 0 0 00 k 0 , M3 (k ) = 01 0 P13 = 0 1 1 M2 (k ) = 0 0 0 1 0 0 0 1 0 1 . 0 1 0 0 00 1 0 . 0k Row Combinations: 10 A12 (k ) = k 1 00 1k A21 (k ) = 0 1 00 0 0 , 1 0 0 , 1 2. We have 3 5 1 −2 1 ∼ 10 A13 (k ) = 0 1 k0 10 A31 (k ) = 0 1 00 1 −2 3 5 1. P12 2 ∼ 0 0 , 1 k 0 , 1 1 −2 0 11 2. A12 (−3) 100 A23 (k ) = 0 1 0 , 0k1 100 A32 (k ) = 0 1 k . 001 3 ∼ 1 −2 0 1 . 1 3. M2 ( 11 ) 1 Elementary Matrices: M2 ( 11 ), A12 (−3), P12 . 3. We have 5 1 8 2 3 −1 1 ∼ 1 5 3 −1 8 2 1. P12 2 ∼ 1 3 −1 0 −7 7 2. A12 (−5) 3 ∼ 1 3 −1 0 1 −1 . 3. M2 (− 1 ) 7 Elementary Matrices: M2 (− 1 ), A12 (−5), P12 . 7 4. We have 13 3 −1 4 1 32 1 3 2 1 3 2 1 2 3 4 2 1 3 ∼ 2 1 3 ∼ 0 −5 −1 ∼ 0 −5 −1 ∼ 0 1 1 32 3 −1 4 0 −10 −2 0 0 0 00 1. P13 2. A12 (−2), A13 (−3) 3. A23 (−2) 2 1 5 . 0 4. M2 (− 1 ) 5 Elementary Matrices: M2 (− 1 ), A23 (−2), A13 (−3), A12 (−2), P13 . 5 5. We have 1 2 3 2 3 4 3 4 5 4 1 2 3 4 1 2 3 4 1 1 2 3 5 ∼ 0 −1 −2 −3 ∼ 0 1 2 3 ∼ 0 6 0 −2 −4 −6 0 −2 −4 −6 0 2 1 0 3 2 0 4 3 . 0 163 1. A12 (−2), A13 (−3) 2. M2 (−1) 3. A23 (2) Elementary Matrices: A23 (2), M2 (−1), A13 (−3), A12 (−2). 6. We reduce A to the identity matrix: 1 1 2 3 1 0 1 ∼ 2 1 1. A12 (−1) 1 0 2 ∼ 0 1 . 2. A21 (−2) 1 −1 The elementary matrices corresponding to these row operations are E1 = 0 1 1 −2 0 1 and E2 = We have E2 E1 A = I2 , so that 10 11 − − A = E 1 1 E2 1 = 1 0 2 1 , − − which is the desired expression since E1 1 and E2 1 are elementary matrices. 7. We reduce A to the identity matrix: −2 −3 5 7 −2 −3 1 1 1 ∼ 1 1 −2 −3 2 ∼ 1. A12 (2) 2. P12 1 1 0 −1 3 ∼ 3. A12 (2) 4 ∼ 1 0 0 −1 1 0 5 ∼ 0 1 . 5. M2 (−1) 4. A21 (1) The elementary matrices corresponding to these row operations are E1 = 10 21 , E2 = 0 1 1 0 , E3 = 1 2 0 1 , 1 0 E4 = 1 1 , 1 0 0 −1 E5 = . We have E5 E4 E3 E2 E1 A = I2 , so 1 −2 − − − − − A = E 1 1 E2 1 E 3 1 E4 1 E5 1 = 0 1 0 1 1 0 1 −2 0 1 1 −1 0 1 1 0 0 −1 , − which is the desired expression since each Ei 1 is an elementary matrix. 8. We reduce A to the identity matrix: 3 −4 −1 2 1 ∼ −1 2 3 −4 1. P12 1 −2 3 −4 2 ∼ 2. M1 (−1) 1 −2 0 2 3 ∼ 3. A12 (−3) 4 ∼ 4. M2 ( 1 ) 2 1 −2 0 1 5 ∼ 1 0 0 1 . 5. A21 (2) The elementary matrices corresponding to these row operations are E1 = 0 1 1 0 , E2 = −1 0 0 1 , E3 = 1 −3 0 1 −1 0 0 1 , E4 = 1 0 0 1 2 , E5 = 1 0 1 −2 0 1 , We have E5 E4 E3 E2 E1 A = I2 , so − − − − − A = E1 1 E 2 1 E3 1 E4 1 E 5 1 = 0 1 1 0 1 3 0 1 1 0 0 2 2 1 . . 164 − which is the desired expression since each Ei 1 is an elementary matrix. 9. We reduce A to the identity matrix: 4 −5 1 4 1 4 4 −5 1 ∼ 1. P12 1 4 0 −21 2 ∼ 1 0 3 ∼ 4 1 1 3. M2 (− 21 ) 2. A12 (−4) 4 ∼ 1 0 0 1 . 4. A21 (−4) The elementary matrices corresponding to these row operations are E1 = 0 1 1 0 , E2 = 10 −4 1 , E3 = 1 0 1 0 − 21 , 1 −4 0 1 E4 = . We have E4 E3 E2 E1 A = I2 , so 01 10 − − − − A = E1 1 E2 1 E 3 1 E4 1 = 1 4 0 1 1 0 0 −21 1 0 4 1 , − which is the desired expression since each Ei 1 is an elementary matrix. 10. We reduce A to the identity matrix: 1 −1 0 1 −1 0 1 −1 0 1 −1 0 1 2 3 2 2 2 ∼ 0 4 2 ∼ 0 4 2 ∼ 0 4 2 3 13 3 13 0 43 0 01 1 −1 1 ∼ 0 0 0 4 1. A12 (−2) 2. A13 (−3) 1 −1 0 1 6 1 1 0 ∼ 0 ∼ 0 2 0 01 0 1 0 3. A23 (−1) 4. M2 ( 1 ) 4 0 0 . 1 0 1 0 5 5. A32 (− 1 ) 2 6. A21 (1) The elementary matrices corresponding to these row operations are 100 100 1 00 1 0 , E1 = −2 1 0 , E2 = 0 1 0 , E3 = 0 001 −3 0 1 0 −1 1 1 E4 = 0 0 0 1 4 0 0 0 , 1 10 0 E5 = 0 1 − 1 , 2 00 1 11 E6 = 0 1 00 0 0 . 1 We have E6 E5 E4 E3 E2 E1 A = I3 , so − − − − − − A = E1 1 E 2 1 E3 1 E4 1 E 5 1 E6 1 100 100 1 = 2 1 0 0 1 0 0 001 301 0 0 1 1 0 1 0 0 1 0 0 4 0 10 0 0 0 1 1 00 − which is the desired expression since each Ei 1 is an elementary matrix. 1 −1 0 1 0 1 0 , 2 0 01 1 0 165 11. We reduce A to the identity matrix: 0 −4 −2 1 1 1 −1 3 ∼ 0 −2 2 2 −2 1 −1 3 1 4 5 ∼ 0 −4 0 ∼ 0 0 01 0 1. P12 2. A13 (2) −1 3 2 −4 −2 ∼ 2 2 −1 0 6 −4 0 ∼ 01 3. M3 ( 1 ) 8 The elementary matrices corresponding to 010 1 E1 = 1 0 0 , E 2 = 0 001 2 1 0 −3 0 , E5 = 0 1 00 1 4. A32 (2) 1 −1 3 1 −1 3 3 0 −4 −2 ∼ 0 −4 −2 0 0 8 0 0 1 1 −1 0 100 7 0 1 0 ∼ 0 1 0 . 0 01 001 6. M2 (− 1 ) 4 5. A31 (−3) 7. A21 (1) these row operations are 100 100 00 1 0 , E3 = 0 1 0 , E4 = 0 1 2 , 01 001 001 8 1 00 110 E6 = 0 − 1 0 , E7 = 0 1 0 . 4 001 0 01 We have E7 E6 E5 E4 E3 E2 E1 A = I3 , so − − − − − − − A = E1 1 E2 1 E 3 1 E4 1 E 5 1 E6 1 E7 1 010 100 100 1 = 1 0 0 0 1 0 0 1 0 0 001 −2 0 1 008 0 0 0 1 1 −2 0 0 1 0 0 1 0 3 1 00 1 −1 0 0 0 −4 0 0 1 0 , 1 0 01 0 01 − which is the desired expression since each Ei 1 is an elementary matrix. 12. We reduce A to the 1 0 3 1. M2 ( 1 ) 8 identity matrix: 23 12 1 8 0 ∼ 0 1 45 34 10 3 4 0 ∼ 0 1 0 0 −4 2. A13 (−3) 3 2 0 ∼ 5 1 5 ∼ 0 0 3. A21 (−2) 1 2 3 3 0 1 0 ∼ 0 −2 −4 03 10 6 1 0 ∼ 0 1 00 01 4. A23 (2) 1 0 3 0 1 0 0 −2 −4 0 0 . 1 5. M3 (− 1 ) 4 6. A31 (−3) The elementary matrices corresponding to these row operations are 100 100 1 −2 0 1 0 , E1 = 0 1 0 , E 2 = 0 1 0 , E 3 = 0 8 −3 0 1 0 01 001 10 0 100 1 0 −3 0 , E6 = 0 1 0 . E4 = 0 1 0 , E 5 = 0 1 1 021 00 1 0 0 −4 166 We have E6 E5 E4 E3 E2 E1 A = I3 , so − − − − − − A = E1 1 E 2 1 E3 1 E4 1 E 5 1 E6 1 100 100 1 = 0 8 0 0 1 0 0 001 301 0 0 1 00 1 0 0 1 0 0 1 0 −2 1 0 2 1 0 0 0 1 1 0 0 0 −4 0 0 1 0 3 0 , 1 − which is the desired expression since each Ei 1 is an elementary matrix. 13. We reduce A to the identity matrix: 2 −1 1 3 1 3 2 −1 1 ∼ 1 3 0 −7 2 ∼ 3. M2 (− 1 ) 7 2. A12 (−2) 1. P12 1 0 3 ∼ 3 1 1 0 4 ∼ 0 1 . 4. A21 (−3) The elementary matrices corresponding to these row operations are E1 = 0 1 1 0 , 1 −2 E2 = 0 1 , 1 0 1 0 −7 E3 = , 1 −3 0 1 E4 = . Direct multiplication verifies that E4 E3 E2 E1 A = I2 . 14. We have 3 −2 −1 5 3 −2 0 13 3 1 ∼ = U. 1. A12 ( 1 ) 3 10 −1 1 3 − 1 Hence, E1 = A12 ( 3 ). Then Equation (2.7.3) reads L = E1 1 = A12 (− 1 ) = 3 . Verifying Equation (2.7.2): LU = 3 1 3 −2 0 13 3 10 1 −3 1 2 3 0 − 13 2 3 −2 −1 5 = 15. We have 2 5 1 ∼ = U =⇒ m21 = = A. 5 =⇒ L = 2 1 0 1 5 2 . Then 1 LU = 0 1 5 2 2 3 0 − 13 2 23 51 = = A. 1. A12 (− 5 ) 2 16. We have 3 5 1 2 1 ∼ 3 0 1 = U =⇒ m21 = 1 3 5 =⇒ L = 3 Then LU = 1 5 3 0 1 3 0 1 1 3 = 3 5 1 2 = A. 1 5 3 0 1 . 167 1. A12 (− 5 ) 3 17. We have 3 −1 2 3 −1 2 3 −1 2 1 2 6 −1 1 ∼ 0 1 −3 ∼ 0 1 −3 = U =⇒ m21 = 2, m31 = −1, m32 = 4. −3 52 0 4 4 0 0 16 Hence, 1 L= 2 −1 0 1 4 0 0 1 and 1 LU = 2 −1 00 3 −1 2 3 −1 2 1 0 0 1 −3 = 6 −1 1 = A. 41 0 0 16 −3 52 1. A12 (−2), A13 (1) 18. We have 5 2 1 5 2 1 5 1 2 −10 −2 3 ∼ 0 2 5 ∼ 0 15 2 −3 0 −4 −6 0 2. A23 (−4) 21 2 5 = U =⇒ m21 = −2, m31 = 3, m32 = −2. 04 Hence, 1 00 1 0 L = −2 3 −2 1 and 1 00 5 1 0 0 LU = −2 3 −2 1 0 1. A12 (2), A13 (−3) 19. We have 1 1 2 3 2 0 3 −4 1 ∼ 3 −1 7 8 1 34 5 21 5 2 1 2 5 = −10 −2 3 = A. 04 15 2 −3 2. A23 (2) 1 −1 2 3 1 −1 2 3 0 2 −1 −10 2 0 2 −1 −10 ∼ 0 2 1 −1 0 0 2 9 0 4 2 2 0 0 4 22 1. A12 (−2), A13 (−3), A14 (−1) 1 −1 2 3 30 2 −1 −10 ∼ = U. 0 0 2 9 0 0 0 4 2. A23 (−1), A24 (−2) 3. A34 (−2) Hence, m21 = 2, Hence, 1 2 L= 3 1 0 1 1 2 0 0 1 2 0 0 0 1 20. We have 2 −3 1 2 4 −1 1 1 −8 2 2 −5 6 15 2 m31 = 3, m41 = 1, m32 = 1, m42 = 2, m43 = 2. and 1 2 LU = 3 1 0 1 1 2 0 0 1 2 0 1 −1 2 3 0 0 2 −1 −10 0 0 0 2 9 1 0 0 0 4 1 −1 2 3 2 0 3 −4 = = A. 3 −1 7 8 1 34 5 2 −3 1 2 2 −3 1 2 10 5 −1 −3 2 0 5 −1 −3 3 ∼ ∼ ∼ 0 −10 0 6 3 0 4 −3 0 10 2 −4 0 0 4 2 2 −3 1 2 0 5 −1 −3 = U. 0 0 4 −3 0 0 0 5 168 1. A12 (−2), A13 (4), A14 (−3) 2. A23 (2), A24 (−2) 3. A34 (−1) Hence, m31 = −4, m21 = 2, Hence, 1 0 2 1 L= −4 −2 3 2 m41 = 3, 00 1 00 2 0 0 10 and LU = −4 −2 1 1 0 11 3 21 m32 = −2, m42 = 2, m43 = 1. 0 2 −3 1 2 2 −3 1 2 0 0 5 −1 −3 4 −1 1 1 = 0 0 0 4 −3 −8 2 2 −5 1 0 0 0 5 6 15 2 = A. 21. We have 12 23 1 ∼ 1 2 0 −1 = U =⇒ m21 = 2 =⇒ L = 10 21 . 1. A12 (−2) We now solve the triangular systems Ly = b and U x = y. From Ly = b, we obtain y = U x = y yields x = −11 7 3 −7 . Then . 22. We have 1 −3 5 1 −3 5 1 −3 5 1 2 3 2 2 ∼ 0 11 −13 ∼ 0 11 −13 = U =⇒ m21 = 3, m31 = 2, m32 = 1. 2 52 0 11 −8 0 0 5 1. A12 (−3), A13 (−2) 1 Hence, L = 3 2 1 obtain y = 2 −5 2. A23 (−1) 00 1 0 . We now solve the triangular systems Ly = b and U x = y. From Ly = b, we 1 1 3 . Then U x = y yields x = −1 . −1 23. We have 22 1 2 2 1 2 2 1 1 2 6 3 −1 ∼ 0 −3 −4 ∼ 0 −3 −4 = U =⇒ m21 = 3, m31 = −2, m32 = −2. −4 2 2 0 0 −4 0 0 −4 1. A12 (−3), A13 (2) 1 Hence, L = 3 −2 1 obtain y = −3 −2 2. A23 (2) 00 1 0 . We now solve the triangular systems Ly = b and U x = y. From Ly = b, we −2 1 −1/12 . Then U x = y yields x = 1/3 . 1/2 169 24. We have 43 00 8 1 20 0 5 36 0 0 −5 7 4 3 00 4 3 00 4 30 1 0 −5 2 0 2 0 −5 2 0 3 0 −5 2 ∼ ∼ ∼ 0 5 3 6 0 0 5 6 0 05 0 0 −5 7 0 0 −5 7 0 00 1. A12 (−2) 2. A23 (1) 0 0 = U. 6 13 3. A34 (1) 1 0 00 2 1 0 0 . We The only nonzero multipliers are m21 = 2, m32 = −1, and m43 = −1. Hence, L = 0 −1 1 0 0 0 −1 1 2 −1 now solve the triangular systems Ly = b and U x = y. From Ly = b, we obtain y = −1 . Then U x = y 4 677/1300 −9/325 yields x = −37/65 . 4/13 25. We have 2 −1 −8 3 1 ∼ 2 −1 0 −1 = U =⇒ m21 = −4 =⇒ L = 1 −4 0 1 1. A12 (4) We now solve the triangular systems Lyi = bi , U xi = yi for i = 1, 2, 3. We have Ly1 = b1 =⇒ y1 = Ly2 = b2 =⇒ y2 = Ly3 = b3 =⇒ y3 = 26. We have 3 11 2 15 5 11 . Then U x1 = y1 =⇒ x1 = . Then U x2 = y2 =⇒ x2 = . Then U x3 = y3 =⇒ x3 = −1 42 −1 1 3 1 4 ∼ 0 5 −7 1 0 4 13 13 −4 ; −1 −6.5 ; −15 −3 . −11 2 −1 2 10 ∼ 0 11 0 1. A12 (3), A13 (5) 4 2 13 10 = U. 0 1 2. A23 (−1) Thus, m21 = −3, m31 = −5, and m32 = 1. We now solve the triangular systems Lyi = bi , for i = 1, 2, 3. We have U xi = yi . 170 −29/13 1 Ly1 = e1 =⇒ y1 = 3 . Then U x1 = y1 =⇒ x1 = −17/13 2 2 0 18/13 Ly2 = e2 =⇒ y2 = 1 . Then U x2 = y2 =⇒ x2 = 11/13 −1 −1 0 −14/13 Ly3 = e3 =⇒ y3 = 0 . Then U x3 = y3 =⇒ x3 = −10/13 1 1 ; ; . 27. Observe that if Pi is an elementary permutation matrix, then Pi−1 = Pi = PiT . Therefore, we have − −1 − − TT T T P −1 = (P1 P2 . . . Pk )−1 = Pk 1 Pk−1 . . . P2 1 P1 1 = Pk Pk−1 . . . P2 . . . P1 = (P1 P2 . . . Pk )T = P T . 28. (a): Let A be an invertible upper triangular matrix with inverse B . Therefore, we have AB = In . Write A = [aij ] and B = [bij ]. We will show that bij = 0 for all i > j , which shows that B is upper triangular. We have n aik bkj = δij . k=1 Since A is upper triangular, aik = 0 whenever i > k . Therefore, we can reduce the above summation to n aik bij = δij . k=i Let i = n. Then the above summation reduces to ann bnj = δnj . If j = n, we have ann bnn = 1, so ann = 0. For j < n, we have ann bnj = 0, and therefore bnj = 0 for all j < n. Next let i = n − 1. Then we have an−1,n−1 bn−1,j + an−1,n bnj = δn−1,j . Setting j = n−1 and using the fact that bn,n−1 = 0 by the above calculation, we obtain an−1,n−1 bn−1,n−1 = 1, so an−1,n−1 = 0. For j < n − 1, we have an−1,n−1 bn−1,j = 0 so that bn−1,j = 0. Next let i = n − 2. Then we have an−2,n−2 bn−2,j + an−2,n−1 bn−1,j + an−2,n bnj = δn−2,j . Setting j = n − 2 and using the fact that bn−1,n−2 = 0 and bn,n−2 = 0, we have an−2,n−2 bn−2,n−2 = 1, so that an−2,n−2 = 0. For j < n − 2, we have an−2,n−2 bn−2,j = 0 so that bn−2,j = 0. Proceeding in this way, we eventually show that bij = 0 for all i > j . For an invertible lower triangular matrix A with inverse B , we can either modify the preceding argument, or we can proceed more briefly as follows: Note that AT is an invertible upper triangular matrix with inverse B T . By the preceding argument, B T is upper triangular. Therefore, B is lower triangular, as required. (b): Let A be an invertible unit upper triangular matrix with inverse B . Use the notations from (a). By (a), we know that B is upper triangular. We simply must show that bjj = 0 for all j . From ann bnn = 1 (see proof of (a)), we see that if ann = 1, then bnn = 1. Moreover, from an−1,n−1 bn−1,n−1 = 1, the fact that an−1,n−1 = 1 proves that bn−1,n−1 = 1. Likewise, the fact that an−2,n−2 bn−2,n−2 = 1 implies that if an−2,n−2 = 1, then bn−2,n−2 = 1. Continuing in this fashion, we prove that bjj = 1 for all j . For the last part, if A is an invertible unit lower triangular matrix with inverse B , then AT is an invertible unit upper triangular matrix with inverse B T , and by the preceding argument, B T is a unit upper triangular matrix. This implies that B is a unit lower triangular matrix, as desired. 171 29. (a): Since A is invertible, Corollary 2.6.12 implies that both L2 and U1 are invertible. Since L1 U1 = L2 U2 , − − we can left-multiply by L−1 and right-multiply by U1 1 to obtain L−1 L1 = U2 U1 1 . 2 2 − (b): By Problem 28, we know that L−1 is a unit lower triangular matrix and U1 1 is an upper triangular 2 −1 −1 matrix. Therefore, L2 L1 is a unit lower triangular matrix and U2 U1 is an upper triangular matrix. Since − these two matrices are equal, we must have L−1 L1 = In and U2 U1 1 = In . Therefore, L1 = L2 and U1 = U2 . 2 30. The system Ax = b can be written as QRx = b. If we can solve Qy = b for y and then solve Rx = y for x, then QRx = b as desired. Multiplying Qy = b by QT and using the fact that QT Q = In , we obtain y = QT b. Therefore, Rx = y can be replaced by Rx = QT b. Therefore, to solve Ax = b, we first determine y = QT b and then solve the upper triangular system Rx = QT b by back-substitution. Solutions to Section 2.8 True-False Review: 1. FALSE. According to the given information, part (c) of the Invertible Matrix Theorem fails, while part (e) holds. This is impossible. 2. TRUE. This holds by the equivalence of parts (d) and (f) of the Invertible Matrix Theorem. 3. FALSE. Part (d) of the Invertible Matrix Theorem fails according to the given information, and therefore part (b) also fails. Hence, the equation Ax = b does not have a unique solution. But it is not valid to conclude 100 that the equation has infinitely many solutions; it could have no solutions. For instance, if A = 0 1 0 000 0 and b = 0 , there are no solutions to Ax = b, although rank(A) = 2. 1 4. FALSE. An easy counterexample is the matrix 0n , which fails to be invertible even though it is upper triangular. Since it fails to be invertible, it cannot e row-equivalent to In , by the Invertible Matrix Theorem. Problems: 1. Since A is an invertible matrix, the only solution to Ax = 0 is x = 0. However, if we assume that AB = AC , then A(B − C ) = 0. If xi denotes the ith column of B − C , then xi = 0 for each i. That is, B − C = 0, or B = C , as required. 2. If rank(A) = n, then the augmented matrix A# for the system Ax = 0 can be reduced to REF such that each column contains a pivot except for the right-most column of all-zeros. Solving the system by back-substitution, we find that x = 0, as claimed. 3. Since Ax = 0 has only the trivial solution, REF(A) contains a pivot in every column. Therefore, the linear system Ax = b can be solved by back-substitution for every b in Rn . Therefore, Ax = b does have a solution. Now suppose there are two solutions y and z to the system Ax = b. That is, Ay = b and Az = b. Subtracting, we find A(y − z) = 0, and so by assumption, y − z = 0. That is, y = z. Therefore, there is only one solution to the linear system Ax = b. 172 4. If A and B are each invertible matrices, then A and B can each be expressed as a product of elementary matrices, say A = E 1 E2 . . . E k and B = E1 E2 . . . E l . Then AB = E1 E2 . . . Ek E1 E2 . . . El , so AB can be expressed as a product of elementary matrices. Thus, by the equivalence of (a) and (e) in the Invertible Matrix Theorem, AB is invertible. 5. We are assuming that the equations Ax = 0 and B x = 0 each have only the trivial solution x = 0. Now consider the linear system (AB )x = 0. Viewing this equation as A(B x) = 0, we conclude that B x = 0. Thus, x = 0. Hence, the linear equation (AB )x = 0 has only the trivial solution. Solutions to Section 2.9 Problems: 1. We have (−4)A − B T = 8 −16 −8 −24 4 4 −20 0 − 2. We have −3 0 2 10 2 −3 1 11 −18 −9 −24 4 2 −17 −1 AB = −2 426 −1 −1 5 0 −3 0 2 2 1 −3 = 0 1 = 16 8 6 −17 . Moreover, tr(AB ) = −1. 3. We have (AC )(AC )T = −2 26 −2 26 = 4 −52 −52 676 . 4. We have 12 0 −8 −8 (−4B )A = −4 12 0 −4 −2 426 −1 −1 5 0 −24 48 24 72 24 −24 −56 −48 . = −4 −28 52 −24 4 4 −20 0 5. Using Problem 2, we find that (AB )−1 = 16 8 6 −17 −1 =− 1 320 −17 −8 −6 16 . . 173 6. We have CT C = −5 −6 3 1 −5 −6 3 = [71], 1 and tr(C T C ) = 71. 7. (a): We have AB = 1 2 2 5 3 7 3b −4 a = ab 3a − 5 2a + 4b 7a − 14 5a + 9b . In order for this product to equal I2 , we require 3a − 5 = 1, 7a − 14 = 0, 2a + 4b = 0, 5a + 9b = 1. We quickly solve this for the unique solution: a = 2 and b = −1. (b): We have 3 −1 2 BA = −4 2 −1 1 1 2 2 2 . = 0 0 −1 −1 1 2 2 5 3 7 8. We compute the (i, j )-entry of each side of the equation. We will denote the entries of AT by aT , which ij equals aji . On the left side, note that the (i, j )-entry of (AB T )T is the same as the (j, i)-entry of AB T , and n n (j, i)-entry of AB T = ajk bT = ki k=0 n bik aT , kj ajk bik = k=0 k=0 and the latter expression is the (i, j )-entry of BAT . Therefore, the (i, j )-entries of (AB T )T and BAT are the same, as required. 9. (a): The (i, j )-entry of A2 is n aik akj . k=1 (b): Assume that A is symmetric. That means that AT = A. We claim that A2 is symmetric. To see this, note that (A2 )T = (AA)T = AT AT = AA = A2 . Thus, (A2 )T = A2 , and so A2 is symmetric. 10. We are assuming that A is skew-symmetric, so AT = −A. To show that B T AB is skew-symmetric, we observe that (B T AB )T = B T AT (B T )T = B T AT B = B T (−A)B = −(B T AB ), as required. 174 11. We have 3 9 −1 −3 A2 = 2 0 0 = 0 0 , so A is nilpotent. 12. We have 00 A2 = 0 0 00 and 0 A3 = A2 A = 0 0 1 0 0 1 01 0 0 0 0 00 0 0 0 1 0 1 = 0 0 0 0 0 0 0 0 , 0 so A is nilpotent. 13. We have 14. We have −3e−3t 6t2 A (t) = 6/t −7t 1 6t − t2 /2 B (t) dt = t + t2 /2 0 et −2 sec2 t tan t . − sin t −5 t3 /3 3 t 4 / 4 + 2 t3 2 π sin(πt/2) 4 t − t /4 1 0 −7 11/2 = 3/2 e−1 1/3 11/4 . 2/π 3/4 15. Since A(t) is 3 × 2 and B (t) is 4 × 2, it is impossible to perform the indicated subtraction. 16. Since A(t) is 3 × 2 and B (t) is 4 × 2, it is impossible to perform the indicated subtraction. 17. From the last equation, we see that x3 = 0. Substituting this into the middle equation, we find that x2 = 0.5. Finally, putting the values of x2 and x3 into the first equation, we find x1 = −6 − 2.5 = −8.5. Thus, there is a unique solution to the linear system, and the solution set is {(−8.5, 0.5, 0)}. 18. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to row-echelon form. This gives us 5 −1 2 7 1 −2 6 9 0 ∼ −7 5 −3 −7 1 4 ∼ 0 0 11 20 7 1 2 6 9 0 ∼ 0 5 −3 −7 0 11 20 7 1 5 1 7/4 1/2 ∼ 0 0 −13/2 1 0 1 −2 −7 11 20 7 1 3 28 49 14 ∼ 0 82 137 42 0 11 20 7 1 7/4 1/2 . 0 1 −2/13 11 20 7 1 7/4 1/2 82 137 42 From the last row, we conclude that x3 = −2/13, and using the middle row, we can solve for x2 : we have 2 20 2 x2 + 7 · − 13 = 1 , so x2 = 26 = 10 . Finally, from the first row we can get x1 : we have x1 +11· 10 +20· − 13 = 4 2 13 13 21 7, and so x1 = 13 . So there is a unique solution: 21 10 2 , ,− 13 13 13 . 175 1. A21 (2) 2. A12 (2), A13 (7) 3. M2 (1/28) 4. A23 (−82) 5. M3 (−2/13) 19. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to row-echelon form. This gives us 1 1 4 1 2 −1 1 1 2 −1 1 1 2 −1 1 2 −1 1 3 1 2 1 −1 −2 ∼ 0 1 −1 −2 . 2 4 ∼ 0 0 1 5 ∼ 0 −2 0 8 00 0 0 −4 4 0 −4 48 4 0 12 From this row-echelon form, we see that z is a free variable. Set z = t. Then from the middle row of the matrix, y = t − 2, and from the top row, x + 2(t − 2) − t = 1 or x = −t + 5. So the solution set is {(−t + 5, t − 2, t) : t ∈ R} = {(5, −2, 0) + t(−1, 1, 1) : t ∈ R}. 1. A12 (−1), A13 (−4) 2. M2 (−1/2) 3. A23 (4) 20. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to row-echelon form. This gives us 1 −2 −1 3 0 1 −2 −1 3 0 1 −2 −1 30 1 −2 −1 30 3 2 1 −2 0 1 1 /3 1 . 0 3 1 3 ∼ 0 0 3 1 3 ∼ 0 4 5 −5 3 ∼ 0 1 0 0 0 0 0 0 005 0 0 −3 −1 2 3 −6 −6 82 The bottom row of this matrix shows that this system has no solutions. 1. A12 (2), A13 (−3) 2. A23 (1) 3. M2 (1/3), M3 (1/3) 21. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to row-echelon form. This gives us 3 0 −1 2 −1 1 1 3 1 3 1 −3 2 −1 1 3 0 ∼ 4 −2 −3 6 −1 5 4 −2 0 0 0 1 4 −2 0 0 1 3 1 −3 2 33 −21 3 0 −27 −12 ∼ 0 28 14 −36 18 0 0 0 1 4 1 3 1 −3 2 −1 1 1 2 −3 −3 −6 6 0 50 ∼ 0 −27 −12 33 −21 12 ∼ 0 0 0 0 1 4 −2 0 1 −3 2 −1 1 3 1 −3 2 −1 −1 2 −1 1 2 0 −9 −4 11 −7 4 ∼ −3 6 −1 5 0 −14 −7 18 −9 9 0 1 4 −2 0 0 0 1 4 −2 −1 1 3 1 −3 2 −1 12 4 0 −27 −12 33 −21 12 ∼ −18 0 1 2 −3 −3 −6 −2 0 0 0 1 4 −2 1 3 1 −3 2 31 −3 2 −1 −3 12 −3 −3 −6 7 0 1 2 −3 ∼ 0 42 −48 −102 −150 0 0 1 − 8 − 17 7 7 00 1 4 −2 000 1 4 We see that x5 = t is the only free variable. Back substitution yields the remaining values: x5 = t, x4 = −4t − 2, x3 = − 41 15 − t, 7 7 2 33 x2 = − − t, 7 7 2 16 x1 = − + t. 7 7 −1 −6 . − 25 7 −2 176 So the solution set is 2 16 2 33 41 15 − + t, − − t, − − t, −4t − 2, t 7 7 7 7 7 7 = 1. P12 t :t∈R 2 2 41 16 33 15 , − , − , −4, 1 + − , − , − , −2, 0 7 7 7 77 7 2. A12 (−3), A13 (−4) 3. M2 (3), M3 (−2) 4. A23 (1) :t∈R . 5. P23 6. A23 (27) 7. M3 (1/42) 22. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to row-echelon form. This gives us 11 1 1 −3 6 11 1 1 −3 6 1 1 1 1 −3 6 1 1 1 2 −5 8 1 0 0 0 1 −2 2 0 1 −2 220 0 ∼ ∼ 2 3 1 4 −9 17 0 1 −1 2 −3 5 0 1 −1 2 −3 5 00 00 00 00 0 −1 2 −2 2 2 2 3 −8 14 11 1 1 −3 6 3 0 1 −1 2 −3 5 . ∼ 0 0 0 1 −2 2 00 00 00 From this row-echelon form, we see that x5 = t and x3 = s are free variables. Furthermore, solving this system by back-substitution, we see that x5 = t, x4 = 2t + 2, x3 = s, x2 = s − t + 1, x1 = 2t − 2s + 3. So the solution set is {(2t − 2s + 3, s − t + 1, s, 2t + 2, t) : s, t ∈ R} = {t(2, −1, 0, 2, 1) + s(−2, 1, 1, 0, 0) + (3, 1, 0, 2, 0) : s, t ∈ R}. 1. A12 (−1), A13 (−2), A14 (−2) 2. A24 (1) 3. P23 23. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to row-echelon form. This gives us 1 −3 2i 1 −2i 6 2 −2 1 ∼ 1 0 −3 2i 1 6 − 6i −2 −2 + 2i 1. A12 (2i) 2 ∼ 1 −3 2i 1 0 1 − 1 (1 + i) − 1 3 6 . 1 2. M2 ( 6−6i ) From the last augmented matrix above, we see that x3 is a free variable. Let us set x3 = t, where t is a complex number. Then we can solve for x2 using the equation corresponding to the second row of the 1 row-echelon form: x2 = − 3 + 1 (1+ i)t. Finally, using the first row of the row-echelon form, we can determine 6 1 that x1 = 2 t(1 − 3i). Therefore, the solution set for this linear system of equations is 1 11 {( t(1 − 3i), − + (1 + i)t, t) : t ∈ C}. 2 36 177 24. We reduce the corresponding linear system as follows: 1 −k 6 2 3k 1 ∼ 1 −k 0 3 + 2k 6 k − 12 . If k = − 3 , then each column of the row-reduced coefficient matrix will contain a pivot, and hence, the linear 2 system will have a unique solution. If, on the other hand, k = − 3 , then the system is inconsistent, because 2 the last row of the row-echelon form will have a pivot in the right-most column. Under no circumstances will the linear system have infinitely many solutions. 25. First observe that if k = 0, then the second equation requires that x3 = 2, and then the first equation requires x2 = 2. However, x1 is a free variable in this case, so there are infinitely many solutions. Now suppose that k = 0. Then multiplying each row of the corresponding augmented matrix for the linear system by 1/k yields a row-echelon form with pivots in the first two columns only. Therefore, the third variable, x3 , is free in this case. So once again, there are infinitely many solutions to the system. We conclude that the system has infinitely many solutions for all values of k . 26. Since this linear system is homogeneous, it already has at least one solution: (0, 0, 0). Therefore, it only remains to determine the values of k for which this will be the only solution. We reduce the corresponding matrix as follows: 1 1/2 −1/2 0 10 k −1 0 10k k 2 −k 0 2 1 k 1 −1 0 ∼ 10k 10 −10 0 −10 0 ∼ 10k 10 2 10k k −k 0 1 1/2 −1/2 0 2 1 −1 0 1 1/2 −1/2 0 1 1/2 −1/2 0 1 1/2 −1/2 0 3 4 5 1 −1 0 ∼ 0 1 −1 0 . ∼ 0 10 − 5k 5k − 10 0 ∼ 0 2 2 0 k − 5k 4k 0 0 k − 5k 4k 0 0 0 k2 − k 0 1. M1 (k ), M2 (10), M3 (1/2) 2. P13 3. A12 (−10k ), A13 (−10k ) 1 4. M2 ( 10−5k ) 5. A23 (5k − k 2 ) Note that the steps above are not valid if k = 0 or k = 2 (because Step 1 is not valid with k = 0 and Step 4 is not valid if k = 2). We will discuss those special cases individually in a moment. However if k = 0, 2, then the steps are valid, and we see from the last row of the last matrix that if k = 1, we have infinitely many solutions. Otherwise, if k = 0, 1, 2, then the matrix has full rank, and so there is a unique solution to the linear system. If k = 2, then the last two rows of the original matrix are the same, and so the matrix of coefficients of the linear system is not invertible. Therefore, the linear system must have infinitely many solutions. If k = 0, we reduce the original linear system as follows: 10 0 2 0 −1 0 1 0 −1/10 0 1 0 −1/10 0 1 0 −1/10 0 1 2 3 1 −1 0 ∼ 0 1 −1 0 ∼ 0 1 −1 0 ∼ 0 1 −1 0 . 1 −1 0 21 −1 0 0 1 −4/5 0 00 1/5 0 The last matrix has full rank, so there will be a unique solution in this case. 1. M1 (1/10) 2. A13 (−2) 3. A23 (−1) To summarize: The linear system has infinitely many solutions if and only if k = 1 or k = 2. Otherwise, the system has a unique solution. 178 27. To solve this system, we need to reduce the corresponding augmented matrix for the linear system to row-echelon form. This gives us 1 −k k 2 0 1 −k k2 0 1 −k k2 0 1 −k k2 0 1 2 3 1 0 k 0 ∼ 0 k k − k2 0 ∼ 0 1 −1 1 ∼ 0 1 −1 1 . 2 0 1 −1 1 0 1 −1 1 0 k k−k 0 0 0 2k − k 2 −k 1. A12 (−1) 2. P23 3. A23 (−k ) Now provided that 2k − k 2 = 0, the system can be solved without free variables via back-substitution, and therefore, there is a unique solution. Consider now what happens if 2k − k 2 = 0. Then either k = 0 or k = 2. If k = 0, then only the first two columns of the last augmented matrix above are pivoted, and we have a free variable corresponding to x3 . Therefore, there are infinitely many solutions in this case. On the other hand, if k = 2, then the last row of the last matrix above reflects an inconsistency in the linear system, and there are no solutions. To summarize, the system has no solutions if k = 2, a unique solution if k = 0 and k = 2, and infinitely many solutions if k = 0. 28. No, there are no common points of intersection. A common point of intersection would be indicated by a solution to the linear system consisting of the equations of the three planes. However, the corresponding augmented matrix can be row-reduced as follows: 12 1 12 1 12 14 4 4 1 2 0 1 −1 1 ∼ 0 1 −1 1 ∼ 0 1 −1 1 . 13 00 0 1 −1 −4 00 0 −5 The last row of this matrix shows that the linear system is inconsistent, and so there are no points common to all three planes. 1. A13 (−1) 2. A23 (−1) 29. (a): We have 4 −2 7 5 1 ∼ 1 −2 7/4 5 1. M1 (1/4) 1 0 2 ∼ 2. A12 (2) 7/4 17/2 1 0 3 ∼ 7/4 1 . 3. M2 (2/17) (b): We have: rank(A) = 2, since the row-echelon form of A in (a) consists two nonzero rows. (c): We have 471 −2 5 0 0 1 1 ∼ 1 7/4 1/4 −2 5 0 4 ∼ 0 1 1 0 2 ∼ 1 0 7/4 1/4 17/2 1/2 0 5/34 −7/34 1 1/17 2/17 . 0 1 3 ∼ 1 0 7/4 1/4 1 1/17 0 2/17 179 1. M1 (1/4) 2. A12 (2) 3. M2 (2/17) 4. A21 (−7/4) Thus, 5 34 1 17 A−1 = 7 − 34 2 17 . 30. (a): We have 2 −7 −4 14 1 ∼ 2 −7 0 0 1. A12 (2) 2 ∼ 1 −7/2 0 0 . 2. M1 (1/2) (b): We have: rank(A) = 1, since the row-echelon form of A in (a) has one nonzero row. (c): Since rank(A) < 2, A is not invertible. 31. (a): We have 3 −1 6 1 −1/3 1 0 2 3 ∼ 0 2 3 −5 0 1 −5/3 2 1 −1/3 2 1 −1/3 2 3 3 ∼ 0 2 3 ∼ 0 2 0 0 −4/3 −2 0 0 2. A13 (−1) 1. M1 (1/3), M3 (1/3) 2 1 −1/3 4 3 ∼ 0 1 0 0 0 3. A23 (2/3) 4. M2 (1/2) (b): We have: rank(A) = 2, since the row-echelon form of A in (a) consists of two nonzero rows. (c): Since rank(A) < 3, A is not invertible. 32. (a): We have 2 1 0 0 1 2 0 0 0 0 3 4 0 1 012 ∼ 4 0 3 0 2 1 0 0 1 40 ∼ 0 0 1. P12 0 0 3 4 2 1 0 0 0 1 20 0 0 2 0 −3 0 0 ∼ 4 0 03 4 3 0 0 1 −1 1 20 0 3 0 −3 0 0 ∼ 0 0 1 −1 0 03 4 0 0 120 0 0 050 1 0 0 ∼ 1 −1 0 0 1 −1 0 7 000 1 2. A12 (−2), A34 (−1) 3. P34 . 4. M2 (−1/3), A34 (−3) 5. M4 (1/7) (b): We have: rank(A) = 4, since the row-echelon form of A in (a) consists of four nonzero rows. 2 3/2 . 0 180 (c): We have 210 1 2 0 0 0 3 004 0 0 4 3 1 0 0 0 0 1 0 0 1 2 0 −3 ∼ 0 0 0 0 3 1 0 ∼ 0 0 5 1. P12 00 0 10 0 0 1 −1 00 1 0 12000100 012 1 0 0 1 0 0 02 ∼ ∼ 0 0 0 3 4 0 0 1 0 1 00430001 1 00 0 00 120 0 0 040 1 0 0 0 1 −2 0 ∼ 1 −1 0 0 −1 1 0 0 1 −1 3 40 0 10 000 7 1000 2/3 −1/3 0 0 −1/3 2/3 0 0 60 1 0 0 ∼ 0 0 −1 1 0 0 1 0 0001 0 0 4/7 −3/7 0 0 1 0 2. A12 (−2), A34 (−1) 3. P34 4. A34 (−3), M2 (−1/3) 0 1 00 1 −2 0 0 0 0 1 0 0 0 −1 1 0 1 0 0 −1/3 2/3 0 0 0 0 −1 1 0 0 4 −3 2/3 −1/3 0 0 −1/3 2/3 0 0 . 0 0 −3/7 4/7 0 0 4/7 −3/7 1 20 0 0 −3 0 0 0 03 4 0 0 1 −1 5. M4 (1/7), A21 (−2) 6. A43 (1) Thus, A−1 2/3 −1/3 0 0 −1/3 2/3 0 0 . = 0 0 −3/7 4/7 0 0 4/7 −3/7 33. (a): We have 3 0 0 1 0 0 1 0 0 1 0 0 1 00 1 1 2 3 4 5 0 2 −1 ∼ 0 2 −1 ∼ 0 2 −1 ∼ 0 −1 2 ∼ 0 −1 2 ∼ 0 1 −1 2 1 −1 2 0 −1 2 0 2 −1 0 03 0 1. M1 (1/3) 2. A13 (−1) 3. P23 4. A23 (2) 5. M2 (−1), M3 (1/3) (b): We have: rank(A) = 3, since the row-echelon form of A in (a) has 3 nonzero rows. (c): We have 3 0 0100 1 0 1 0 2 −1 0 1 0 ∼ 0 2 1 −1 2001 1 −1 1 0 0 1/3 3 2 −1/3 ∼ 0 −1 0 2 −1 0 10 0 1/3 0 5 0 ∼ 0 1 −2 1/3 00 1 −2/9 1/3 0 1/3 0 −1 0 1 20 0 00 1 4 0 1 ∼ 0 10 0 0 1 6 −1 ∼ 0 2/3 0 0 1 0 0 1/3 2 0 ∼ 0 2 −1 0 1 0 −1 2 −1/3 0 0 1/3 0 0 −1 2 −1/3 0 1 0 3 −2/3 1 2 0 0 1/3 0 0 1 0 −1/9 2/3 1/3 . 0 1 −2/9 1/3 2/3 00 1 0 01 0 0 1 −2 . 0 1 181 2. A13 (−1) 1. M1 (1/3) 3. P23 5. M2 (−1), M3 (1/3) 4. A23 (2) 6. A32 (2) Hence, A−1 1/3 = −1/9 −2/9 0 1/3 . 2/3 0 2/3 1/3 34. (a): We have −2 −3 1 1 42 1 1 2 1 4 2 ∼ −2 −3 1 ∼ 0 0 53 0 53 0 1. P12 2. A12 (2) 42 1 3 5 5 ∼ 0 53 0 3. A23 (−1) 4 2 1 4 5 5 ∼ 0 0 −2 0 4 1 0 2 1 . 1 4. M2 (1/5), M3 (−1/2) (b): We have: rank(A) = 3, since the row-echelon form of A in (a) consists of 3 nonzero rows. (c): We have 1 420 −2 −3 1 1 0 0 1 1 4 2 0 1 0 ∼ −2 −3 1 1 0 530 0 53001 1 14 2 0 10 3 4 5 1 2 0 ∼ 0 ∼ 0 5 0 0 −2 −1 −2 1 0 1 1 0 −2 −4/5 −3/5 0 6 5 2/5 0 ∼ 0 1 1/5 ∼ 0 1 0 00 1 1/2 1 −1/2 1. P12 2. A12 (2) 3. A23 (−1) 1 0 0 4 1 0 14 0 2 0 ∼ 0 5 05 1 20 1 1/5 1 1/2 201 512 300 1 0 2/5 0 1 −1/2 0 0 1 00 1/5 7/5 −1 1 0 −3/10 −3/5 1/2 . 01 1/2 1 −1/2 4. M2 (1/5), M3 (−1/2) 5. A21 (−4) 6. A31 (2), A32 (−1) Thus, A−1 35. We use the Gauss-Jordan 1 −1 3 1 0 4 −3 13 0 1 1 1400 1 −1 3 10 3 1 1 −4 1 ∼ 0 0 0 1 −7 2 1/5 7/5 −1 = −3/10 −3/5 1/2 . 1/2 1 −1/2 method to find A−1 : 0 1 −1 3 1 0 ∼ 0 11 1 0 21 0 104 4 0 ∼ 0 1 1 −1 001 1. A12 (−4), A13 (−1) 2. A23 (−2) 00 1 2 1 0 ∼ 0 01 0 −3 1 0 5 −4 1 0 ∼ −7 2 −1 1 −4 −1 3. M3 (−1) −1 3 1 00 1 1 −4 1 0 0 −1 7 −2 1 1 0 0 25 −7 4 010 3 −1 1 . 0 0 1 −7 2 −1 4. A21 (1) 5. A31 (−4), A32 (−1) 182 Thus, 25 −7 4 1 . = 3 −1 −7 2 −1 A−1 Now xi = A−1 ei for each i. So 25 x1 = A−1 e1 = 3 , −7 −7 x2 = A−1 e2 = −1 , 2 4 x3 = A−1 e3 = 1 . −1 36. We have xi = A−1 bi , where A−1 = − 1 39 −2 −5 −7 2 . Therefore, x1 = A−1 b1 = − 1 39 −2 −5 −7 2 x2 = A−1 b2 = − and x3 = A−1 b3 = − 1 39 1 39 1 2 =− −2 −5 −7 2 −2 −5 −7 2 −2 5 1 39 4 3 =− = 1 39 −23 −22 =− 1 39 1 39 −12 −3 −21 24 = 12 3 = 1 13 4 1 1 39 23 22 , 21 −24 = 1 13 = 1 39 , 7 −8 . 37. (a): We have (A−1 B )(B −1 A) = A−1 (BB −1 )A = A−1 In A = A−1 A = In and (B −1 A)(A−1 B ) = B −1 (AA−1 )B = B −1 In B = B −1 B = In . Therefore, (B −1 A)−1 = A−1 B. (b): We have (A−1 B )−1 = B −1 (A−1 )−1 = B −1 A, as required. 38. We prove this by induction on k . If k = 0, then Ak = A0 = In and S −1 D0 S = S −1 In S = S −1 S = In . Thus, Ak = S −1 Dk S when k = 0. Now assume that Ak−1 = S −1 Dk−1 S . We wish to show that Ak = S −1 Dk S . We have Ak = AAk−1 = (S −1 DS )(S −1 Dk−1 S ) = S −1 D(SS −1 )Dk−1 S = S −1 DIn Dk−1 S = S −1 DDk−1 S = S −1 Dk S, as required. 39. (a): We reduce A to the identity matrix: 47 −2 5 1 ∼ 1 −2 7 4 5 2 ∼ 1 0 7 4 17 2 3 ∼ 1 0 7 4 1 4 ∼ 1 0 0 1 . 183 1. M1 ( 1 ) 4 2 3. M2 ( 17 ) 2. A12 (2) 4. A21 (− 7 ) 4 The elementary matrices corresponding to these row operations are E1 = 1 4 0 0 1 , E2 = 1 2 0 1 , E3 = 1 0 0 , 2 17 1 −7 4 0 1 E4 = . We have E4 E3 E2 E1 A = I2 , so that − − − − A = E 1 1 E2 1 E3 1 E 4 1 = 4 0 0 1 1 −2 1 0 0 1 0 1 0 17 2 7 4 1 , − which is the desired expression since Ei 1 is an elementary matrix for each i. (b): We can reduce A to upper triangular form by the following elementary row operation: 4 −2 7 5 1 ∼ 4 0 7 17 2 . 1. A12 ( 1 ) 2 Therefore we have the multiplier m12 = − 1 . Hence, setting 2 L= 10 −1 1 2 and U= 4 0 7 , 17 2 we have the LU factorization A = LU , which can be easily verified by direct multiplication. 40. (a): We reduce 210 1 2 0 0 0 3 004 A to the identity 0 12 012 1 ∼ 4 0 0 3 00 100 50 1 0 ∼ 0 0 1 004 matrix: 00 0 02 ∼ 3 4 43 0 06 4 ∼ 3 3 1 200 1 0 −3 0 0 3 0 ∼ 0 0 3 4 0 0 043 0 100 0 100 010 070 1 0 4 ∼ 001 001 3 000 0 0 0 −7 3 2. A12 (−2) 3. M2 (− 1 ) 3 6. A34 (−4) 1. P12 M4 (− 3 ) 7 The elementary matrices corresponding to 0100 10 1 0 0 0 −2 1 E1 = 0 0 1 0 , E2 = 0 0 0001 00 7. 2 1 0 0 0 08 4 ∼ 3 1 4. A21 (−2) 8. 0 0 3 4 0 1 040 ∼ 4 0 3 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 3 4 0 0 4 3 0 0 . 0 1 5. M3 ( 1 ) 3 A43 (− 4 ) 3 these row operations are 1 000 00 1 0 0 , E3 = 0 − 3 0 0 , 0 10 0 1 0 01 0 001 1 −2 0 0 0 1 0 0 E4 = 0 0 1 0 0 001 184 10 0 1 E5 = 00 00 0 0 , 0 1 0 0 1 3 0 0 00 1 0 0 , 0 1 0 0 −4 1 1 0 E6 = 0 0 1 0 E7 = 0 0 0 0 0 0 , 1 0 0 −3 7 0 1 0 0 1 0 E8 = 0 0 0 1 0 0 0 0 0 0 4 . 1 −3 0 1 We have E8 E7 E6 E5 E4 E3 E2 E1 A = I4 so that − − − − − − − − A = E1 1 E2 1 E 3 1 E4 1 E 5 1 E6 1 E7 1 E 8 1 0100 1000 1 0 0 0 2 1 0 0 = 0 0 1 0 0 0 1 0 0001 0001 1000 10 0 1 0 0 0 1 ··· 0 0 3 0 0 0 0001 00 1 00 0 −3 0 0 01 0 00 00 0 0 1 0 41 0 12 0 0 1 0 0 0 1 00 1 0 0 0 0 1 0 0 0 0 1 0 00 0 0 ··· 1 0 01 10 0 0 0 1 0 0 0 00 −7 3 0 0 1 0 0 0 4 , 3 1 − which is the desired expression since Ei 1 is an elementary matrix for each i. (b): We can reduce A to upper 21 1 2 0 0 00 triangular form 2 00 0 010 ∼ 3 4 0 43 0 by the following elementary row operations: 210 0 100 3 0 0 020 3 0 2 2 ∼ . 0 0 3 4 034 043 0 0 0 −7 3 1. A12 (− 1 ) 2 Therefore, the nonzero multipliers are m12 = L= 1 1 2 0 0 0 1 0 0 0 0 1 4 3 1 2 0 0 0 1 2. A34 (− 4 ) 3 and m34 = 4 . Hence, setting 3 and 2 0 U = 0 0 1 3 2 0 0 0 0 0 0 , 3 4 0 −7 3 we have the LU factorization A = LU , which can be easily verified by direct multiplication. 41. (a): We reduce A to the identity matrix: 1 −1 2 1 −1 2 3 0 0 1 −1 2 1 −1 2 4 1 2 3 0 1 −1 2 −1 ∼ 0 2 −1 ∼ 0 2 −1 ∼ 0 1 −1 ∼ 0 2 2 1 −1 2 3 0 0 0 3 −6 0 3 −6 0 0 −9 2 1 −1 2 1 6 1 1 −2 ∼ 0 ∼ 0 0 0 1 0 5 3 10 0 0 1 2 7 8 1 −1 ∼ 0 1 −1 ∼ 0 2 2 0 00 1 0 1 0 1 0 0 0 . 1 185 1. P13 3. M2 ( 1 ) 2 2. A13 (−3) 7. A31 (− 3 ) 2 6. A21 (1) The elementary matrices corresponding to 001 1 E1 = 0 1 0 , E 2 = 0 100 −3 10 0 1 0 , E6 = 0 E5 = 0 1 0 0 0 −2 9 4. A23 (−3) 5. M3 (− 2 ) 9 8. A32 ( 1 ) 2 these row operations are 100 00 1 1 0 , E3 = 0 1 0 , E4 = 0 2 01 0 001 1 10 1 0 −3 2 1 0 , E7 = 0 1 0 , E8 = 0 01 00 1 0 00 1 0 −3 1 00 1 1 . 2 01 We have E8 E7 E6 E5 E4 E3 E2 E1 A = I3 so that − − − − − − − − A = E1 1 E2 1 E 3 1 E4 1 E 5 1 E6 1 E7 1 E 8 1 001 100 100 = 0 1 0 0 1 0 0 2 0 100 301 001 10 0 1 −1 0 0 1 ··· 0 1 0 0 0 0 −9 2 100 0 1 0 ··· 031 1 0 103 2 0 0 1 0 0 1 0 001 0 0 1 −1 , 2 0 1 − which is the desired expression since Ei 1 is an elementary matrix for each i. (b): We can reduce A to upper triangular form by the following elementary row operations: 30 0 3 0 0 3 0 0 1 2 0 2 −1 ∼ 0 2 −1 ∼ 0 2 −1 . 3 1 −1 2 0 −1 2 00 2 1. A13 (− 1 ) 3 2. A23 ( 1 ) 2 Therefore, the nonzero multipliers are m13 = 1 and m23 = − 1 . Hence, setting 3 2 30 0 1 00 1 0 and U = 0 2 −1 , L= 0 1 1 3 −2 1 00 3 2 we have the LU factorization A = LU , which can be verified by direct multiplication. 42. (a): We reduce A to the identity matrix: −2 −3 1 1 42 1 1 4 2 ∼ −2 −3 1 0 53 0 53 14 2 14 2 5 6 ∼ 0 1 −8 ∼ 0 1 −8 0 0 45 00 1 14 2 14 2 14 2 3 4 ∼ 0 5 5 ∼ 0 5 5 ∼ 0 1 −8 0 5 −3 0 1 −8 05 5 1 0 34 1 0 34 100 7 8 9 ∼ 0 1 −8 ∼ 0 1 0 ∼ 0 1 0 . 00 1 001 001 2 186 1. P12 6. 2. A12 (2) 1 M3 ( 45 ) 3. A23 (−1) 7. A21 (−4) The elementary matrices corresponding to these row 010 1 E1 = 1 0 0 , E 2 = 2 001 0 10 E4 = 0 0 01 1 −4 1 E7 = 0 0 0 4. P23 8. A32 (8) 5. A23 (−5) 9. A31 (−34) operations are 00 1 00 1 0 , E3 = 0 1 0 , 01 0 −1 1 0 1 1 , E5 = 0 0 0 0 1 0 , E8 = 0 1 0 100 00 1 0 , E6 = 0 1 0 , 1 −5 1 0 0 45 00 1 0 −34 1 8 , E9 = 0 1 0 . 01 00 1 We have E9 E8 E7 E6 E5 E4 E3 E2 E1 A = I3 so that − − − − − − − − − A = E1 1 E 2 1 E3 1 E 4 1 E5 1 E6 1 E 7 1 E8 1 E 9 1 010 100 100 = 1 0 0 −2 1 0 0 1 0 001 001 011 100 100 1 ··· 0 1 0 0 1 0 0 051 0 0 45 0 00 0 1 ··· 10 40 10 0 1 1 0 0 1 −8 0 01 00 1 0 1 0 0 0 1 0 34 0 , 1 − which is the desired expression since Ei 1 is an elementary matrix for each i. (b): We can reduce A to upper triangular form by the following elementary row operations: −2 −3 1 −2 −3 1 −2 −3 1 2 1 5 5 5 5 1 4 2 ∼ 0 ∼ 0 . 2 2 2 2 0 53 0 53 0 0 −2 Therefore, the nonzero multipliers are m12 = − 1 and m23 = 2. Hence, setting 2 00 L = −1 1 0 2 021 1 and −2 −3 1 5 5 U = 0 , 2 2 0 0 −2 we have the LU factorization A = LU , which can be verified by direct multiplication. 43. (a): Note that (A+B )3 = (A+B )2 (A+B ) = (A2 +AB +BA+B 2 )(A+B ) = A3 +ABA+BA2 +B 2 A+A2 B +AB 2 +BAB +B 3 . 187 (b): We have (A−B )3 = (A−B )2 (A−B ) = (A2 −AB −BA+B 2 )(A−B ) = A3 −ABA−BA2 +B 2 A−A2 B +AB 2 +BAB −B 3 . (c): The answer is 2k , because each term in the expansion of (A + B )k consists of a string of k matrices, each of which is either A or B (2 possibilities for each matrix in the string). Multiplying the possibilities for each position in the string of length k , we get 2k different strings, and hence 2k different terms in the expansion of (A + B )k . 44. We claim that A 0 0 B −1 −1 A−1 0 = 0 B . To see this, simply note that A 0 0 B −1 0 B = In 0 0 Im = In+m A−1 0 and A−1 0 A 0 0 B −1 = In 0 0 Im = In+m . 0 B 45. For a 2 × 4 matrix, the leading ones can occur in 6 different positions: 1∗∗∗ 01∗∗ , 1∗∗∗ 001∗ For a 3 × 4 matrix, 1 0 0 , 1∗∗∗ 0001 the leading ones can ∗∗∗ 1∗ 1 ∗ ∗ , 0 1 01∗ 00 , 0 0 1∗∗ 01∗ , 0 0 1∗∗ 001 , 0 0 0 0 1∗ 01 occur in 4 different positions: ∗∗ 1∗∗∗ 01∗∗ ∗ ∗ , 0 0 1 ∗ , 0 0 1 ∗ 01 0001 0001 For a 4 × 6 matrix, the leading ones can occur in 15 different positions: 1∗∗∗∗∗ 1∗∗∗ 0 1 ∗ ∗ ∗ ∗0 1 ∗ ∗ 0 0 1 ∗ ∗ ∗ , 0 0 1 ∗ 0001∗∗ 0000 1∗∗ 0 1 ∗ 0 0 0 000 1∗∗ 0 0 1 0 0 0 000 ∗∗∗ ∗ ∗ ∗ , 1 ∗ ∗ 001 ∗∗∗ ∗ ∗ ∗ , 0 1 ∗ 001 ∗ ∗ ∗ 1 ∗ 1∗∗∗∗ ∗0 1 ∗ ∗ ∗ , ∗0 0 1 ∗ ∗ ∗ 00000 1∗∗∗∗∗ ∗ ∗0 1 ∗ ∗ ∗ ∗ , ∗0 0 0 1 ∗ ∗ 00001∗ 1 1∗∗∗∗∗ 1∗∗∗∗∗ 0 1 ∗ ∗ ∗ ∗0 0 1 ∗ ∗ ∗ , 0 0 0 0 1 ∗0 0 0 1 ∗ ∗ 000001 00001∗ 01∗∗∗∗ 1∗∗∗∗∗ 0 0 0 1 ∗ ∗0 0 1 ∗ ∗ ∗ , 0 0 0 0 1 ∗0 0 0 1 ∗ ∗ 00001∗ 000001 1∗∗∗∗ 0 0 1 ∗ ∗ , 0 0 0 1 ∗ 00000 01∗∗∗ 0 0 1 ∗ ∗ , 0 0 0 1 ∗ 00000 , ∗ ∗ , ∗ 1 ∗ ∗ , ∗ 1 188 0 0 0 0 1∗∗∗∗ 0 1 ∗ ∗ ∗ , 0 0 0 1 ∗ 00001 0 0 0 0 1∗ 00 00 00 ∗∗∗ 1 ∗ ∗ , 0 1 ∗ 001 001∗∗∗ 0 0 0 1 ∗ ∗ 0 0 0 0 1 ∗ 000001 For an m × n matrix with m ≤ n, the answer is the binomial coefficient C (n, m) = n m = n! . m!(n − m)! This represents n “choose” m, which is the number of ways to choose m columns from the n columns of the matrix in which to put the leading ones. This choice then determines the structure of the matrix. 46. The inverse of A10 is B 5 . To see this, we use the fact that A2 B = In = BA2 as follows: A10 B 5 = A8 (A2 B )B 4 = A8 In B 4 = A8 B 4 = A6 (A2 B )B 3 = A6 In B 3 = A6 B 3 = A4 (A2 B )B 2 = A4 In B 2 = A4 B 2 = A2 (A2 B )B = A2 In B = A2 B = In and B 5 A10 = B 4 (BA2 )A8 = B 4 In A8 = B 4 A8 = B 3 (BA2 )A6 = B 3 In A6 = B 3 A6 = B 2 (BA2 )A4 = B 2 In A4 = B 2 A4 = B (BA2 )A2 = BIn A2 = BA2 = In . Solutions to Section 3.1 True-False Review: 1. TRUE. Let A = a0 bc . Then det(A) = ac − b0 = ac, which is the product of the elements on the main diagonal of A. abc 2. TRUE. Let A = 0 d e . Using the schematic of Figure 3.1.1, we have 00f det(A) = adf + be0 + c00 − 0dc − 0ea − f 0b = adf, which is the product of the elements on the main diagonal of A. 3. FALSE. The volume of this parallelepiped is determined by the absolute value of det(A), since det(A) could very well be negative. 4. TRUE. There are 12 of each. The ones of even parity are (1, 2, 3, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 4, 3), (2, 4, 3, 1), (2, 3, 1, 4), (3, 1, 2, 4), (3, 2, 4, 1), (3, 4, 1, 2), (4, 1, 3, 2), (4, 2, 1, 3), (4, 3, 2, 1), and the others are all of odd parity. 5. FALSE. Many examples are possible here. If we take A = det(B ) = 0, but A + B = I2 , and det(I2 ) = 1 = 0. 1 0 0 0 and B = 0 0 0 1 , then det(A) = 189 6. FALSE. Many examples are possible here. If we take A = 12 34 , for example, then det(A) = 1 · 4 − 2 · 3 = −2 < 0, even though all elements of A are positive. 7. TRUE. In the summation that defines the determinant, each term in the sum is a product consisting of one element from each row and each column of the matrix. But that means one of the factors in each term will be zero, since it must come from the row containing all zeros. Hence, each term is zero, and the summation is zero. 8. TRUE. If the determinant of the 3 × 3 matrix [v1 , v2 , v3 ] is zero, then the volume of the parallelepiped determined by the three vectors is zero, and this means precisely that the three vectors all lie in the same plane. Problems: 1. σ (2, 1, 3, 4) = (−1)N (2,1,3,4) = (−1)1 = −1, odd. 2. σ (1, 3, 2, 4) = (−1)N (1,3,2,4) = (−1)1 = −1, odd. 3. σ (1, 4, 3, 5, 2) = (−1)N (1,4,3,5,2) = (−1)4 = 1, even. 4. σ (5, 4, 3, 2, 1) = (−1)N (5,4,3,2,1) = (−1)10 = 1, even. 5. σ (1, 5, 2, 4, 3) = (−1)N (1,5,2,4,3) = (−1)4 = 1, even. 6. σ (2, 4, 6, 1, 3, 5) = (−1)N (2,4,6,1,3,5) = (−1)6 = 1, even. 7. det(A) = a11 a21 a12 a22 8. det(A) = 1 −1 2 3 = 1 · 3 − (−1)2 = 5. 9. det(A) = 2 −1 6 −3 = 2(−3) − (−1)6 = 0. 10. det(A) = −4 −1 11. det(A) = 1 −1 0 2 3 6 0 2 −1 12. det(A) = 2 4 9 13. det(A) = 0 0 2 0 −4 1 −1 5 −7 = σ (1, 2)a11 a22 + σ (2, 1)a12 a21 = a11 a22 − a12 a21 . 10 8 1 2 5 5 3 1 = −4 · 8 − 10(−1) = −22. = 1 · 3(−1) + (−1)6 · 0 + 0 · 2 · 2 − 0 · 3 · 0 − 6 · 2 · 1 − (−1)(−1)2 = −17. = 2 · 2 · 1 + 1 · 3 · 9 + 5 · 4 · 5 − 5 · 2 · 9 − 3 · 5 · 2 − 1 · 1 · 4 = 7. = 0(−4)(−7) + 0 · 1(−1) + 2 · 0 · 5 − 2(−4)(−1) − 1 · 5 · 0 − (−7)0 · 0 = −8. 123 4 056 7 = 400, since of the 24 terms in the expression (3.1.3) for the determinant, 14. det(A) = 008 9 0 0 0 10 only the term σ (1, 2, 3, 4)a11 a22 a33 a44 = 400 contains all nonzero entries. 190 0020 5000 15. det(A) = = −60, since of the 24 terms in the expression (3.1.3) for the determinant, 0003 0200 only the term σ (3, 1, 4, 2)a13 a21 a34 a42 contains all nonzero entries, and since σ (3, 1, 4, 2) = −1, we obtain σ (3, 1, 4, 2)a13 a21 a34 a42 = (−1) · 2 · 5 · 3 · 2 = −60. 16. π √ 2 17. 2 1 3 18. 3 2 −1 π2 2π = 2π 2 − 3 −1 4 1 1 6 2 6 1 −1 1 4 √ 2π 2 = (2 − √ 2)π 2 . = (2)(4)(6) + (3)(1)(3) + (−1)(1)(1) − (3)(4)(−1) − (1)(1)(2) − (6)(1)(3) = 48. = (3)(1)(4) + (2)(−1)(−1) + (6)(2)(1) − (−1)(1)(6) − (1)(−1)(3) − (4)(2)(2) = 19. 236 0 1 2 = (2)(1)(0) + (3)(2)(1) + (6)(0)(5) − (1)(1)(6) − (5)(2)(2) − (0)(0)(3) = −20. 150 √ e2 e−1 √ √π √ √ 20. 67 1/30 2001 = ( π )(1/30)(π 3 )+(e2 )(2001)(π )+(e−1 )( 67)(π 2 )−(π )(1/30)(e−1 )−(π 2 )(2001)( π )− π π2 π3 √ (π 3 )( 67)(e2 ) ≈ 9601.882. 19. e2t e3t e−4t 2t 3t 3e −4e−4t = (e2t )(3e3t )(16e−4t )+(e3t )(−4e−4t )(4e2t )+(e−4t )(2e2t )(9e3t )−(4e2t )(3e3t )(e−4t )− 21. 2e 2t 3t 4e 9e 16e−4t 3t −4t 2t (9e )(−4e )(e ) − (16e−4t )(2e2t )(e3t ) = 42et . 22. y1 − y1 + 4y1 − 4y1 = 8 sin 2x + 4 cos 2x − 8 sin 2x − 4 cos 2x = 0, y2 − y2 + 4y2 − 4y2 = −8 cos 2x + 4 sin 2x + 8 cos 2x − 4 sin 2x = 0, y3 − y3 + 4y3 − 4y3 = ex − ex + 4ex − 4ex = 0. y 1 y2 y3 cos 2x sin 2x ex y1 y2 y3 = −2 sin 2x 2 cos 2x ex y 1 y2 y 3 −4 cos 2x −4 sin 2x ex = 2ex cos2 2x − 4ex sin 2x cos 2x + 8ex sin2 2x + 8ex cos2 2x + 4ex sin 2x cos 2x + 2ex sin2 2x = 10ex . 23. (a): y1 − y1 − y1 + y1 = ex − ex − ex + ex = 0, y2 − y2 − y2 + y2 = sinh x − cosh x − sinh x + cosh x = 0, y3 − y3 − y3 + y3 = cosh x − sinh x − cosh x + sinh x = 0. y1 y1 y1 y2 y2 y2 y3 y3 y3 = ex ex ex cosh x sinh x sinh x cosh x cosh x sinh x 191 = ex sinh2 x + ex cosh2 x + ex sinh x cosh x − ex sinh2 x − ex cosh2 x − ex sinh x cosh x = 0. (b): The formulas we need are cosh x = ex + e−x 2 and sinh x = ex − e−x . 2 Adding the two equations, we find that cosh x + sinh x = ex , so that −ex + cosh x + sinh x = 0. Therefore, we may take d1 = −1, d2 = 1, and d3 = 1. 24. (a): S4 = {1, 2, 3, 4} (1, 2, 3, 4) (1, 2, 4, 3) (1, 3, 2, 4) (1, 3, 4, 2) (1, 4, 2, 3) (1, 4, 3, 2) (2, 1, 3, 4) (2, 1, 4, 3) (2, 3, 1, 4) (2, 3, 4, 1) (2, 4, 1, 3) (2, 4, 3, 1) (3, 1, 2, 4) (3, 1, 4, 2) (3, 2, 1, 4) (3, 2, 4, 1) (3, 4, 1, 2) (3, 4, 2, 1) (4, 1, 2, 3) (4, 1, 3, 2) (4, 2, 1, 3) (4, 2, 3, 1) (4, 3, 1, 2) (4, 3, 2, 1) (b): N (1, 2, 3, 4) = 0, σ (1, 2, 3, 4) = 1, even; N (1, 2, 4, 3) = 1, σ (1, 2, 4, 3) = −1, odd; N (1, 3, 2, 4) = 1, σ (1, 3, 2, 4) = −1, odd; N (1, 3, 4, 2) = 2, σ (1, 3, 4, 2) = 1, even; N (1, 4, 2, 3) = 2, σ (1, 4, 2, 3) = 1, even; N (1, 4, 3, 2) = 3, σ (1, 4, 3, 2) = −1, odd; N (2, 1, 3, 4) = 1, σ (2, 1, 3, 4) = −1, odd; N (2, 1, 4, 3) = 2, σ (2, 1, 4, 3) = 1, even; N (2, 3, 1, 4) = 2, σ (2, 3, 1, 4) = 1, even; N (2, 3, 4, 1) = 3, σ (2, 3, 4, 1) = −1, odd; N (2, 4, 1, 3) = 3, σ (2, 4, 1, 3) = −1, odd; N (2, 4, 3, 1) = 4, σ (2, 4, 3, 1) = 1, even; N (3, 1, 2, 4) = 2, σ (3, 1, 2, 4) = 1, even; N (3, 1, 4, 2) = 3, σ (3, 1, 4, 2) = −1, odd; N (3, 2, 1, 4) = 3, σ (3, 2, 1, 4) = −1, odd; N (3, 2, 4, 1) = 4, σ (3, 2, 4, 1) = 1, even; N (3, 4, 1, 2) = 4, σ (3, 4, 1, 2) = 1, even; N (3, 4, 2, 1) = 5, σ (3, 4, 2, 1) = −1, odd; N (4, 1, 2, 3) = 3, σ (4, 1, 2, 3) = −1, odd; N (4, 1, 3, 2) = 4, σ (4, 1, 3, 2) = 1, even; N (4, 2, 1, 3) = 4, σ (4, 2, 1, 3) = 1, even; N (4, 2, 3, 1) = 5, σ (4, 2, 3, 1) = −1, odd; N (4, 3, 1, 2) = 5, σ (4, 3, 1, 2) = −1, odd; N (4, 3, 2, 1) = 6, σ (4, 3, 2, 1) = 1, even. (c): det(A) = a11 a22 a33 a44 − a11 a22 a34 a43 − a11 a23 a32 a44 + a11 a23 a34 a42 + a11 a24 a32 a43 − a11 a24 a33 a42 − a12 a21 a33 a44 + a12 a21 a34 a43 + a12 a23 a31 a44 − a12 a23 a34 a41 − a12 a24 a31 a43 + a12 a24 a33 a41 + a13 a21 a32 a44 − a13 a21 a34 a42 − a13 a22 a31 a44 + a13 a22 a34 a41 + a13 a24 a31 a42 − a13 a24 a32 a41 − a14 a21 a32 a43 + a14 a21 a33 a42 + a14 a22 a31 a43 − a14 a22 a33 a41 − a14 a23 a31 a42 + a14 a23 a32 a41 192 25. det(A) = 1 −1 0 1 3 025 2 103 9 −1 2 1 1 · 0 · 0 · 1 − 1 · 0 · 2 · 3 − 1 · 1 · 2 · 1 + 1 · 1 · 2 · 5 + 1(−1)2 · 3 − 1(−1)0 · 5 −3(−1)0 · 1 + 3(−1)2 · 3 + 3 · 1 · 0 · 1 − 3 · 1 · 2 · 1 − 3(−1)0 · 3 + 1(−1)0 · 1 +2(−1)2 · 1 − 2(−1)2 · 5 − 2 · 0 · 0 · 1 + 2 · 0 · 2 · 1 + 2(−1)0 · 5 − 2(−1)2 · 1 −9(−1)2 · 3 + 9(−1)0 · 5 + 9 · 0 · 0 · 3 − 9 · 0 · 0 · 1 − 9 · 1 · 0 · 5 + 9 · 1 · 2 · 1 = 70. = 26. det(A) = 1 3 2 −2 1 0 1 1 −2 3 3 1 2 3 5 −2 1 · 1 · 1(−2) − 1 · 1 · 5 · 2 − 1 · 3(−2)(−2) + 1 · 3 · 5 · 3 + 1 · 3(−2)2 − 1 · 3 · 1 · 3 −3 · 1 · 1(−2) + 3 · 1 · 5 · 2 + 3 · 3 · 0(−2) − 3 · 3 · 5 · 1 − 3 · 3 · 0 · 2 + 3 · 3 · 1 · 1 +2 · 1(−2)(−2) − 2 · 1 · 5 · 3 − 2 · 1 · 0(−2) + 2 · 1 · 5 · 1 + 2 · 3 · 0 · 3 − 2 · 3(−2)1 −(−2)1(−2)2 + (−2)1 · 1 · 3 + (−2)1 · 0 · 2 − (−2) · 1 · 1 · 1 − (−2)3 · 0 · 3 + (−2)3(−2)1 = 0. = 27. 0123 2034 det(A) = 3405 4560 = 0·0·0·0−0·0·6·5−0·4·3·0+0·4·6·4+0·5·3·5−0·5·0·4 −2 · 1 · 0 · 0 + 2 · 1 · 6 · 5 + 2 · 4 · 2 · 0 − 2 · 4 · 6 · 3 − 2 · 5 · 2 · 5 + 2 · 5 · 0 · 3 +3 · 1 · 3 · 0 − 3 · 1 · 6 · 4 − 3 · 0 · 2 · 0 + 3 · 0 · 6 · 3 + 3 · 5 · 2 · 4 − 3 · 5 · 3 · 3 −4 · 1 · 3 · 5 + 4 · 1 · 0 · 4 + 4 · 0 · 2 · 5 − 4 · 0 · 0 · 3 − 4 · 4 · 2 · 4 + 4 · 4 · 3 · 3 = −315. 28. In evaluating det(A) with the expression (3.1.3), observe that the only nonzero terms in the summation occur when p5 = 5. Such terms include the factor a55 = 7, which is multiplied by each corresponding term from the 4 × 4 determinant calculated in Problem 27. Therefore, by factoring a55 out of the expression for the determinant, we are left with the determinant of the corresponding 4 × 4 matrix appearing in Problem 27. Therefore, the answer here is 7 · (−315) = −2205. 29. (a): det(cA) = ca11 ca21 ca12 ca22 = (ca11 )(ca22 ) − (ca12 )(ca21 ) = c2 a11 a22 − c2 a12 a21 = c2 (a11 a22 − a12 a21 ) = c2 det(A). 193 (b): det(cA) σ (p1 , p2 , p3 , . . . , pn )ca1p1 ca2p2 ca3p3 · · · canpn = = cn σ (p1 , p2 , p3 , . . . , pn )a1p1 a2p2 a3p3 · · · anpn n = c det(A), where each summation above runs over all permutations σ of {1, 2, 3, . . . , n}. 30. a11 a25 a33 a42 a54 . All row and column indices are distinct, so this is a term of an order 5 determinant. Further, N (1, 5, 3, 2, 4) = 4, so that σ (1, 5, 3, 2, 4) = (−1)4 = +1. 31. a11 a23 a34 a43 a52 . This is not a possible term of an order 5 determinant, since the column indices are not distinct. 32. a13 a25 a31 a44 a42 . This is not a possible term of an order 5 determinant, since the row indices are not distinct. 33. a11 a32 a24 a43 a55 . This is a possible term of an order 5 determinant. N (1, 4, 2, 3, 5) = 2 =⇒ σ (1, 4, 2, 3, 5) = (−1)2 = +1. 34. a13 ap4 a32 a2q = a13 a2q a32 ap4 . We must choose p = 4 and q = 1 in order for the row and column indices to be distinct. N (3, 1, 2, 4) = 2 so that σ (3, 1, 2, 4) = (−1)2 = +1. 35. a21 a3q ap2 a43 = ap2 a21 a3q a43 . We must choose p = 1 and q = 4. N (2, 1, 4, 3) = 2 and σ (2, 1, 4, 3) = (−1)2 = +1. 36. a3q ap4 a13 a42 . We must choose p = 2 and q = 1. N (3, 4, 1, 2) = 4 and σ (3, 4, 1, 2) = (−1)4 = +1. 37. apq a34 a13 a42 . We must choose p = 2 and q = 1. N (3, 1, 4, 2) = 3 and σ (3, 1, 4, 2) = (−1)3 = −1. 38. (a): 123 = 1, 132 = 3 3 −1, 213 = −1, 231 = 1, 312 = 1, 321 = −1. 3 (b): Consider ijk a1i a2j a3k . The only nonzero terms arise when i, j, and k are distinct. Conse- i=1 j =1 k=1 quently, 3 3 3 ijk a1i a2j a3k = 123 a11 a22 a33 + 132 a11 a23 a32 + 213 a12 a21 a33 i=1 j =1 k=1 + 231 a12 a23 a31 + 312 a13 a21 a32 + 321 a13 a22 a31 = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a11 a23 a32 − a12 a21 a33 − a13 a22 a31 = det(A). 39. From the given term, we have N (n, n − 1, n − 2, . . . , 1) = 1 + 2 + 3 + · · · + (n − 1) = n(n − 1) , 2 because the series of (n − 1) terms is just an arithmetic series which has a first term of one, common difference of one, and last term (n − 1). Thus, σ (n, n − 1, n − 2, . . . , 1) = (−1)n(n−1)/2 . 194 Solutions to Section 3.2 True-False Review: 1. FALSE. The determinant of the matrix will increase by a factor of 2n . For instance, if A = I2 , then 20 det(A) = 1. However, det(2A) = det = 4, so the determinant in this case increases by a factor of 02 four. 2. TRUE. In both cases, the determinant of the matrix is multiplied by a factor of c. 3. TRUE. This follows by repeated application of Property (P8): det(A5 ) = det(AAAAA) = (detA)(detA)(detA)(detA)(detA) = (detA)5 . 4. TRUE. Since det(A2 ) = (detA)2 and since det(A) is a real number, det(A2 ) must be nonnegative. 5. FALSE. The matrix is not invertible if and only if its determinant, x2 y − xy 2 = xy (x − y ), is zero. For example, if x = y = 1, the matrix is not invertible since x = y ; however, neither x nor y is zero in this case. We conclude that the statement is false. 6. TRUE. We have det(AB ) = (detA)(detB ) = (detB )(detA) = det(BA). Problems: From this point on, P2 will denote an application of the property of determinants that states that if every element in any row (column) of a matrix A is multiplied by a number c, then the determinant of the resulting matrix is c det(A). 1. 1 23 2 64 3 −5 2 1 = 1 2 3 0 2 −2 0 −11 −7 2 =2 1. A12 (−2), A13 (−3) 2. 2 −1 4 3 21 −2 14 1 = 2 −1 4 3 21 0 08 2 −1 4 1 2 1 3 6 12 1 =− 4 −1 2 4 = −45 2. P2 1 −1/2 2 2 21 =2 3 0 08 1. A13 (1) 3. 2 1 1 −1 0 0 1 3 =2 0 0 1 2 3 0 1 −1 0 −11 −7 6 3 12 2 1 0 2. P2 2 =− 6 3 1 2 3 1 −1 0 −18 = 2(−18) = −36. 3. A23 (11) 1 −1/2 2 3 7/2 −5 =2 0 0 0 8 = 2 · 28 = 56. 3. A12 (−3) −1 0 0 2 5 9 6 15 36 3 = −5 · 9 = (−45)(−1) = 45. −1 0 0 26 13 14 195 1. P12 4. 0 1 −2 −1 0 3 2 −3 0 2. A12 (2), A13 (4) 1 0 −3 0 1 −2 2 −3 0 1 = 5. 3 5 2 7 1 9 −6 1 3 1 = 6 −2 9 −6 1 3 1 5 2 1. A31 (−1) 1 0 0 3 = 0 −3 1 −2 0 0 = 0. 3. A23 (3) 1 6 −2 0 −21 4 0 −11 7 2 = 6 −2 1 −10 0 −103 1 0 0 4 = 1 0 −3 0 1 −2 0 −3 6 2 = 2. A13 (−2) 1. P12 , P2 4. A23 (−1) 3. P2 3 = 1 6 −2 0 1 −10 0 −11 7 = −103. 2. A12 (−5), A13 (−2) 3. A32 (−2) 4. A24 (11) 6. 1 −1 2 4 3 124 −1 132 2 142 1 = 1 −1 2 4 0 4 −4 −8 0 0 5 6 0 3 0 −6 1 −1 2 4 0 1 −1 −2 = −12 0 0 1 0 0 0 5 6 4 1 −1 2 4 0 1 −1 −2 =3·4 0 0 5 6 0 1 0 −2 2 1 −1 2 4 0 1 −1 −2 = −12 0 0 1 0 0 0 0 6 1. A12 (−3), A13 (1), A14 (−2) 5 2. P2 3. P34 1 −1 2 4 0 1 −1 −2 = −12 0 1 0 −2 0 0 5 6 3 = −12 · 6 = −72. 4. A23 (−1) 5. A34 (−5) 7. Note that in the first step below, we extract a factor of 13 from the second row of the matrix, and we also extract a factor of 8 from the second column of the matrix. 2 26 2 1 32 104 56 40 1 4 26 −13 2 7 1 5 24 1 4 21 2 −1 = 13 · 8 27 2 7 1 5 −1 5 1 1 5 1 5 0 −9 0 −11 = −104 0 −3 0 −3 0 −6 −1 −6 3 5 = 312 1 0 0 0 5 1 5 1 0 1 0 0 −2 0 −1 0 1 2 = −104 2 2 2 5 −1 5 1 2 −1 7 2 7 4 1 4 1 5 1 5 0 1 0 1 = −104(−3) 0 −9 0 −11 0 −6 −1 −6 4 6 = 312 15 1 5 01 0 1 0 0 −1 0 00 0 −2 = 312(−1)(−2) = 624. 196 1. P2 2. P14 3. A12 (−2), A13 (−2), A14 (−2) 4. P2, P23 5. A23 (9), A24 (6) 6. P34 8. 0 1 −1 1 −1 0 11 1 −1 01 −1 −1 −1 0 1 −1 01 −1 0 11 =− 0 1 −1 1 −1 −1 −1 0 2 1 −1 0 1 0 −1 1 2 =− 0 0 −3 −3 0 0 0 3 1 −1 0 1 0 −1 1 2 =− 0 0 0 3 0 0 −3 −3 3 1. P13 1 −1 01 0 −1 12 =− 0 1 −1 1 0 −2 −1 1 1 4 3. A23 (1), A24 (−2) 2. A12 (1), A14 (1) = 9. 4. P34 9. 2 3 4 5 1 0 1 2 3 1 4 5 5 2 3 3 1 0 =− 1 2 1 2 3 4 5 12 3 4 0 1 −1 −7 =− 02 1 −2 03 1 2 3 1 0 =3 0 0 5 2 3 4 1 −1 −7 0 1 4 0 4 23 3 1 4 5 5 2 3 3 4 = 6 =3 1 0 =− 0 0 2 2 3 5 3 1 2 2 1 −2 1 −1 −7 12 3 4 0 1 −1 −7 00 3 12 00 4 23 1 0 0 0 2 3 4 1 −1 −7 0 1 4 0 0 7 = 3 · 1 · 1 · 1 · 7 = 21. 1. CP12 2. A13 (−1), A14 (−2) 3. P24 4. A23 (−2), A24 (−3) 5. P2 6. A34 (−4) 10. 2 −1 3 4 7 1 2 3 −2 4 8 6 6 −6 18 −24 −1 23 4 1 72 3 = −6 4 −2 8 6 −1 1 3 −4 =6 1 −2 −3 −4 0 9 5 7 =6 0 6 20 22 0 −1 0 −8 1 −2 −3 −4 0 −1 0 −8 = −6 0 6 20 22 0 9 5 7 1 −2 −3 −4 0 −1 0 −8 = −6 0 0 20 −26 0 0 5 −65 3 1 −2 −3 −4 0 −1 0 −8 =6 0 0 5 −65 0 0 20 −26 6 1 4 7 =6 1 −2 −3 −4 0 −1 0 −8 0 0 5 −65 0 0 0 234 2 1 −2 −3 −4 1 7 2 3 4 −2 8 6 −1 1 3 −4 5 = 6 · 1(−1) · 5 · 234 = −7020. 197 1. CP12 , P2 2. M1 (−1) 5. A23 (6), A24 (9) 6. P34 3. A12 (−1), A13 (−4), A14 (1) 4. P24 7. A34 (−4) 11. 7 −1 3 4 14 246 21 134 −7 458 1 −1 3 2 2 243 =7·2 3 132 −1 454 1 1 −1 3 2 0 4 −2 −1 = 14 0 0 −4 −3 0 3 8 6 3 6 = −56 1 −1 3 2 0 1 −10 −7 = 14 0 0 −4 −3 0 3 8 6 4 1 −1 3 2 0 1 −10 7 0 0 1 3/4 0 0 38 27 5 = 14 1 −1 3 2 0 1 −10 −7 = −56 0 0 1 3/4 0 0 0 −3/2 7 2. A12 (−2), A13 (−3), A14 (1) 1. P2 1 −1 3 2 0 4 −2 −1 = 14 0 4 −6 −4 0 3 8 6 2 5. A24 (−3) 6. P2 1 −1 3 2 0 1 −10 −7 0 0 −4 −3 0 0 38 27 = −56 · (−3/2) = 84. 3. A23 (−1) 4. A42 (−1) 7. A34 (−38) 12. 3 1 4 3 8 7 123 1 −1 0 1 8 −1 6 6 7 094 16 −1 8 12 1 3 1 =− 4 3 8 1 −1 0 1 7 123 8 −1 6 6 7 094 16 −1 8 12 1 0 2 =− 0 0 0 −1 4 −1 −1 −1 1 0 4 =− 0 0 0 1 −1 0 1 4 42 0 0 −1 4 2 0 0 3 −1 0 00 2 = −(1)(4)(−1)(3)(2) = 24. 1 0 3 =− 0 0 0 1. P12 1 2 1 4 0 0 0 0 2 4 7 4 1 0 2 1 4 2. A12 (−3), A13 (−4), A14 (−3), A15 (−8) 13. 2 3 14. −1 1 1 −1 15. 2 3 2 16. −1 23 5 −2 1 8 −2 5 = 1; invertible. = 0; not invertible. 6 −1 5 1 0 1 = 14; invertible. = −8; invertible. 1 −1 0 1 4 420 4 362 4 391 8 784 3. A23 (−1), A24 (−1), A25 (−2) 4. A34 (−1), A35 (−1) 198 17. 1 0 2 −1 3 −2 1 4 2 16 2 1 −3 4 0 1 0 2 −1 0 −2 −5 7 0 1 2 4 0 −3 2 1 = = −2 −5 7 1 24 −3 21 = 133; invertible. 11 1 1 02 0 2 0 0 −2 −2 02 2 0 = 2 0 2 0 −2 −2 2 2 0 = 16; invertible. 18. 11 1 1 −1 1 −1 1 1 1 −1 −1 −1 1 1 −1 19. 1 2 −3 5 −1 2 −3 6 2 3 −1 4 1 −2 3 −6 = 1 2 −3 5 1 −2 3 −6 =− 2 3 −1 4 1 −2 3 −6 1 2 = 0; not invertible. 1. M2 (−1) 2. P7 1k x1 b1 ,x= , and b = . According to Corollary k4 x2 b2 3.2.5, the system has unique solution if and only if det(A) = 0. But, det(A) = 4 − k 2 , so that the system has a unique solution if and only if k = ±2. 20. The system is Ax = b, where A = 1 2 k 1 2k 2 −k 1 = 2 −k 1 = (3k − 1)(k +4). Consequently, the system has an infinite 0 0 1 − 3k 3 61 number of solutions if and only if k = −4 or k = 1/3 (Corollary 3.2.5). 21. det(A) = 22. The given system is (1 − k )x1 + 2x2 + 2x1 + (1 − k )x2 + x1 + x2 + x3 x3 (2 − k )x3 = 0, = 0, = 0. The determinant of the coefficients as a function of k is given by det(A) = 1−k 2 1 2 1−k 1 1 1 2−k = −(1 + k )(k − 1)(k − 4). Consequently, the system has an infinite number of solutions if and only if k = −1, k = 1, or ,k = 4. 1k k1 11 and only if k = 0, 1. 23. det(A) = 0 1 1 = 1 + k − 1 − k 2 = k (1 − k ). Consequently, the system has a unique solution if 1 −1 2 3 1 4 = 1 · 1 · 3 + (−1)4 · 0 + 2 · 3 · 1 − 2 · 1 · 0 − 1 · 4 · 1 − (−1)3 · 3 = 14. 0 13 det(AT ) = det(A) = 14. det(−2A) = (−2)3 det(A) = −8 · 14 = −112. 24. det(A) = 199 1 −1 12 and B = . det(A) det(B ) = [3 · 1 − (−1)2][1 · 4 − 2(−2)] = 5 · 8 = 40. 2 3 −2 4 3 −2 det(AB ) = = 3 · 16 − (−2)(−4) = 40. Hence, det(AB ) = det(A) det(B ). −4 16 25. A = cosh x sinh x cosh y sinh y and B = . sinh x cosh x sinh y cosh y det(AB ) = det(A) det(B ) = (cosh2 x − sinh2 x)(cosh2 y − sinh2 y ) = 1 · 1 = 1. 26. A = 27. 32 1 6 4 −1 96 2 1 1 =3 2 3 28. 1 −3 2 −1 3 1 29. 1 + 3a 1 3 1 + 2a 1 2 2 20 1 7 13 2 1 4 −1 6 2 11 1 2 = 3 · 2 2 2 −1 33 2 1 −3 + 4 · 1 2 −1 + 4 · 2 3 1+4·3 1 = 1 = 1 1 2 1 1 2 1 7 13 = 11 27 3 13 3 3a 1 3 2 + 2a 1 2 0 020 3 1. P2 = 0. 1 7 13 2. P2 3. P7 2 = 0. 3 2 = 0+a 2 0 1 1 2 3 2 0 3 = a·0 = 0. 1. P5 2. P2, P7 30. B is obtained from A by the following elementary row operations: (1) M1 (4), (2) M2 (3), (3) P12 . Thus, det(B ) = det(A) · 4 · 3 · (−1) = −12. 31. B is obtained from AT by the following elementary row operations: (1) A13 (3), (2) M1 (−2). Since det(AT ) = det(A) = 1, and (1) leaves the determinant unchanged, we have det(B ) = −2. 32. B is obtained from A by the following operations: (1) Interchange the two columns, (2) M1 (−1), (3) A12 (−4). Now (3) leaves the determinant unchanged, and (1) and (2) each change the sign of the determinant. Therefore, det(B ) = det(A) = 1. 33. B is obtained from A by the following row operations: (1) A13 (5), (2) M2 (−4), (3) P12 , (4) P23 . Thus, det(B ) = det(A) · (−4) · (−1) · (−1) = (−6) · (−4) = 24. 34. B is obtained from A by the following operations: (1) M1 (−3), (2) A23 (−4), (3) P12 . Thus, det(B ) = det(A) · (−3) · (−1) = (−6) · (3) = −18. 35. B is obtained from AT by the following row operations: (1) M1 (2), (2) A32 (−1), (3) A13 (−1). Thus, det(B ) = det(AT ) · 2 = (−6) · (2) = 12. 36. We have det(AB T ) = det(A)det(B T ) = 5 · 3 = 15. 37. We have det(A2 B 5 ) = (detA)2 (detB )5 = 52 · 35 = 6075. 38. We have det((A−1 B 2 )3 ) = (det(A−1 B 2 ))3 = (detA−1 )(detB )2 3 = 12 ·3 5 3 = 9 5 3 = 729 = 5.832. 125 39. We have det((2B )−1 (AB )T ) = (det((2B )−1 ))(det(AB )T ) = 40. We have det((5A)(2B )) = (5 · 54 )(3 · 24 ) = 150, 000. 41. det(A)det(B ) det(2B ) = 5·3 3 · 24 = 5 . 16 3. P7 200 (a): The volume of the parallelepiped is given by |det(A)|. In this case, we have |det(A)| = |2 + 12k + 36 − 4k − 18 − 12| = |8 + 8k |. (b): NO. The answer does not change because the determinants of A and AT are the same, so the volume determined by the columns of A, which is the same as the volume determined by the rows of AT , is |det(AT )| = |det(A)|. 42. 1 −1 x 2 1 x2 4 −1 x3 = 1 −1 x 0 3 x2 − 2x 0 3 x3 − 4x = 1 −1 x 0 3 x(x − 2) . 0 0 x(x − 2)(x + 1) 1 −1 x 0 3 x2 − 2x 0 0 x3 − x2 − 2x = We see directly that the determinant will be zero if and only if x ∈ {0, −1, 2}. 43. αx − βy βx + αy βx − αy αx + βy = αx βx − αy βx αx + βy = αx βx βx αx = x2 α β β α + −βy αy βx − αy αx + βy αx −αy βx βy + + xy α β −βy αy + −α β − xy β α βx αx + −β α −βy αy + y2 = (x2 + y 2 ) α β β α + xy α β −α β + xy β α α β β α + xy α β −α β − xy α β −α β = (x2 + y 2 ) α β β α β α β −α = (x2 + y 2 ) α β −αy βy . 44. a1 + βb1 a2 + βb2 a3 + βb3 b1 + γc1 b2 + γc2 b3 + γc3 c1 + αa1 c2 + αa2 c3 + αa3 = a1 a2 a3 b1 + γc1 b2 + γc2 b3 + γc3 = a1 a2 a3 b1 b2 b3 c1 c2 c3 b1 = (1 + αβγ ) b2 b3 c1 + αa1 c2 + αa2 c3 + αa3 + β b1 βb2 βb3 b1 b2 b3 c1 c2 c3 a1 a2 a3 + αβγ c1 c2 c3 a1 a2 a3 b1 + γc1 b2 + γc2 b3 + γc3 c1 + αa1 c2 + αa2 c3 + αa3 . Now if the last expression is to be zero for all ai , bi , and ci , then it must be the case that 1 + αβγ = 0; hence, αβγ = −1. 45. Suppose A is a matrix with a row of zeros. We will use (P3) and (P7) to justify the fact that det(A) = 0. If A has more than one row of zeros, then by (P7), since two rows of A are the same, det(A) = 0. Assume 201 instead that only one row of A consists entirely of zeros. Adding a nonzero row of A to the row of zeros yields a new matrix B with two equal rows. Thus, by (P7), det(B ) = 0. However, B was obtained from A by adding a multiple of one row to another row, and by (P3), det(B ) = det(A). Hence, det(A) = 0, as required. 46. A is orthogonal, so AT = A−1 . Using the properties of determinants, it follows that 1 = det(In ) = det(AA−1 ) = det(A) det(A−1 ) = det(A) det(AT ) = det(A) det(A). Therefore, det(A) = ±1. 47. (a): From the definition of determinant we have: σ (p1 , p2 , p3 . . . , pn )a1p1 a2p2 a3p3 · · · anpn . det(A) = (47.1) n! If A is lower triangular, then aij = 0 whenever i < j , and therefore the only nonzero terms in (47.1) are those with pi ≤ i for all i. Since all the pi must be distinct, the only possibility is pi = i for all i with 1 ≤ i ≤ n, and so (47.1) reduces to the single term: det(A) = σ (1, 2, 3, . . . , n)a11 a22 a33 · · · ann . (b): det(A) = 2 −1 3 5 1 221 3 014 1 201 1 = 2 16 0 0 0 13 0 0 =− −1 −8 1 0 1 201 3 1. A41 (−5), A42 (−1), A43 (−4) −3 −11 3 0 0 020 −1 −8 1 0 1 201 2 0 0 13 =− −1 −8 1 2 4 0 13 0 2 16 0 −1 −8 1 1 20 2 = 00 00 10 01 2. A31 (−3), A32 (−2) 0 0 0 1 = −26. 3. P12 4. A21 (−16/13) 48. The problem stated in the text is wrong. It should state: “Use determinants to prove that if A is invertible and B and C are matrices with AB = AC , then det(B ) = det(C ).” To prove this, take the determinant of both sides of AB = AC to get det(AB ) = det(AC ). Thus, by Property P8, we have det(A)det(B ) = det(A)det(C ). Since A is invertible, det(A) = 0. Thus, we can cancel det(A) from the last equality to obtain det(B ) = det(C ). 49. det(S −1 AS ) = det(S −1 ) det(A) det(S ) = det(S −1 ) det(S ) det(A) = det(S −1 S ) det(A) = det(In ) det(A) = det(A). 50. No. If A were invertible, then det(A) = 0, so that det(A3 ) = det(A) det(A) det(A) = 0. 51. Let E be an elementary matrix. There are three different possibilities for E . (a): E permutes two rows: Then E is obtained from In by interchanging two rows of In . Since det(In ) = 1 and using Property P1, we obtain det(E ) = −1. 202 (b): E adds a multiple of one row to another: Then E is obtained from In by adding a multiple of one row of In to another. Since det(In ) = 1 and using Property P3, we obtain det(E ) = +1. (c): E scales a row by k : Then E is obtained from In by multiplying a row of In by k . Since det(In ) = 1 and using Property P2, we obtain det(E ) = k . 52. We have x x1 x2 0= y y1 y2 1 1 1 = xy1 + yx2 + x1 y2 − x2 y1 − xy2 − yx1 , which can be rewritten as x(y1 − y2 ) + y (x2 − x1 ) = x2 y1 − x1 y2 . Setting a = y1 − y2 , b = x2 − x1 , and c = x2 y1 − x1 y2 , we can express this equation as ax + by = c, the equation of a line. Moreover, if we substitute x1 for x and y1 for y , we obtain a valid identity. Likewise, if we substitute x2 for x and y2 for y , we obtain a valid identity. Therefore, we have a straight line that includes the points (x1 , y1 ) and (x2 , y2 ). 53. 1 x x2 1 y y2 1 z z2 1 = 1 x x2 0 y − x (y − x)(y + x) 0 z − x (z + x)(z − x) 1x x2 = (y − x)(z − x) 0 1 y + x 0 1 z+x 2 1x x2 = (y − x)(z − x) 0 1 y + x 0 0 z−y 3 1. A12 (−1)A13 (−1) = (y − x)(z − x)(z − y ) = (y − z )(z − x)(x − y ). 2. P2 3. A23 (−1) 54. Since A is an n × n skew-symmetric matrix, AT = −A; thus, det(AT ) = det(−A) = (−1)n det(A) = − det(A) since n is given as odd. But by P4, det(AT ) = det(A), so det(A) = − det(A) or det(A) = 0. 55. Solving b = c1 a1 + c2 a2 + · · · + cn an for c1 a1 yields c1 a1 = b − c2 a2 − · · · − cn an . Consequently, det(Bk ) can be written as det(Bk ) = det([a1 , a2 , . . . , ak−1 , b, ak+1 , . . . , an ]) = det([a1 , a2 , . . . , ak−1 , (c1 a1 + c2 a2 + · · · + cn an ), ak+1 , . . . , an ]) = c1 det([a1 , . . . , ak−1 , a1 , ak+1 , . . . , an ]) + c2 det([a1 , . . . , ak−1 , a2 , ak+1 , . . . , an ]) + · · · + ck det([a1 , . . . , ak−1 , ak , ak+1 , . . . , an ]) + · · · + cn det([a1 , . . . , ak−1 , an , ak+1 , . . . , an ]). Now by P7, all except the k th determinant are zero since they have two equal columns, so that we are left with det(Bk ) = ck det(A). 58. Using technology we find that: 1 2 3 4 a 2 1 2 3 4 3 2 1 2 3 4a 34 23 12 21 = −192 + 88a − 8a2 = −8(a − 3)(a − 8). 203 Consequently, the matrix is invertible provided a = 3, 8. 59. Using technology we find that: 1−k 3 3 4 2−k 4 1 1 −1 − k = −(k − 6)(k + 2)2 . Consequently, the system has an infinite number of solutions if and only if = 6, −2. k −5 4 1 1 . This system has solution set k = 6: In this case, the system is B x = 0, where B = 3 −4 3 4 −7 {t(1, 1, 1) : t ∈ R}. 341 k = −2: In this case, the system is C x = 0, where C = 3 4 1 . This system has solution set 341 {(r, s, −3r − 4s) : r, s ∈ R}. 60. Using technology we find that: det(A) = −20; A−1 −2/5 1/2 0 1/10 19/10 1/2 −1 1/2 0 −5 ; x = A−1 b = = 1/2 . 0 1/2 −1 1/2 1/10 0 1/2 −2/5 12/5 Solutions to Section 3.3 True-False Review: 1. FALSE. Because 2 + 3 = 5 is odd, the (2, 3)-cofactor is the negative of the (2, 3)-minor of the matrix. 2. TRUE. This just requires a slight modification of the proof of Theorem 3.3.16. We compute n (A · adj(A))ij = n aik Cjk = δij · det(A). aik adj(A)kj = k=1 k=1 Therefore, A · adj(A) = det(A) · In . 3. TRUE. The Cofactor Expansion Theorem allows for expansion along any row or any column of the matrix, and in all cases, the result is the determinant of the matrix. 123 4. FALSE. For example, let A = 4 5 6 , and let c = 2. It is easy to see that the (1, 1)-entry of 789 adj(A) is −3. But the (1, 1)-entry of adj(2A) is −12, not −6. Therefore, the equality posed in this review item does not generally hold. Many other counterexamples could be given as well. 987 10 10 10 123 5. FALSE. For example, let A = 4 5 6 and B = 6 5 4 . Then A + B = 10 10 10 . 789 321 10 10 10 The (1, 1)-entry of adj(A + B ) is therefore 0. However, the (1, 1)-entry of adj(A) is −3, and the (1, 1)-entry of adj(B ) is also −3. But (−3) + (−3) = 0. Many other examples abound, of course. 204 6. FALSE. Let A = adj(AB ) = ab cd e g and let B = f h cf + dh −(af + bh) −(ce + dg ) ae + bg . Then AB = = d −b −c a ae + bg ce + dg h −g −f e af + bh cf + dh . We compute = adj(A)adj(B ). 7. TRUE. This can be immediately deduced by substituting In for A in Theorem 3.3.16. Problems: From this point on, CET(col#n) will mean that the Cofactor Expansion Theorem has been applied to column n of the determinant, and CET(row#n) will mean that the Cofactor Expansion Theorem has been applied to row n of the determinant. 1. Minors: M11 = 4, M21 = −3, M12 = 2, M22 = 1; Cofactors: C11 = 4, C21 = 3, C12 = −2, C22 = 1. 2. Minors: M11 = −9, M21 = −7, M31 = −2, M12 = 7, M22 = 1, M32 = −2, M13 = 5, M23 = 3, M33 = 2; Cofactors: C11 = −9, C21 = 7, C31 = −2, C12 = −7, C22 = 1, C32 = 2, C13 = 5, C23 = −3, C33 = 2. 3. Minors: M11 = −5, M21 = 47, M31 = 3, M12 = 0, M22 = −2, M32 = 0, M13 = 4, M23 = −38, M33 = −2; Cofactors: C11 = −5, C21 = −47, C31 = 3, C12 = 0, C22 = −2, C32 = 0, C13 = 4, C23 = 38, C33 = −2. 4. M12 = 3 7 5 12 46 12 = −4, M31 = 3 −1 2 4 12 0 12 = 16, M23 = 1 7 5 32 16 02 = 40, 1 −1 2 3 12 7 46 = 12, M42 = C12 = 4, C31 = 16, C23 = −40, C42 = 12. 5. 1 −2 1 3 6. −1 1 3 7. 2 7 1 1 −4 1 3 5 −2 = −7 · 8. 3 7 2 1 4 1 2 3 −5 =3· 9. 0 2 −3 −2 0 5 3 −5 0 = 1 · |3| + 2 · |1| = 5. 2 3 4 −2 1 4 =3· 1 3 4 1 1 −4 5 −2 1 2 3 −5 +2· + − −1 3 2 1 2 −4 1 −2 1 4 3 −5 +4· 21 15 −3· +2· = 3 · 10 + 5(−6) + 0 · 4 = 0. −1 1 1 1 4 2 2 4 = 3(−11) + 2(−7) + 4(−6) = −71. = −7 · 18 + 0 − 3 · 9 = −153. = 3(−11) − 7(−17) + 2(−2) = 82. 205 10. 1 −2 3 0 4 0 7 −2 0 1 3 4 1 5 −2 0 1 = −2 · 1 −2 3 0 7 = −2 · (−26) − 4 · (−5) = 72. −4· 4 1 5 −2 1 −2 3 0 1 3 1 5 −2 1. CET(col#4) 0 −2 1 −1 2 5 11. 1 3 7 12. −1 23 0 14 2 −1 3 13. 2 −1 3 5 21 3 −3 7 1 −1 2 5 1 = 1 = −1 · 1 =2· 2 −3 −2· 1 −1 1 7 4 3 3 7 1 2 +2· −1 −3 −5· = 7 − 2(−1) = 9. 2 1 3 4 3 7 1. CET(row#1) = −7 + 10 = 3. +3· −1 2 3 1 1. CET(col#1) = 2 · 17 − 5 · 2 + 3(−7) = 3. 1. CET(col#1) 14. 0 −2 1 2 0 −3 −1 3 0 1 =0· 0 −3 3 0 −2· −2 3 1 0 −1· −2 1 0 −3 = 0 + 6 − 6 = 0. 1. CET(col#1) 15. 1 0 −1 0 0 −1 0 1 0 −1 0 −1 0 1 0 1 1 =1· 1 0 −1 0 −1 0 1 0 1 0 −1 0 1 0 −1 1 0 1 −1· = −2 − 2 = −4. 1. CET(col#1) 16. 2 −1 31 1 4 −2 3 0 2 −1 0 1 3 −2 4 = 2 5 31 1 0 −2 3 0 0 −1 0 1 −1 −2 4 = 5 −1 1 1. CA32 (2) 1 4 +3· 2 51 2 03 =− 1 1 −1 4 2 5 1 −1 2. CET(row#3) = 21 − 21 = 0. 206 17. 3 5 2 6 2 3 5 −5 7 5 −3 −16 9 −6 27 −12 3 = 1 = −1 11 −27 −9 18 −93 −24 54 −111 1. A21 (−1) 1 2 −3 11 2 3 5 −5 7 5 −3 −16 9 −6 27 −12 −1 11 −27 0 −81 150 0 −210 537 4 = 1 2 −3 11 0 −1 11 −27 0 −9 18 −93 0 −24 54 −111 2 = −81 150 −210 537 5 =− 2. A12 (−2), A13 (−7), A14 (−9) 4. A12 (−9), A13 (−24) = 11997. 3. CET(col#1) 5. CET(col#1) 18. 2 −7 43 5 5 −3 7 6 2 63 4 2 −4 5 2 −7 43 1 3 12 6 2 63 4 2 −4 5 = 4 = 1. A42 (−1) = 13 −2 1 −16 0 −9 −10 −8 −3 1 = 2 5 0 −13 2 −1 1 3 1 2 0 −16 0 −9 0 −10 −8 −3 13 −2 1 101 −18 0 29 −14 0 2. A21 (−2), A23 (−6), A24 (−4) 5. A12 (9), A13 (3) 6 = −13 2 −1 3 0 −9 = − −16 −10 −8 −3 101 −18 29 14 = −892. 3. CET(col#1) 4. P2 0 −3 5 3 01 =− 1 30 0 −1 3 0 2 4 5 6. CET(col#3) 19. 2 0 −1 3 03 0 1 01 3 0 10 1 −1 30 2 0 0 2 4 0 5 1 = 0 −3 5 0 0 −9 1 −10 =− 1 30 4 0 −1 3 5 3 6 = 0 −3 50 3 0 12 1 3 04 0 1 −1 0 0 −1 35 −3 4 = − −9 −1 42 0 50 −9 1 −10 −26 0 −35 1. A41 (−2),A45 (−3) 5. P2 0 0 0 1 0 7 = 5 0 1 −10 3 5 42 50 −26 −35 2. CET(col#1) 6. A21 (−5), A23 (3) 2 5 = −3 5 0 −9 1 −10 1 −3 −5 = −170. 3. A12 (−3) 4. CET(col#1) 7. CET(col#2) 20. 0 x y z −x 0 1 −1 −y −1 0 1 −z 1 −1 0 1 = xz −x −y − z −z 0 x+y z 0 1 −1 0 −1 1 1 −1 0 2 = xz x + y z −x 1 −1 −y − z −1 1 207 xz − yz + z 2 −x + y − z −y − z 3 = x+y+z 0 00 −1 1 1. A41 (−x), A43 (1) xz − yz + z 2 −x − y − z 4 = 2. CET(col#2) x+y+z 0 = (x + y + z )2 . 3. A32 (1), A31 (−z ) 4. CET(col#3) 21. (a): V (r1 , r2 , r3 ) = 2 = 1 r1 2 r1 1 r2 2 r2 r 2 − r1 2 2 r 2 − r1 1 r3 2 r3 1 = 1 r1 2 r1 r3 − r1 2 2 r3 − r1 0 r 2 − r1 2 2 r 2 − r1 0 r 3 − r1 2 2 r3 − r1 2 2 2 2 = (r2 − r1 )(r3 − r1 ) − (r3 − r1 )(r2 − r1 ) = (r3 − r1 )(r2 − r1 )[(r3 + r1 ) − (r2 + r1 )] = (r2 − r1 )(r3 − r1 )(r3 − r2 ). 1. CA12 (−1), CA13 (−1) 2. CET(row#1) (b): We use mathematical induction. The result is vacuously true when n = 1 and quickly verified when n = 2. Suppose that the result is true when n = k − 1, for k ≥ 3, and consider 1 r2 2 r2 . . . 1 r3 2 r3 . . . ··· ··· ··· 1 rk 2 rk . . . k r1 −1 V (r1 , r2 , . . . , rk ) = 1 r1 2 r1 . . . k r2 −1 k r3 −1 ··· k rk −1 . The determinant vanishes when rk = r1 , r2 , . . . , rk−1 , so we can write n−1 (rk − ri ), V (r1 , r2 , . . . , rk ) = a(r1 , r2 , . . . , rk ) i=1 k where a(r1 , r2 , . . . , rk ) is the coefficient of rk −1 in the expansion of V (r1 , r2 , . . . , rk ). However, using the Cofactor Expansion Theorem along column k , we see that this coefficient is just V (r1 , r2 , . . . , rk−1 ), so by hypothesis, (rm − ri ). a(r1 , r2 , . . . , rk ) = V (r1 , r2 , . . . , rk−1 ) = 1≤i<m≤n−1 Thus, n−1 (rm − ri ) V (r1 , r2 , . . . , rk ) = 1≤i<m≤n−1 (rk − ri ) = i=1 (rm − ri ). 1≤i<m≤n Hence the result is true for n = k , so, by induction, is true for all non-negative integers n. 22. (a): det(A) = 11; 208 5 −4 −1 3 (b): MC = 5 −1 −4 3 (c): adj(A) = (d): A−1 = ; 1 11 ; 5 −1 −4 3 . 23. (a): det(A) = 7; 1 −4 2 −1 (b): MC = ; (c): adj(A) = 1 2 −4 −1 ; 1 7 1 2 −4 −1 . (d): A−1 = 24. (a): det(A) = 0; −6 −2 (b): MC = (c): adj(A) = 15 5 ; −6 −2 15 5 ; (d): A−1 does not exist because det(A) = 0. 25. (a): det(A) = 2 −3 0 2 15 0 −1 2 = 2 −3 0 0 45 0 −1 2 7 −4 −2 6 4 2 ; (b): MC = −15 −10 8 7 6 −15 (c): adj(A) = −4 4 −10 ; −2 2 8 7 1 1 −4 adj(A) = (d): A−1 = det(A) 26 −2 = 2 · 13 = 26; 26. (a): det(A) = −2 2 0 3 −1 1 5 2 3 = 6 −15 4 −10 . 2 8 −2 3 −1 04 4 02 3 = −2 · 4 = −8; 209 −7 −6 4 4 ; (b): MC = −11 −6 16 8 −8 −7 −11 16 8 ; (c): adj(A) = −6 −6 4 4 −8 −7 −11 16 1 1 8 . (d): A−1 = adj(A) = − −6 −6 det(A) 8 4 4 −8 27. (a): det(A) = 1 −1 2 3 −1 4 5 17 = 6 8 5 0 0 1 9 11 7 = 6; −11 −1 8 9 −3 −6 ; (b): MC = −2 2 2 −11 9 −2 2 ; (c): adj(A) = −1 −3 8 −6 2 −11 9 −2 −11/6 3/2 −1/3 1 1 −1 −3 2 = −1/6 −1/2 1/3 . (d): A−1 = adj(A) = det(A) 6 8 −6 2 4/3 −1 1/3 28. 0 12 −1 −1 3 = 1 −2 1 5 43 (b): MC = −5 −2 1 ; 5 −2 1 5 −5 5 (c): adj(A) = 4 −2 −2 ; 3 1 1 1 1 adj(A) = (d): A−1 = det(A) 10 (a): det(A) = 29. (a): det(A) = 2 −3 5 1 2 1 0 7 −1 = −9 1 7 (b): MC = 32 −2 −14 ; −13 3 7 0 12 0 −3 4 1 −2 1 = 10; 5 −5 5 4 −2 −2 . 3 1 1 0 −7 3 1 2 1 0 7 −1 = 14; 210 −9 32 −13 3 ; (c): adj(A) = 1 −2 7 −14 7 (d): A−1 −9 32 −13 −9/14 16/7 −13/14 1 1 1 −2 3 = 1/14 −1/7 3/14 . = adj(A) = det(A) 14 7 −14 7 1/2 −1 1/2 30. 1 −1 1 −1 (a): det(A) = 1 1 1 1 −1 1 1 −1 −1 1 1 −1 = 1 0 0 0 1 1 1 2 0 2 2 −2 0 2 0 −2 = 2 0 2 2 −2 0 2 0 −2 = 16; 44 4 4 −4 4 −4 4 (b): MC = 4 4 −4 −4 ; −4 4 4 −4 4 −4 4 −4 4 4 4 4 ; (c): adj(A) = 4 −4 −4 4 4 4 −4 −4 (d): A−1 4 −4 4 −4 14 1 4 4 4 . adj(A) = = 4 −4 −4 4 det(A) 16 4 4 −4 −4 31. 1 −2 3 2 (a): det(A) = 0 1 9 0 3 5 1 3 0 2 3 −1 = 11 4 7 0 0 18 5 1 10 3 96 2 0 0 −1 84 −46 −29 81 −162 60 99 −27 ; (b): MC = 18 38 −11 3 −30 26 130 −72 84 −162 18 −30 −46 60 38 26 ; (c): adj(A) = −29 99 −11 130 81 −27 3 −72 =− 11 4 7 0 18 1 10 96 =− 11 0 18 41 10 −29 0 −84 = 402; 211 (d): A−1 84 −162 18 −30 1 1 −46 60 38 26 = adj(A) = −29 99 −11 130 det(A) 402 81 −27 3 −72 14/67 −27/67 3/67 −5/67 −23/201 10/67 19/201 13/201 = −29/402 33/134 −11/402 65/201 27/134 −9/134 1/134 −12/67 32. (a): Using the Cofactor Expansion Theorem along row 1 yields det(A) = (1 + 2x2 ) + 2x(2x + 4x3 ) + 2x2 (2x2 + 4x4 ) = 8x6 + 12x4 + 6x2 + 1 = (1 + 2x2 )3 . (b): 1 + 2x2 −(2x + 4x3 ) 2x2 + 4x4 1 − 4x4 −(2x + 4x3 ) MC = 2x + 4x3 2 4 3 2x + 4x 2x + 4x 1 + 2x2 1 + 2x2 −2x(1 + 2x2 ) 2x2 (1 + 2x2 ) = 2x(1 + 2x2 ) (1 + 2x2 )(1 − 2x2 ) −2x(1 + 2x2 ) , 2x2 (1 + 2x2 ) 2x(1 + 2x2 ) 1 + 2x2 so that A−1 33. det(A) = 1 1 1 1 2 2 1 2 3 1 2x 1 1 −2x 1 − 2x2 = adj(A) = det(A) (1 + 2x2 )2 2x2 −2x = 1 0 0 1 1 1 1 1 2 = 1 1 1 2 2x2 2x . 1 = 2 − 1 = 1, and C23 = − 1 1 1 2 = −(2 − 1) = −1. Thus, (A−1 )32 = 34. det(A) = 2 0 −1 2 1 1 3 −1 0 = 4 10 2 11 3 −1 0 (adj(A))32 C23 = = −1. det(A) det(A) =− 4 1 3 −1 = −(−4 − 3) = 7, and C13 = 2 1 3 −1 = −2 − 3 = −5. . 212 Thus, (A−1 )31 = (adj(A))31 C13 5 = =− . det(A) det(A) 7 35. 1 0 10 2 −1 13 0 1 −1 2 −1 1 20 = = det(A) = 1 0 00 2 −1 −1 3 0 1 −1 2 −1 1 30 −1 23 1 −4 2 1 00 = 2 −4 = −1 −1 3 1 −1 2 1 30 3 2 = 4 + 12 = 16, and 1 2 −1 C32 = − 1 1 2 0 3 0 = −(−3) 1 −1 1 2 = 3(2 + 1) = 9. Thus, (A−1 )23 = 2e2t −e2t 36. MC = −2et 3et 2e2t −e2t , and det(A) = 4e3t , so −2et 3et 1 1 2e2t −e2t = adj(A) = 3t . −2et 3et det(A) 4e , adj(A) = A−1 e−t sin(2t) −et cos(2t) −et cos(2t) et sin(2t) 37. MC = C32 9 (adj(A))23 = . = det(A) det(A) 16 e−t sin(2t) et cos(2t) −et cos(2t) et sin(2t) , adj(A) = , and det(A) = sin2 (2t) + cos2 (2t) = 1, so 1 e−t sin(2t) et cos(2t) adj(A) = . −et cos(2t) et sin(2t) det(A) −e−t −te2t 3te−t −te−t −te−t e−t 0 , adj(A) = −e−t e−t 0 , and 2t 2t 2t 0 te −te 0 te A−1 = 3te−t −te−t 38. MC = −te−t det(A) = et et et 2tet 2tet tet e−2t e−2t 2e−2t = et 0 0 tet tet 0 e−2t 0 e−2t = et (te−t ) = t, so A−1 3te−t 1 1 −e−t = adj(A) = det(A) t −te2t −te−t −te−t e−t 0 . 2t 0 te 213 −1 3 −2 −1 2 −1 4 . Hence, 3 , adj(A) = 2 −6 39. MC = 3 −6 −2 4 −2 −1 3 −2 1 A · adj(A) = 3 4 2 4 5 3 −1 3 −2 0 5 2 −6 4 = 0 6 −1 3 −2 0 00 0 0 = 03 . 00 From Equation (3.3.4) of the text we have that, in general, A · adj(A) = det(A) · In . Since, for the given matrix, A · adj(A) = 03 , we must have det(A) = 0. 40. det(A) = 2 −3 1 2 2 −3 4 2 = 7, det(B1 ) = x1 = = 16, and det(B2 ) = 16 det(B1 ) = and det(A) 7 x2 = 22 14 = 6. Thus, det(B2 ) 6 =. det(A) 7 Solution: (16/7, 6/7). 41. det(A) = 3 −2 1 1 1 −1 2 0 1 = 1 −2 1 3 1 −1 0 0 1 = 7, det(B1 ) = 4 −2 1 2 1 −1 1 0 1 = 4 −2 −3 2 1 −3 1 0 0 = 9, det(B2 ) = 3 1 2 4 1 2 −1 1 1 = 4 1 3 det(B3 ) = 3 −2 4 1 12 2 01 = −5 −2 4 −3 12 0 01 6 0 2 −1 3 0 = −6, = −11. 9 det(B2 ) 6 det(B3 ) 11 det(B1 ) = , x2 = = − , and x3 = =− . det(A) 7 det(A) 7 det(A) 7 Solution: (9/7, −6/7, −11/7). Thus, x1 = 1 −3 1 1 −3 1 1 4 −1 = 0 7 −2 = 2 1 −3 0 7 −5 Thus, the system has only the trivial solution. 42. det(A) = 7 −2 7 −5 = −35 + 14 = −21 = 0. 214 43. −5 −2 3 −1 0 0 1 0 1 1 0 −1 4 1 −2 1 1 −2 3 −1 2 0 1 0 1 1 0 −1 0 1 −2 1 det(A) = = −5 −2 −1 1 1 −1 4 1 1 =− =− det(B1 ) = 1 −2 3 −1 2 0 1 0 0 1 0 −1 3 1 −2 1 det(B2 ) = 11 3 −1 22 1 0 10 0 −1 0 3 −2 1 det(B3 ) = 1 −2 1 −1 2 02 0 1 1 0 −1 0 13 1 = −2 −1 −1 0 5 20 4 11 1 −3 3 −1 2 0 1 0 0 0 0 −1 3 2 −2 1 = = 1 2 1 0 1 30 2 12 0 00 3 −2 1 = = −3 −1 0 20 = −2 −2 −3 11 1 −2 31 2 0 12 1 1 00 0 1 −2 3 =− = 1 −3 3 2 0 1 3 2 −2 = −5 −3 3 0 0 1 7 2 −2 0 −2 1 −1 0 02 0 1 1 0 −1 −3 13 1 0 −2 −1 1 1 −1 −3 1 1 det(B4 ) = = −3, −3 0 3 3 −5 1 0 2 7 1 30 2 12 3 −2 1 =− 1 30 −4 50 3 −2 1 = 17, = −2(−8) = 16, 3 −2 31 2 0 12 0 1 00 −1 1 −2 3 = = −3 −5 3 7 =− 3 31 2 12 −1 −2 3 = 6. Therefore, x1 = 17 16 11 , x2 = − , x3 = − , and x4 = −2. 3 3 3 Solution: (11/3, −17/3, −16/3, −2). 44. det(A) = et et e−2t −2e−2t det(B1 ) = 3 sin t 4 cos t det(B2 ) = et et e−2t −2e−2t 3 sin t 4 cos t 1 1 1 −2 = et e−2t = et = e−2t 1 1 3 sin t 4 cos t 3 sin t 4 cos t = −3e−t , 1 −2 = −2e−2t (3 sin t + 2 cos t), = et (4 cos t − 3 sin t). = −11, 215 Thus, x1 = Solution: det(B1 ) 2(3 sin t + 2 cos t) det(B2 ) e2t (3 sin t − 4 cos t) = and x2 = = . t det(A) 3e det(A) 3 1 2 −t e (3 sin t + 2 cos t), e2t (3 sin t − 4 cos t) . 3 3 45. det(A) = det(B2 ) = = 1 4 −2 1 2 9 −3 −2 15 0 −1 3 14 7 −2 1 2 1 3 2 −2 1 5 −3 −2 3 0 −1 6 7 −2 5 −16 −1 −3 1 −1 −2 2 2 −1 −3 0 1 0 00 3 −1 71 = = 1 −1 −2 2 2 −1 −3 0 1 0 00 3 −3 71 = −31. Therefore x2 = = = −1 −2 2 −1 −3 0 −1 71 −1 −2 2 −1 −3 0 −3 71 1 −16 0 −1 −3 0 −1 71 = = = −19, 5 −16 0 −1 −3 0 −3 71 31 det(B2 ) = . det(A) 19 46. det(A) = b+c b c = −c det(B1 ) = a a a+c b c a+b −a + b + c a −a + b − c b a a a b a+c b c c a+b = (a − b + c) det(B2 ) = b+c b c det(B3 ) = b+c b c = (−a + b + c) = = b b c a+b a a a+c b c c −a + b + c a a −a + b − c a + c b 0 c a+b + (a + b) a a c a+b a a b b c a+b = −(a − b − c) = = a+c b c c −a + b + c −a + b − c a a+c = 4abc, a 0 a b a−b+c b c 0 a+b = a(a − b + c)(a + b − c), b+c a−b−c a b 0 b c 0 a+b = −b(a − b − c)(a + b − c), −a + b + c a a 0 a+c b 0 c c = −c(a − b − c)(a − b + c). Case 1: If a = 0, b = 0, and c = 0, then det(A) = 0, so it follows by Theorem 3.2.4 that the given system of equations has a unique solution. The solution in this case is given by: 216 det(B1 ) (a − b + c)(a + b − c) = , det(A) 4bc det(B2 ) −(a − b − c)(a + b − c) x2 = = , det(A) 4ac det(B3 ) −(a − b − c)(a − b + c) x3 = = . det(A) 4ab x1 = Case 2: If a = b = 0 and c = 0, then it is easy to see from the system that the system is inconsistent. By the symmetry of the equations, the same will be true if a = c = 0 with b = 0, or b = c = 0 with a = 0. Case 3: Suppose a = 0 where matrix: b+c 0 c b b c b = 0 and c = 0, and consider the reduced row echelon form of the augmented 10 1 0 0 0 00 b b ∼ 0 b−c b−c b−c ∼ 0 1 00 0 c−b c−b c−b cc 00 1 1 . 00 From the last matrix we see that the system has an infinite number of solutions of the form {(0, 1 − r, r) : r ∈ R}. By the symmetry of the three equations, it follows that the other two cases: b = 0 with a = 0 and c = 0, and c = 0 with a = 0 and b = 0 have similar forms of solutions. 47. Let B be the matrix obtained from A by adding column i to column j (i = j ) in the matrix A. By the property for columns corresponding to Property P3, we have det(B ) = det(A). Cofactor expansion of B along column j gives n det(A) = det(B ) = n (akj + aki )Ckj = k=1 n akj Ckj + k=1 aki Ckj . k=1 That is, n det(A) = det(A) + aki Ckj , k=1 since by the Cofactor Expansion Theorem the first summation on the right-hand side is simply det(A). It follows immediately that n aki Ckj = 0, i = j. k=1 51. 1.21 3.42 A = 5.41 2.32 21.63 3.51 1.21 3.25 B2 = 5.41 4.61 21.63 9.93 2.15 3.25 3.42 2.15 7.15 , B1 = 4.61 2.32 7.15 , 9.22 9.93 3.51 9.22 2.15 1.21 3.42 3.25 7.15 , B3 = 5.41 2.32 4.61 . 9.22 21.63 3.51 9.93 From Cramer’s Rule, x1 = det(B1 ) det(B2 ) det(B3 ) ≈ 0.25, x2 = ≈ 0.72, x3 = ≈ 0.22. det(A) det(A) det(A) 217 52. det(A) = 32, det(B1 ) = −3218, det(B2 ) = 3207, det(B3 ) = 2896, det(B4 ) = −9682, det(B5 ) = 2414. So, x1 = − 3207 181 4841 1207 1609 , x2 = , x3 = , x4 = − , x5 = . 16 32 2 32 16 53. We have n n bjk aki = (BA)ji = k=1 k=1 1 1 · adj(A)jk · aki = det(A) det(A) n Ckj aki = δij , k=1 where we have used Equation (3.3.4) in the last step. Solutions to Section 3.4 Problems: 1. 5 −1 3 7 2. 3 −1 6 5 7 2 4 3 −2 3. 5 6 14 14 13 27 = 5 · 7 − 3(−1) = 38. = −43. = −3. 4. 2.3 1.5 7.9 4.2 3.3 5.1 6.8 3.6 5.7 = 2.3 3.3 5.1 3.6 5.7 − 1.5 4.2 6.8 5.1 5.7 + 7.9 4.2 3.3 6.8 3.6 = 1.035 + 16.11 − 57.828 = −40.683. 5. abc bca cab =a ca ab −b ba cb +c bc ca = a(bc − a2 ) − b(b2 − ac) + c(ab − c2 ) = 3abc − a3 − b3 − c3 . 6. 3 5 −1 2 2 1 52 3 2 57 1 −1 21 = 0 8 −7 −1 0 3 1 0 0 5 −1 4 1 −1 2 1 = −1 29 −7 −1 0 1 0 8 −1 4 8 −7 −1 1 0 = −1 3 5 −1 4 = −1 29 −1 8 4 = −124. 218 7. 7 12 2 −2 4 3 −1 5 18 9 27 3 6 4 54 =9 7 1 2 −2 3 −1 2 1 = −36 7 −2 −5 3 423 10 7 7 −5 1 3 14 −3 45 −14 = 36 8. det(A) = 11. MC = 23 46 54 36 = −36 3 12 7 3 = −9 14 0 −3 45 0 −14 −5 1 3 = −2196. =⇒ adj(A) = A−1 = 712 16 0 8 =9 10 0 7 −5 0 1 7 −5 −2 3 1 adj(A) = det(A) 7/11 −2/11 , so that −5/11 3/11 . 123 1 2 3 2 3 1 = 0 −1 −5 = −18. 312 0 −5 −7 5 −1 −7 5 −1 −7 5 =⇒ adj(A) = −1 −7 5 , so that MC = −1 −7 −7 5 −1 −7 5 −1 −5/18 1/18 7/18 1 7/18 −5/18 . adj(A) = 1/18 A−1 = det(A) 7/18 −5/18 1/18 9. det(A) = 4 7 −11 −38 7 −11 −38 6 1= 0 0 1 =− = 30. 5 20 14 −1 5 20 −1 −20 5 10 −20 102 −38 5 −24 11 , so that MC = 102 −24 −30 =⇒ adj(A) = −38 11 10 10 −30 10 −2/3 17/5 −19/15 1 11/30 . A−1 = adj(A) = 1/6 −4/5 det(A) 1/3 −1 1/3 10. det(A) = 3 2 3 2 57 4 −3 2 = 116. 6 9 11 −51 −32 54 −51 8 31 8 −20 12 =⇒ adj(A) = −32 −20 24 , so that MC = 31 24 −26 54 12 −26 −51/116 2/29 31/116 1 6/29 . A−1 = adj(A) = −8/29 −5/29 det(A) 27/58 3/29 −13/58 11. det(A) = 16 8 12 10 7 7 −5 1 3 219 12. det(A) = −38 0 MC = 38 0 5 −1 2 3 −1 4 1 −1 2 5 9 −3 32 34 28 44 −124 −222 −24 −16 A−1 3 6 13. A = 5 2 5 5 1 2 = −152. 2 −38 0 38 0 −60 28 −124 −24 =⇒ adj(A) = 32 , so that 34 130 44 −222 −16 8 2 −60 130 8 1/4 0 −1/4 0 −4/19 −7/38 1 31/38 3/19 . = adj(A) = −17/76 −11/38 111/76 2/19 det(A) −1/76 15/38 −65/76 −1/19 , B1 = 4 9 5 2 x1 = cos t sin t 14. A = sin t − cos t x1 = , B1 = det(B1 ) 37 det(B2 ) 1 = , x2 = =− . det(A) 24 det(A) 8 e−t 3e−t 4 13 5 1 15. A = 2 −1 5 , B1 = 7 −1 2 31 2 3 det(B1 ) 1 =, x1 = det(A) 4 5 16. A = 2 2 , so that sin t − cos t , B2 = cos t sin t e−t 3e−t , so that det(B1 ) det(B2 ) = e−t [cos t + 3 sin t], x2 = = e−t [sin t − 3 cos t]. det(A) det(A) 34 69 , B2 = 3 45 5 , B2 = 2 7 1 22 det(B2 ) 1 x2 = = , det(A) 16 3 4 15 5 , B3 = 2 −1 7 , so that 1 2 32 det(B3 ) 21 x3 = = . det(A) 16 3 6 33 6 5 3 6 53 3 4 −7 , B1 = −1 4 −7 , B2 = 2 −1 −7 , B3 = 2 4 −1 , so that 5 9 45 9 2 4 9 25 4 det(B1 ) 30 det(B2 ) 59 det(B3 ) 81 x1 = = , x2 = = , x3 = = . det(A) 271 det(A) 271 det(A) 271 3.1 3.5 7.1 3.6 3.5 7.1 3.1 17. A = 2.2 5.2 6.3 , B1 = 2.5 5.2 6.3 , B2 = 2.2 1.4 8.1 0.9 9.3 8.1 0.9 1.4 so that x1 = 3.6 2.5 9.3 7.1 3.1 3.5 6.3 , B3 = 2.2 5.2 0.9 1.4 8.1 3.6 2.5 , 9.3 det(B1 ) det(B2 ) det(B3 ) = 3.77, x2 = = 0.66, x3 = = −1.46. det(A) det(A) det(A) 18. Since A is invertible, A−1 exists. Then, AA−1 = In =⇒ det(AA−1 ) = det(In ) = 1 =⇒ det(A) det(A−1 ) = 1, so that det(A−1 ) = 1 . det(A) 220 19. det(2A) = 23 det(A) = 24. 1 1 =. det(A) 3 From the result of the preceding problem, det(A−1 ) = det(AT B ) = det(AT ) det(B ) = det(A) det(B ) = −12. det(B 5 ) = [det(B )]5 = −1024. 1 det(B −1 AB ) = det(B −1 ) det(A) det(B ) = det(A) det(B ) = det(A) = 3. det(B ) Solutions to Section 3.5 Additional Problems: 1. (a): We have det(A) = (−7)(−5) − (1)(−2) = 37. (b): We have 1 −5 −7 −2 1 A∼ 1. P12 1 −5 0 −37 2 ∼ = B. 2. A12 (7) Now, det(A) = −det(B ) = −(−37) = 37. (c): Let us use the Cofactor Expansion Theorem along the first row: det(A) = a11 C11 + a12 C12 = (−7)(−5) + (−2)(−1) = 37. 2. (a): We have det(A) = (6)(1) − (−2)(6) = 18. (b): We have 1 A∼ 1 −2 1 1 2 ∼ 1 0 1 3 = B. 1. M1 (1/6) 2. A12 (2) Now, det(A) = 6 det(B ) = 6 · 3 = 18. (c): Let us use the Cofactor Expansion Theorem along the first row: det(A) = a11 C11 + a12 C12 = (6)(1) + (6)(2) = 18. 3. (a): Using Equation (3.1.2), we have det(A) = (−1)(2)(−3)+(4)(2)(2)+(1)(0)(2)−(−1)(2)(2)−(4)(0)(−3)−(1)(2)(2) = 6+16+0−(−4)−0−4 = 22. (b): We have −1 A∼ 0 0 1 4 1 −1 2 2 2 ∼ 0 10 −1 0 4 1 2 2 = B. 0 −11 221 1. A13 (2) 2. A23 (−5) Now, det(A) = det(B ) = (−1)(2)(−11) = 22. (c): Let us use the Cofactor Expansion Theorem along the first column: det(A) = a11 C11 + a21 C21 + a31 C31 = (−1)(−10) + (0)(14) + (2)(6) = 22. 4. (a): Using Equation (3.1.2), we have det(A) = (2)(0)(3)+(3)(2)(6)+(−5)(−4)(−3)−(−6)(0)(−5)−(−3)(2)(2)−(3)(−4)(3) = 0+36−60+−0+12+36 = 24. (b): We have 2 2 3 −5 1 2 6 −8 ∼ 0 A∼ 0 0 −12 18 0 1. A12 (2), A13 (−3) 3 −5 6 −8 = B. 0 2 2. A23 (2) Now, det(A) = det(B ) = (2)(6)(2) = 24. (c): Let us use the Cofactor Expansion Theorem along the second row: det(A) = (−4)(6) + (0)(36) + (2)(24) = −24 + 0 + 48 = 24. 5. (a): Of the 24 terms appearing in the determinant expression (3.1.3), only terms containing the factors a11 and a44 will be nonzero (all other entries in the first column and fourth row of A are zero). Looking at entries in the second and third rows and columns of A, we see that only the product a23 a32 is nonzero. Therefore, the only nonzero term in the summation (3.1.3) is a11 a23 a32 a44 = (3)(1)(2)(−4) = −24. The permutation associated with this term is (1, 3, 2, 4) which contains one inversion. Therefore, σ (1, 3, 2, 4) = −1, and so the determinant is (−24)(−1) = 24. (b): We have 3 −1 −2 1 2 1 −1 10 = B. A∼ 0 0 1 4 0 0 0 −4 1. P23 Now, det(A) = −det(B ) = −(3)(2)(1)(−4) = 24. 01 4 (c): Cofactor expansion along the first column yields: det(A) = 3 · 2 1 −1 . This latter determinant can 0 0 −4 be found by cofactor expansion along the last column: (−4)[(0)(1) − (2)(1)] = 8. Thus, det(A) = 3 · 8 = 24. 6. 222 (a): Of the 24 terms in the determinant expression (3.1.3), the only nonzero term is a41 a32 a23 a14 , which has a positive sign: σ (4, 3, 2, 1) = +1. Therefore, det(A) = (−3)(1)(−5)(−2) = −30. (b): By permuting the first and last rows, and by permuting the middle two rows, we bring A to upper triangular form. The two permutations introduce two sign changes, so the resulting upper triangular matrix has the same determinant as A, and it is (−3)(1)(−5)(−2) = −30. 0 0 −2 0 −5 1 . This latter 1 −4 1 determinant can be found by cofactor expansion along the first column: (1)[(0)(1) − (−5)(−2)] = −10. Hence, det(A) = 3 · (−10) = −30. (c): Cofactor expansion along the first column yields: det(A) = −(−3) · 7. To obtain the given matrix from A, we perform two row permutations, multiply a row through by −4, and multiply a row through by 2. The combined effect of these operations is to multiply the determinant of A by (−1)2 · (−4) · (2) = −8. Hence, the given matrix has determinant det(A) · (−8) = 4 · (−8) = −32. 8. To obtain the given matrix from A, we add 5 times the middle row to the top row, we multiply the last row by 3, we multiply the middle row by −1, we add a multiple of the last row to the middle row, and we perform a row permutation. The combined effect of these operations is to multiply the determinant of A by (1) · (3) · (−1) · (−1) = 3. Hence, the given matrix has determinant det(A) · (3) = 4 · (3) = 12. 9. To obtain the given matrix from AT , we do two row permutations, multiply a row by −1, multiply a row by 3, and add 2 times one row to another. The combined effect of these operations is to multiply the determinant of A by (−1)(−1)(−1)(3)(1) = −3. Hence, the given matrix has determinant det(A) · (−3) = (4) · (−3) = −12. 10. To obtain the given matrix from A, we permute the bottom two rows, multiply a row by 2, multiply a row by −1, add −1 times one row to another, and then multiply each row of the matrix by 3. The combined effect of these operations is to multiply the determinant of A by (−1)(2)(−1)(1)(3)3 = 54. Hence, the given matrix has determinant det(A) · 54 = (4)(54) = 216. 11. We have det(AB ) = det(A)det(B ) = (−2) · 3 = −6. 12. We have det(B 2 A−1 ) = (det(B ))2 det(A−1 ) = 32 −2 = −4.5. 1 13. We have det((A−1 B )T (2B −1 )) = det(A−1 B ) · det(2B −1 ) = − 2 · 3 · 24 · 1 3 = −8. 14. We have det((−A)3 (2B 2 )) = det(−A)3 det(2B 2 ) = (−1)3 (detA)3 · 24 · (detB )2 = (−1)3 (−2)3 · 24 · 32 = 1152. 15. Since A and B are not square matrices, det(A) and det(B ) are not possible to compute. We have 8 −10 det(C ) = −18, and therefore, det(C T ) = −18. Now, AB = , and so det(AB ) = 474. Since 25 28 45 2 BA = 1 8 −13 , we have det(BA) = 0. Next, det(B T AT ) = det((AB )T ) = det(AB ) = 474. Next, 18 15 24 38 54 det(BAC ) = det(BA)det(C ) = 0 · (−18) = 0. Finally, we have det(ACB ) = det = 4104. 133 297 16. The matrix of cofactors of B is MC = 1 −1 −4 5 , and so adj(B ) = 1 −4 −1 5 . Since det(B ) = 1, 223 we have B −1 = 1 −4 −1 5 . Therefore, (A−1 B T )−1 = (B T )−1 A = (B −1 )T A = 17. 16 −1 −5 5 −3 ; MC = 4 −4 2 10 16 4 −4 5 2 adj(A) = −1 −5 −3 10 det(A) 28; = 4/7 1/7 A−1 = −1/28 5/28 −5/28 −3/28 ; −1/7 1/14 . 5/14 18. 5 12 −11 −1 65 −24 −23 −13 ; MC = −25 0 −5 5 −35 0 5 −5 5 65 −25 −35 12 −24 0 0 ; adj(A) = −11 −23 −5 5 −1 −13 5 −5 det(A) = −60; 5 65 −25 −35 1 12 −24 0 0 . A−1 = − −11 −23 −5 5 60 −1 −13 5 −5 19. 88 −24 −40 −48 32 12 −20 0 ; MC = 16 12 −4 0 −4 6 −2 0 88 32 16 −4 −24 12 12 6 adj(A) = −40 −20 −4 −2 ; −48 0 0 0 det(A) = −48; 88 32 16 −4 1 −24 12 12 6 . A−1 = − 48 −40 −20 −4 −2 −48 0 0 0 20. 1 −1 −4 5 1 3 2 4 = −2 −2 11 12 . 224 21 12 MC = 24 9 48 24 21 adj(A) = 12 −12 det(A) 9; = 7/3 A−1 = 4/3 −4/3 21. −12 −12 ; −27 24 48 9 24 ; −12 −27 8/3 1 −4/3 16/3 8/3 . −3 7 −2 0 2 −2 ; MC = 0 −6 0 2 7 0 −6 2 0 ; adj(A) = −2 0 −2 2 det(A) 2; = 3.5 0 −3 1 0 . A−1 = −1 0 −1 1 4 −1 0 1 4 . 22. There are many ways to do this. One choice is to let B = 5 0 0 10 9 123 23. FALSE. For instance, if one entry of the matrix A = 1 2 3 is changed, two of the rows of A 123 will still be identical, and therefore, the determinant of the resulting matrix must be zero. It is not possible to force the determinant to equal r. 24. Note that det(A) = 2 + 12k + 36 − 4k − 18 − 12 = 8k + 8 = 8(k + 1). (a): Based on the calculation above, we see that A fails to be invertible if and only if k = −1. (b): The volume of the parallelepiped determined by the row vectors of A is precisely |det(A)| = |8k + 8|. The volume of the parallelepiped determined by the column vectors of A is the same as the volume of the parallelepiped determined by the row vectors of AT , which is |det(AT )| = |det(A)| = |8k + 8|. Thus, the volume is the same. 25. Note that det(A) = 3(k + 1) + 2k + 0 − 3 − k (k + 1) − 0 = −k 2 + 4k = k (4 − k ). (a): Based on the calculation above, we see that A fails to be invertible if and only if k = 0 or k = 4. (b): The volume of the parallelepiped determined by the row vectors of A is precisely |det(A)| = | − k 2 + 4k |. The volume of the parallelepiped determined by the column vectors of A is the same as the volume of the parallelepiped determined by the row vectors of AT , which is |det(AT )| = |det(A)| = | − k 2 + 4k |. Thus, the volume is the same. 225 26. Note that det(A) = 0 + 4(k − 3) + 2k 3 − k 2 − 8k − 0 = 2k 3 − k 2 − 4k − 12. (a): Based on the calculation above, we see that A fails to be invertible if and only if 2k 3 − k 2 − 4k − 12 = 0, and the only real solution to this equation for k is k ≈ 2.39 (from calculator). (b): The volume of the parallelepiped determined by the row vectors of A is precisely |det(A)| = |2k 3 − k 2 − 4k − 12|. The volume of the parallelepiped determined by the column vectors of A is the same as the volume of the parallelepiped determined by the row vectors of AT , which is |det(AT )| = |det(A)| = |2k 3 − k 2 − 4k − 12|. Thus, the volume is the same. 27. From the assumption that AB = −BA, we can take the determinant of each side: det(AB ) = det(−BA). Hence, det(A)det(B ) = (−1)n det(B )det(A) = −det(A)det(B ). From this, it follows that det(A)det(B ) = 0, and therefore, either det(A) = 0 or det(B ) = 0. Thus, either A or B (or both) fails to be invertible. 28. Since AAT = In , we can take the determinant of both sides to get det(AAT ) = det(In ) = 1. Hence, det(A)det(AT ) = 1. Therefore, we have (det(A))2 = 1. We conclude that det(A) = ±1. −3 1 29. The coefficient matrix of this linear system is A = det(A) = −3 1 1 2 = −7, det(B1 ) = 31 12 1 2 . We have = 5, and det(B2 ) = −3 1 3 1 Thus, x1 = det(B1 ) 5 det(B2 ) 6 =− and x2 = =. det(A) 7 det(A) 7 Solution: (−5/7, 6/7). 2 −1 1 5 3 . We have 30. The coefficient matrix of this linear system is A = 4 4 −3 3 det(A) = det(B2 ) = 2 −1 1 4 53 4 −3 3 2 4 4 2 0 2 1 3 3 = 16, det(B1 ) = = −4, det(B3 ) = 2 −1 1 0 53 2 −3 3 2 −1 2 4 50 4 −3 2 = 32, = −36. Thus, x1 = det(B1 ) det(B2 ) 1 det(B3 ) 9 = 2, x2 = = − , and x3 = =− . det(A) det(A) 4 det(A) 4 Solution: (2, −1/4, −9/4). 3 12 31. The coefficient matrix of this linear system is A = 2 −1 1 . We have 0 55 det(A) = 3 12 2 −1 1 0 55 = −20, det(B1 ) = −1 12 −1 −1 1 −5 55 = −10, = −6. 226 det(B2 ) = 3 −1 2 2 −1 1 0 −5 5 = −10, det(B3 ) = 3 1 −1 2 −1 −1 0 5 −5 = 30. Thus, x1 = 1 det(B2 ) 1 det(B3 ) 3 det(B1 ) = , x2 = = , and x3 = =− . det(A) 2 det(A) 2 det(A) 2 Solution: (1/2, 1/2, −3/2). Solutions to Section 4.1 True-False Review: 1. FALSE. The vectors (x, y ) and (x, y, 0) do not belong to the same set, so they are not even comparable, let alone equal to one another. 2. TRUE. The unique additive inverse of (x, y, z ) is (−x, −y, −z ). 3. TRUE. The solution set refers to collections of the unknowns that solve the linear system. Since this system has 6 unknowns, the solution set will consist of vectors belonging to R6 . 4. TRUE. The vector (−1) · (x1 , x2 , . . . , xn ) is precisely (−x1 , −x2 , . . . , −xn ), and this is the additive inverse of (x1 , x2 , . . . , xn ). 5. FALSE. There is no such name for a vector whose components are all positive. 6. FALSE. The correct result is (s + t)(x + y) = (s + t)x + (s + t)y = sx + tx + sy + ty, and in this item, only the first and last of these four terms appear. 7. TRUE. When the vector x is scalar multiplied by zero, each component becomes zero: 0x = 0. This is the zero vector in Rn . 8. TRUE. This is seen geometrically from addition and subtraction of geometric vectors. 9. FALSE. If k < 0, then k x is a vector in the third quadrant. For instance, (1, 1) lies in the first quadrant, but (−2)(1, 1) = (−2, −2) lies in the third quadrant. 10. TRUE. Recalling that i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1), we have √ √ √ 5i − 6j + 2k = 5(1, 0, 0) − 6(0, 1, 0) + 2(0, 0, 1) = (5, −6, 2), as stated. 11. FALSE. If the three vectors lie on the same line or the same plane, the resulting object may determine a one-dimensional segment or two-dimensional area. For instance, if x = y = z = (1, 0, 0), then the vectors x,y, and z rest on the segment from (0, 0, 0) to (1, 0, 0), and do not determine a three-dimensional solid region. 12. FALSE. The components of k x only remain even integers if k is an integer. But, for example, if k = π , then the components of k x are not even integers at all, let alone even integers. Problems: 227 y (3, 8) v3 (-3, 6) v2 (6, 2) v1 x Figure 0.0.61: Figure for Exercise 1 y x v2 v1 (20, -4) v3 (-3, -12) (17, -16) Figure 0.0.62: Figure for Exercise 2 1. v1 = (6, 2), v2 = (−3, 6), v3 = (6, 2) + (−3, 6) = (3, 8). 2. v1 = (−3, −12), v2 = (20, −4), v3 = (−3, −12) + (20, −4) = (17, −16). 3. v = 5(3, −1, 2, 5) − 7(−1, 2, 9, −2) = (15, −5, 10, 25) − (−7, 14, 63, −14) = (22, −19, −53, 39). Additive inverse: −v = (−1)v = (−22, 19, 53, −39). 4. We have y= 2 1 2 1 x + z = (1, 2, 3, 4, 5) + (−1, 0, −4, 1, 2) = 3 3 3 3 142 , , , 3, 4 . 333 5. Let x = (x1 , x2 , x3 , x4 ), y = (y1 , y2 , y3 , y4 ) be arbitrary vectors in R4 . Then x+y = (x1 , x2 , x3 , x4 ) + (y1 , y2 , y3 , y4 ) = (x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4 ) = (y1 + x1 , y2 + x2 , y3 + x3 , y4 + x4 ) = (y1 , y2 , y3 , y4 ) + (x1 , x2 , x3 , x4 ) = y + x. 228 6. Let x = (x1 , x2 , x3 , x4 ), y = (y1 , y2 , y3 , y4 ), z = (z1 , z2 , z3 , z4 ) be arbitrary vectors in R4 . Then x + (y + z) = = = = = (x1 , x2 , x3 , x4 ) + (y1 + z1 , y2 + z2 , y3 + z3 , y4 + z4 ) (x1 + (y1 + z1 ), x2 + (y2 + z2 ), x3 + (y3 + z3 ), x4 + (y4 + z4 )) ((x1 + y1 ) + z1 , (x2 + y2 ) + z2 , (x3 + y3 ) + z3 , (x4 + y4 ) + z4 ) (x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4 ) + (z1 , z2 , z3 , z4 ) (x + y) + z. 7. Let x = (x1 , x2 , x3 ) and y = (y1 , y2 , y3 ) be arbitrary vectors in R3 , and let r, s, t be arbitrary real numbers. Then: 1x = 1(x1 , x2 , x3 ) = (1x1 , 1x2 , 1x3 ) = (x1 , x2 , x3 ) = x. (st)x = (st)(x1 , x2 , x3 ) = ((st)x1 , (st)x2 , (st)x3 ) = (s(tx1 ), s(tx2 ), s(tx3 )) = s(tx1 , tx2 , tx3 ) = s(tx). r(x + y) = r(x1 + y1 , x2 + y2 , x3 + y3 ) = (r(x1 + y1 ), r(x2 + y2 ), r(x3 + y3 )) = (rx1 + ry1 , rx2 + ry2 , rx3 + ry3 ) = (rx1 , rx2 , rx3 ) + (ry1 , ry2 , ry3 ) = r x + r y. (s + t)x = (s + t)(x1 , x2 , x3 ) = ((s + t)x1 , (s + t)x2 , (s + t)x3 ) = (sx1 + tx1 , sx2 + tx2 , sx3 + tx3 ) = (sx1 , sx2 , sx3 ) + (tx1 , tx2 , tx3 ) = sx + tx. 8. For example, if x = (2, 2) and y = (−1, −1), then x + y = (1, 1) lies in the first quadrant. If x = (1, 2) and y = (−5, 1), then x + y = (−4, 3) lies in the second quadrant. If x = (1, 1) and y = (−2, −2), then x + y = (−1, −1) lies in the third quadrant. If x = (2, 1) and y = (−1, −5), then x + y = (1, −4) lies in the fourth quadrant. Solutions to Section 4.2 True-False Review: 1. TRUE. This is part 1 of Theorem 4.2.6. 2. FALSE. The statement would be true if it was required that v be nonzero. However, if v = 0, then rv = sv = 0 for all values of r and s, and r and s need not be equal. We conclude that the statement is not true. 3. FALSE. This set is not closed under scalar multiplication. In particular, if k is an irrational number such as k = π and v is an integer, then k v is not an integer. 4. TRUE. We have (x + y) + ((−x) + (−y)) = (x + (−x)) + (y + (−y)) = 0 + 0 = 0, where we have used the vector space axioms in these steps. Therefore, the additive inverse of x + y is (−x) + (−y). 5. TRUE. This is part 1 of Theorem 4.2.6. 6. TRUE. This is called the trivial vector space. Since 0 + 0 = 0 and k · 0 = 0, it is closed under addition and scalar multiplication. Both sides of the remaining axioms yield 0, and 0 is the zero vector, and it is its own additive inverse. 229 7. FALSE. This set is not closed under addition, since 1 + 1 ∈ {0, 1}. Therefore, (A1) fails, and hence, this set does not form a vector space. (It is worth noting that the set is also not closed under scalar multiplication.) 8. FALSE. This set is not closed under scalar multiplication. If k < 0 and x is a positive real number, the result kx is a negative real number, and therefore no longer belongs to the set of positive real numbers. Problems: 1. If x = p/q and y = r/s, where p, q, r, s are integers (q = 0, s = 0), then x + y = (ps + qr)/(qs), which is a rational number. Consequently, the set of all rational numbers is closed under addition. The set is not closed under scalar multiplication since, if we multiply a rational number by an irrational number, the result is an irrational number. 2. Let A = [aij ] and B = [bij ] be upper triangular matrices, and let k be an arbitrary real number. Then whenever i > j , it follows that aij = 0 and bij = 0. Consequently, aij + bij = 0 whenever i > j , and kaij = 0 whenever i > j . Therefore, A + B and kA are upper triangular matrices, so the set of all upper triangular matrices with real elements is closed under both addition and scalar multiplication. 3. V = {y : y + 9y = 4x2 } is not a vector space because it is not closed under vector addition. Let u, v ∈ V . Then u + 9u = 4x2 and v + 9v = 4x2 . It follows that (u + v ) + 9(u + v ) = (u + v ) + 9(u + v ) = u + 9u + v + 9v = 4x2 + 4x2 = 8x2 = 4x2 . Thus, u + v ∈ V . Likewise, V is not closed under scalar / multiplication. 4. V = {y : y + 9y = 0 for all x ∈ I } is closed under addition and scalar multiplication, as we now show: A1: Addition: For u, v ∈ V, u + 9u = 0 and v + 9v = 0, so (u + v ) + 9(u + v ) = u + v + 9u + 9v = u + 9u + v + 9v = 0 + 0 = 0 =⇒ u + v ∈ V , therefore we have closure under addition. A2: Scalar Multiplication: If α ∈ R and u ∈ V , then (αu) + 9(αu) = αu + 9αu = α(u + 9u) = α · 0 = 0, so we also have closure under multiplication. 5. V = {x ∈ Rn : Ax = 0, where A is a fixed matrix} is closed under addition and scalar multiplication, as we now show: Let u, v ∈ V and k ∈ R. A1: Addition: A(u + v) = Au + Av = 0 + 0 = 0, so u + v ∈ V . A2: Scalar Multiplication: A(k u) = kAu = k 0 = 0, thus k u ∈ V . 0 0 6. (a): The zero vector in M2 (R) is 02 = 1 0 (b): Let A = A+B = 1 0 0 1 0 0 and B = 0 0 0 1 0 0 . Since det(02 ) = 0, 02 is an element of S . . Then det(A) = 0 = det(B ), so that A, B ∈ S . However, =⇒ det(A + B ) = 1, so that A + B ∈ S . Consequently, S is not closed under addition. / (c): YES. Note that det(cA) = c2 det(A), so if det(A) = 0, then det(cA) = 0. 7. (1) N is not closed under scalar multiplication, since multiplication of a positive integer by a real number does not, in general, result in a positive integer. 230 (2) There is no zero vector in N. (3) No element of N has an additive inverse in N. 8. Let V be R2 , i.e. {(x, y ) : x, y ∈ R}. With addition and scalar multiplication as defined in the text, this set is clearly closed under both operations. A3: u + v = (u1 , u2 ) + (v1 , v2 ) = (u1 + v1 , u2 + v2 ) = (v1 + u1 , v2 + u2 ) = v + u. A4: [u + v]+ w = [(u1 + v1 , u2 + v2 )]+(w1 , w2 ) = ([u1 + v1 ]+ w1 , [u2 + v2 ]+ w2 ) = (u1 +[v1 + w1 ], u2 +[v2 + w2 ]) = (u1 , u2 ) + [(v1 + w1 , v2 + w2 )] = u + [v + w]. A5: 0 = (0, 0) since (x, y ) + (0, 0) = (x + 0, y + 0) = (x, y ). A6: If u = (a, b), then −u = (−a, −b), a, b ∈ R, since (a, b) + (−a, −b) = (a − a, b − b) = (0, 0). Now, let u = (u1 , u2 ), v = (v1 , v2 ), where u1 , u2 , v1 , v2 ∈ R, and let r, s, t ∈ R. A7: 1 · v = 1(v1 , v2 ) = (1 · v1 , 1 · v2 ) = (v1 , v2 ) = v. A8: (rs)v = (rs)(v1 , v2 ) = ((rs)v1 , (rs)v2 ) = (rsv1 , rsv2 ) = (r(sv1 ), r(sv2 )) = r(sv1 , sv2 ) = r(sv). A9: r(u + v) = r((u1 , u2 ) + (v1 , v2 )) = r(u1 + v1 , u2 + v2 ) = (r(u1 + v1 ), r(u2 + v2 )) = (ru1 + rv1 , ru2 + rv2 ) = (ru1 , ru2 ) + (rv1 , rv2 ) = r(u1 , u2 ) + r(v1 , v2 ) = ru + rv. A10: (r + s)u = (r + s)(u1 , u2 ) = ((r + s)u1 , (r + s)u2 ) = (ru1 + su1 , ru2 + su2 ) = (ru1 , ru2 ) + (su1 , su2 ) = r(u1 , u2 ) + s(u1 , u2 ) = ru + su. Thus, R2 is a vector space. 9. Let A = ab de c f ∈ M2×3 (R). Then we see that the zero vector is 02×3 = A + 02×3 = A. Further, A has additive inverse −A = −a −b −c −d −e −f 0 0 0 0 0 0 , since since A + (−A) = 02×3 . 10. The zero vector of Mm×n (R) is 0m×n , the m × n zero matrix. The additive inverse of the m × n matrix A with (i, j )-element aij is the m × n matrix −A with (i, j )-element −aij . 11. V = {p : p is a polynomial in x of degree 2}. V is not a vector space because it is not closed under addition. For example, x2 ∈ V and −x2 ∈ V , yet x2 + (−x2 ) = 0 ∈ V . / 12. YES. We verify the ten axioms of a vector space. A1: The product of two positive real numbers is a positive real number, so the set is closed under addition. A2: Any power of a positive real number is a positive real number, so the set is closed under scalar multiplication. A3: We have x + y = xy = yx = y + x for all x, y ∈ R+ , so commutativity under addition holds. A4: We have (x + y ) + z = (xy ) + z = (xy )z = x(yz ) = x(y + z ) = x + (y + z ) for all x, y, z ∈ R+ , so that associativity under addition holds. A5: We claim that the zero vector in this set is the real number 1. To see this, note that 1 + x = 1x = x = x1 = x + 1 for all x ∈ R+ . 1 1 1 A6: We claim that the additive inverse of the vector x ∈ R+ is x ∈ R+ . To see this, note that x + x = x x = 1 1 1 = x x = x + x. A7: Note that 1 · x = x1 = x for all x ∈ R+ , so that the unit property holds. A8: For all r, s ∈ R and x ∈ R+ , we have (rs) · x = xrs = xsr = (xs )r = r · xs = r · (s · x), as required for associativity of scalar multiplication. 231 A9: For all r ∈ R and x, y ∈ R+ , we have r · (x + y ) = r · (xy ) = (xy )r = xr y r = xr + y r = r · x + r · y, as required for distributivity of scalar multiplication over vector addition. A10: For all r, s ∈ R and x ∈ R+ , we have (r + s) · x = xr+s = xr xs = xr + xs = r · x + s · x, as required for distributivity of scalar multiplication over scalar addition. The above verification of axioms A1-A10 shows that we have a vector space structure here. 13. Axioms A1 and A2 clearly hold under the given operations. A3: u + v = (u1 , u2 ) + (v1 , v2 ) = (u1 − v1 , u2 − v2 ) = (−(v1 − u1 ), −(v2 − u2 )) = (v1 − u1 , v2 − u2 ) = v + u. Consequently, A3 does not hold. A4: (u+v)+w = (u1 −v1 , u2 −v2 )+(w1 , w2 ) = ((u1 −v1 )−w1 , (u2 −v2 )−w2 ) = (u1 −(v1 +w1 ), u2 −(v2 +w2 )) = (u1 , u2 ) + (v1 + w1 , v2 + w2 ) = (u1 , u2 ) + (v1 − w1 , v2 − w2 ) = u + (v + w). Consequently, A4 does not hold. A5: 0 = (0, 0) since u + 0 = (u1 , u2 ) + (0, 0) = (u1 − 0, u2 − 0) = (u1 , u2 ) = u. A6: If u = (u1 , u2 ), then −u = (u1 , u2 ) since u + (−u) = (u1 , u2 ) + (u1 , u2 ) = (u1 − u1 , u2 − u2 ) = (0, 0) = 0. Each of the remaining axioms do not hold. 14. Axiom A5: The zero vector on R2 with the defined operation of addition is given by (1,1), for if (u1 , u2 ) is any element in R2 , then (u1 , u2 ) + (1, 1) = (u1 · 1, u2 · 1) = (u1 , u2 ). Thus, Axiom A5 holds. Axiom A6: Suppose that (u1 , v1 ) is any element in R2 with additive inverse (a, b). From the first part of the problem, we know that (1, 1) is the zero element, so it must be the case that (u1 , v1 ) + (a, b) = (1, 1) so that (u1 a, v1 b) = (1, 1); hence, it follows that u1 a = 1 and v1 b = 1, but this system is not satisfied for all (u1 , v1 ) ∈ R, namely, (0, 0). Thus, Axiom A6 is not satisfied. 15. Let A, B, C ∈ M2 (R) and r, s, t ∈ R. A3: The addition operation is not commutative since A + B = AB = BA = B + A. A4: Addition is associative since (A + B ) + C = AB + C = (AB )C = A(BC ) = A(B + C ) = A + (B + C ). 10 is the zero vector in M2 (R) because A + I2 = AI2 = A for all A ∈ M2 (R). 01 A6: We wish to determine whether for each matrix A ∈ M2 (R) we can find a matrix B ∈ M2 (R) such that A + B = I2 (remember that we have shown in A5 that the zero vector is I2 ), equivalently, such that AB = I2 . However, this equation can be satisfied only if A is nonsingular, therefore the axiom fails. A7: 1 · A = A is true for all A ∈ M2 (R). A8: (st)A = s(tA) is true for all A ∈ M2 (R) and s, t ∈ R. A9: sA + tA = (sA)(tA) = st(AA) = (s + t)A for all s, t ∈ R and A ∈ M2 (R). Consequently, the axiom fails. A10: rA + rB = (rA)(rB ) = r2 AB = r2 (A + B ). Thus, rA + rB = rA + rB for all r ∈ R, so the axiom fails. A5: I2 = 16. M2 (R) = {A : A is a 2 × 2 real matrix}. Let A, B ∈ M2 (R) and k ∈ R. A1: A ⊕ B = −(A + B ). A2: k · A = −kA. A3 and A4: A ⊕ B = −(A + B ) = −(B + A) = B ⊕ A. Hence, the operation ⊕ is commutative. 232 (A ⊕ B ) ⊕ C = −((A ⊕ B ) + C ) = −(−(A + B ) + C ) = A + B − C , but A ⊕ (B ⊕ C ) = −(A + (B ⊕ C )) = −(A + (−(B + C )) = −A + B + C . Thus the operation ⊕ is not associative. A5: An element B is needed such that A ⊕ B = A for all A ∈ M2 (R), but −(A + B ) = A =⇒ B = −2A. Since this depends on A, there is no zero vector. A6: Since there is no zero vector, we cannot define the additive inverse. Let r, s, t ∈ R. A7: 1 · A = −1A = −A = A. A8: (st) · A = −[(st)A], but s · (t · A) = s · (−tA) = −[s(−tA)] = s(tA) = (st)A, so it follows that (st) · A = s · (t · A). A9: r · (A ⊕ B ) = −r(A ⊕ B ) = −r(−(A + B )) = r(A + B ) = rA + rB = −[(−rA)+(−rB )] = −rA ⊕ (−rB ) = r · A + r · B , whereas rA ⊕ rB = −(rA + rB ) = −rA − rB , so this axiom fails to hold. A10: (s + t) · A = −(s + t)A = −sA +(−tA), but s · A ⊕ t · A = −(sA) ⊕ (−tA) = −[−(sA)+(−tA)] = sA + tA, hence (s + t) · A = s · A ⊕ t · A. We conclude that only the axioms (A1)-(A3) hold. 17. Let C2 = {(z1 , z2 ) : zi ∈ C} under the usual operations of addition and scalar multiplication. A3 and A4: Follow from the properties of addition in C2 . A5: (0, 0) is the zero vector in C2 since (z1 , z2 ) + (0, 0) = (z1 + 0, z2 + 0) = (z1 , z2 ) for all (z1 , z2 ) ∈ C2 . A6: The additive inverse of the vector (z1 , z2 ) ∈ C2 is the vector (−z1 , −z2 ) for all (z1 , z2 ) ∈ C2 . A7-A10: Follows from properties in C2 . Thus, C2 together with its defined operations, is a complex vector space. 18. Let M2 (C) = {A : A is a 2 × 2 matrix with complex entries} under the usual operations of matrix addition and multiplication. A3 and A4: Follows from properties of matrix addition. A5: The zero vector, 0, for M2 (C) is the 2 × 2 zero matrix, 02 . A6: For each vector A = [aij ] ∈ M2 (C), the vector −A = [−aij ] ∈ M2 (C) satisfies A + (−A) = 02 . A7-A10: Follow from properties of matrix algebra. Hence, M2 (C) together with its defined operations, is a complex vector space. 19. Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be vectors in C3 , and let k ∈ R. A1: u + v = (u1 , u2 , u3 ) + (v1 , v2 , v3 ) = (u1 + v1 , u2 + v2 , u3 + v3 ) ∈ C3 . A2: k u = k (u1 , u2 , u3 ) = (ku1 , ku2 , ku3 ) ∈ C3 . A3 and A4: Satisfied by the properties of addition in C3 . A5: (0, 0, 0) is the zero vector in C3 since (0, 0, 0) + (z1 , z2 , z3 ) = (0 + z1 , 0 + z2 , 0 + z3 ) = (z1 , z2 , z3 ) for all (z1 , z2 , z3 ) ∈ C3 . A6: (−z1 , −z2 , −z3 ) is the additive inverse of (z1 , z2 , z3 ) because (z1 , z2 , z3 ) + (−z1 , −z2 , −z3 ) = (0, 0, 0) for all (z1 , z2 , z3 ) ∈ C3 . Let r, s, t ∈ R. A7: 1 · u = 1 · (u1 , u2 , u3 ) = (1u1 , 1u2 , 1u3 ) = (u1 , u2 , u3 ) = u. A8: (st)u = (st)(u1 , u2 , u3 ) = ((st)u1 , (st)u2 , (st)u3 ) = (s(tu1 ), s(tu2 ), s(tu3 )) = s(tu1 , tu2 , tu3 ) = s(t(u1 , u2 , u3 )) = s(tu). A9: r(u + v) = r(u1 + v1 , u2 + v2 , u3 + v3 ) = (r(u1 + v1 ), r(u2 + v2 ), r(u3 + v3 )) = (ru1 + rv1 , ru2 + rv2 , ru3 + rv3 ) = (ru1 , ru2 , ru3 ) + (rv1 , rv2 , rv3 ) = r(u1 , u2 , u3 ) + r(v1 , v2 , v3 ) = ru + rv. A10: (s + t)u = (s + t)(u1 , u2 , u3 ) = ((s + t)u1 , (s + t)u2 , (s + t)u3 ) = (su1 + tu1 , su2 + tu2 , su3 + tu3 ) = (su1 , su2 , su3 ) + (tu1 , tu2 , tu3 ) = s(u1 , u2 , u3 ) + t(u1 , u2 , u3 ) = su + tu. Thus, C3 is a real vector space. 20. NO. If we scalar multiply a vector (x, y, z ) ∈ R3 by a non-real (complex) scalar r, we will obtain the vector (rx, ry, rz ) ∈ R3 , since rx, ry, rz ∈ R. 21. Let k be an arbitrary scalar, and let u be an arbitrary vector in V . Then, using property 2 of Theorem 233 4.2.6, we have k 0 = k (0u) = (k 0)u = 0u = 0. 22. Assume that k is a scalar and u ∈ V such that k u = 0. If k = 0, the desired conclusion is already 1 reached. If, on the other hand, k = 0, then we have k ∈ R and 1 1 · (k u) = · 0 = 0, k k or 1 · k u = 0. k Hence, 1 · u = 0, and the unit property A7 now shows that u = 0, and again, we reach the desired conclusion. 23. We verify the axioms A1-A10 for a vector space. A1: If a0 + a1 x + · · · + an xn and b0 + b1 x + · · · + bn xn belong to Pn , then (a0 + a1 x + · · · + an xn ) + (b0 + b1 x + · · · + bn xn ) = (a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn , which again belongs to Pn . Therefore, Pn is closed under addition. A2: If a0 + a1 x + · · · + an xn and r is a scalar, then r · (a0 + a1 x + · · · + an xn ) = (ra0 ) + (ra1 )x + · · · + (ran )xn , which again belongs to Pn . Therefore, Pn is closed under scalar multiplication. A3: Let p(x) = a0 + a1 x + · · · + an xn and q (x) = b0 + b1 x + · · · + bn xn belong to Pn . Then p(x) + q (x) = (a0 + a1 x + · · · + an xn ) + (b0 + b1 x + · · · + bn xn ) = (a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn = (b0 + a0 ) + (b1 + a1 )x + · · · + (bn + an )xn = (b0 + b1 x + · · · + bn xn ) + (a0 + a1 x + · · · + an xn ) = q (x) + p(x), so Pn satisfies commutativity under addition. A4: Let p(x) = a0 + a1 x + · · · + an xn , q (x) = b0 + b1 x + · · · + bn xn , and r(x) = c0 + c1 x + · · · + cn xn belong to Pn . Then [p(x) + q (x)] + r(x) = [(a0 + a1 x + · · · + an xn ) + (b0 + b1 x + · · · + bn xn )] + (c0 + c1 x + · · · + cn xn ) = [(a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn ] + (c0 + c1 x + · · · + cn xn ) = [(a0 + b0 ) + c0 ] + [(a1 + b1 ) + c1 ]x + · · · + [(an + bn ) + cn ]xn = [a0 + (b0 + c0 )] + [a1 + (b1 + c1 )]x + · · · + [an + (bn + cn )]xn = (a0 + a1 x + · · · + an xn ) + [(b0 + c0 ) + (b1 + c1 )x + · · · + (bn + cn )xn ] = (a0 + a1 x + · · · + an xn ) + [(b0 + b1 x + · · · + bn xn ) + (c0 + c1 x + · · · + cn xn )] = p(x) + [q (x) + r(x)], so Pn satisfies associativity under addition. A5: The zero vector is the zero polynomial z (x) = 0 + 0 · x + · · · + 0 · xn , and it is readily verified that this polynomial satisfies z (x) + p(x) = p(x) = p(x) + z (x) for all p(x) ∈ Pn . A6: The additive inverse of p(x) = a0 + a1 x + · · · + an xn is −p(x) = (−a0 ) + (−a1 )x + · · · + (−an )xn . 234 It is readily verified that p(x) + (−p(x)) = z (x), where z (x) is defined in A5. A7: We have 1 · (a0 + a1 x + · · · + an xn ) = a0 + a1 x + · · · + an xn , which demonstrates the unit property in Pn . A8: Let r, s ∈ R, and p(x) = a0 + a1 x + · · · + an xn ∈ Pn . Then (rs) · p(x) = (rs) · (a0 + a1 x + · · · + an xn ) = [(rs)a0 ] + [(rs)a1 ]x + · · · + [(rs)an ]xn = r[(sa0 ) + (sa1 )x + · · · + (san )xn ] = r[s(a0 + a1 x + · · · + an xn )] = r · (s · p(x)), which verifies the associativity of scalar multiplication. A9: Let r ∈ R, let p(x) = a0 + a1 x + · · · + an xn ∈ Pn , and let q (x) = b0 + b1 x + · · · + bn xn ∈ Pn . Then r · (p(x) + q (x)) = r · ((a0 + a1 x + · · · + an xn ) + (b0 + b1 x + · · · + bn xn )) = r · [(a0 + b0 ) + (a1 + b1 )x + · · · + (an + bn )xn ] = [r(a0 + b0 )] + [r(a1 + b1 )]x + · · · + [r(an + bn )]xn = [(ra0 ) + (ra1 )x + · · · + (ran )xn ] + [(rb0 ) + (rb1 )x + · · · + (rbn )xn ] = [r(a0 + a1 x + · · · + an xn )] + [r(b0 + b1 x + · · · + bn xn )] = r · p(x) + r · q (x), which verifies the distributivity of scalar multiplication over vector addition. A10: Let r, s ∈ R and let p(x) = a0 + a1 x + · · · + an xn ∈ Pn . Then (r + s) · p(x) = (r + s) · (a0 + a1 x + · · · + an xn ) = [(r + s)a0 ] + [(r + s)a1 ]x + · · · + [(r + s)an ]xn = [ra0 + ra1 x + · · · + ran xn ] + [sa0 + sa1 x + · · · + san xn ] = r(a0 + a1 x + · · · + an xn ) + s(a0 + a1 x + · · · + an xn ) = r · p(x) + s · p(x), which verifies the distributivity of scalar multiplication over scalar addition. The above verification of axioms A1-A10 shows that Pn is a vector space. Solutions to Section 4.3 True-False Review: 1. FALSE. The null space of an m × n matrix A is a subspace of Rn , not Rm . 2. FALSE. It is not necessarily the case that 0 belongs to the solution set of the linear system. In fact, 0 belongs to the solution set of the linear system if and only if the system is homogeneous. 3. TRUE. If b = 0, then the line is y = mx, which is a line through the origin of R2 , a one-dimensional subspace of R2 . On the other hand, if b = 0, then the origin does not lie on the given line, and therefore since the line does not contain the zero vector, it cannot form a subspace of R2 in this case. 235 4. FALSE. The spaces Rm and Rn , with m < n, are not comparable. Neither of them is a subset of the other, and therefore, neither of them can form a subspace of the other. 5. TRUE. Choosing any vector v in S , the scalar multiple 0v = 0 still belongs to S . 6. FALSE. In order for a subset of V to form a subspace, the same operations of addition and scalar multiplication must be used in the subset as used in V . 7. FALSE. This set is not closed under addition. For instance, the point (1, 1, 0) lies in the xy -plane, the point (0, 1, 1) lies in the yz -plane, but (1, 1, 0) + (0, 1, 1) = (1, 2, 1) does not belong to S . Therefore, S is not a subspace of V . 8. FALSE. For instance, if we consider V = R3 , then the xy -plane forms a subspace of V , and the x-axis forms a subspace of V . Both of these subspaces contain in common all points along the x-axis. Other examples abound as well. Problems: 1. S = {x ∈ R2 : x = (2k, −3k ), k ∈ R}. (a) S is certainly nonempty. Let x, y ∈ S . Then for some r, s ∈ R, x = (2r, −3r) and y = (2s, −3s). Hence, x + y = (2r, −3r) + (2s, −3s) = (2(r + s), −3(r + s)) = (2k, −3k ), where k = r + s. Consequently, S is closed under addition. Further, if c ∈ R, then cx = c(2r, −3r) = (2cr, −3cr) = (2t, −3t), where t = cr. Therefore S is also closed under scalar multiplication. It follows from Theorem 4.3.2 that S is a subspace of R2 . (b) The subspace S consists of all points lying along the line in the accompanying figure. y y=-3x/2 x Figure 0.0.63: Figure for Exercise 1(b) 2. S = {x ∈ R3 : x = (r − 2s, 3r + s, s), r, s ∈ R}. 236 (a) S is certainly nonempty. Let x, y ∈ S . Then for some r, s, u, v ∈ R, x = (r − 2s, 3r + s, s) and y = (u − 2v, 3u + v, v ). Hence, x + y = (r − 2s, 3r + s, s) + (u − 2v, 3u + v, v ) = ((r + u) − 2(s + v ), 3(r + u) + (s + v ), s + v ) = (a − 2b, 3a + b, b), where a = r + u, and b = s + v . Consequently, S is closed under addition. Further, if c ∈ R, then cx = c(r − 2s, 3r + s, s) = (cr − 2cs, 3cr + cs, cs) = (k − 2l, 3k + l, l), where k = cr and l = cs. Therefore S is also closed under scalar multiplication. It follows from Theorem 4.3.2 that S is a subspace of R3 . (b) The coordinates of the points in S are (x, y, z ) where x = r − 2s, y = 3r + s, z = s. Eliminating r and s between these three equations yields 3x − y + 7z = 0. 3. S = {(x, y ) ∈ R2 : 3x + 2y = 0}. S = ∅ since (0, 0) ∈ S . Closure under Addition: Let (x1 , x2 ), (y1 , y2 ) ∈ S . Then 3x1 + 2x2 = 0 and 3y1 + 2y2 = 0, so 3(x1 + y1 ) + 2(x2 + y2 ) = 0, which implies that (x1 + y1 , x2 + y2 ) ∈ S . Closure under Scalar Multiplication: Let a ∈ R and (x1 , x2 ) ∈ S . Then 3x1 + 2x2 = 0 =⇒ a(3x1 + 2x2 ) = a · 0 =⇒ 3(ax1 ) + 2(ax2 ) = 0, which shows that (ax1 , ax2 ) ∈ S . Thus, S is a subspace of R2 by Theorem 4.3.2. 4. S = {(x1 , 0, x3 , 2) : x1 , x3 ∈ R} is not a subspace of R4 because it is not closed under addition. To see this, let (a, 0, b, 2), (c, 0, d, 2) ∈ S . Then (a, 0, b, 2) + (c, 0, d, 2) = (a + c, 0, b + d, 4) ∈ S . / 5. S = {(x, y, z ) ∈ R3 : x + y + z = 1} is not a subspace of R3 because (0, 0, 0) ∈ S since 0 + 0 + 0 = 1. / 6. S = {u ∈ R2 : Au = b, A is a fixed m × n matrix} is not a subspace of Rn since 0 ∈ S . / 7. S = {(x, y ) ∈ R2 : x2 − y 2 = 0} is not a subspace of R2 , since it is not closed under addition, as we now observe: If (x1 , y1 ), (x2 , y2 ) ∈ S , then (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ). 2 2 (x1 + x2 )2 − (y1 + y2 )2 = x2 + 2x1 x2 + x2 − (y1 + 2y1 y2 + y2 ) 1 2 2 2 = (x2 − y1 ) + (x2 − y2 ) + 2(x1 x2 − y1 y2 ) 1 2 = 0 + 0 + 2(x1 x2 − y1 y2 ) = 0, in general. Thus, (x1 , y1 ) + (x2 , y2 ) ∈ S . / 8. S = {A ∈ M2 (R) : det(A) = 1} is not a subspace of M2 (R). To see this, let k ∈ R be a scalar and let A ∈ S . Then det(kA) = k 2 det(A) = k 2 · 1 = k 2 = 1, unless k = ±1. Note also that det(A) = 1 and det(B ) = 1 does not imply that det(A + B ) = 1. 9. S = {A = [aij ] ∈ Mn (R) : aij = 0 whenever i < j }. Note that S = ∅ since 0n ∈ S . Now let A = [aij ] and B = [bij ] be lower triangular matrices. Then aij = 0 and bij = 0 whenever i < j . Then aij + bij = 0 237 and caij = 0 whenever i < j . Hence A + B = [aij + bij ] and cA = [caij ] are also lower triangular matrices. Therefore S is closed under addition and scalar multiplication. Consequently, S is a subspace of M2 (R) by Theorem 4.3.2. 10. S = {A ∈ Mn (R) : A is invertible} is not a subspace of Mn (R) because 0n ∈ S . / 11. S = {A ∈ M2 (R) : AT = A}. S = ∅ since 02 ∈ S . Closure under Addition: If A, B ∈ S , then (A + B )T = AT + B T = A + B , which shows that A + B ∈ S . Closure under Scalar Multiplication: If r ∈ R and A ∈ S , then (rA)T = rAT = rA, which shows that rA ∈ S . Consequently, S is a subspace of M2 (R) by Theorem 4.3.2. 12. S = {A ∈ M2 (R) : AT = −A}. S = ∅ because 02 ∈ S . Closure under Addition: If A, B ∈ S , then (A + B )T = AT + B T = −A + (−B ) = −(A + B ), which shows that A + B ∈ S . Closure under Scalar Multiplication: If k ∈ R and A ∈ S , then (kA)T = kAT = k (−A) = −(kA), which shows that kA ∈ S . Thus, S is a subspace of M2 (R) by Theorem 4.3.2. 13. S = {f ∈ V : f (a) = f (b)}, where V is the vector space of all real-valued functions defined on [a, b]. Note that S = ∅ since the zero function O(x) = 0 for all x belongs to S . Closure under Addition: If f, g ∈ S , then (f + g )(a) = f (a) + g (a) = f (b) + g (b) = (f + g )(b), which shows that f + g ∈ S . Closure under Scalar Multiplication: If k ∈ R and f ∈ S , then (kf )(a) = kf (a) = kf (b) = (kf )(b), which shows that kf ∈ S . Therefore S is a subspace of V by Theorem 4.3.2. 14. S = {f ∈ V : f (a) = 1}, where V is the vector space of all real-valued functions defined on [a, b]. We claim that S is not a subspace of V . Not Closed under Addition: If f, g ∈ S , then (f + g )(a) = f (a) + g (a) = 1 + 1 = 2 = 1, which shows that f + g ∈ S. / It can also be shown that S is not closed under scalar multiplication. 15. S = {f ∈ V : f (−x) = f (x) for all x ∈ R}. Note that S = ∅ since the zero function O(x) = 0 for all x belongs to S . Let f, g ∈ S . Then (f + g )(−x) = f (−x) + g (−x) = f (x) + g (x) = (f + g )(x) and if c ∈ R, then (cf )(−x) = cf (−x) = cf (x) = (cf )(x), so f + g and c · f belong to S . Therefore, S is closed under addition and scalar multiplication. Therefore, S is a subspace of V by Theorem 4.3.2. 16. S = {p ∈ P2 : p(x) = ax2 + b, a, b ∈ R}. Note that S = ∅ since p(x) = 0 belongs to S . Closure under Addition: Let p, q ∈ S . Then for some a1 , a2 , b1 , b2 ∈ R, p(x) = a1 x2 + b1 and q (x) = a2 x2 + b2 . 238 Hence, (p + q )(x) = p(x) + q (x) = (a1 + a2 )x2 + b1 + b2 = ax2 + b, where a = a1 + a2 and b = b1 + b2 , so that S is closed under addition. Closure under Scalar Multiplication: If k ∈ R, then (kp)(x) = kp(x) = ka1 x2 + kb1 = cx2 + d, where c = ka1 and d = kb1 , so that S is also closed under scalar multiplication. It follows from Theorem 4.3.2 that S is a subspace of P2 . 17. S = {p ∈ P2 : p(x) = ax2 + 1, a ∈ R}. We claim that S is not closed under addition: Not Closed under Addition: Let p, q ∈ S . Then for some a1 , a2 ∈ R, p(x) = a1 x2 + 1 and q (x) = a2 x2 + 1. Hence, (p + q )(x) = p(x) + q (x) = (a1 + a2 )x2 + 2 = ax2 + 2 where a = a1 + a2 . Consequently, p + q ∈ S , and therefore S is not closed under addition. It follows that S / is not a subspace of P2 . 18. S = {y ∈ C 2 (I ) : y + 2y − y = 0}. Note that S is nonempty since the function y = 0 belongs to S . Closure under Addition: Let y1 , y2 ∈ S . (y1 + y2 ) + 2(y1 + y2 ) − (y1 + y2 ) = y1 + y2 + 2(y1 + y2 ) − y1 − y2 = y1 + 2y1 − y1 + y2 + 2y2 − y2 Thus, y1 + y2 ∈ S . = (y1 + 2y1 − y1 ) + (y2 + 2y2 − y2 ) = 0 + 0 = 0. Closure under Scalar Multiplication: Let k ∈ R and y1 ∈ S . (ky1 ) + 2(ky1 ) − (ky1 ) = ky1 + 2ky1 − ky1 = k (y1 + 2y1 − y1 ) = k · 0 = 0, which shows that ky1 ∈ S . Hence, S is a subspace of V by Theorem 4.3.2. 19. S = {y ∈ C 2 (I ) : y + 2y − y = 1}. S is not a subspace of V . We show that S fails to be closed under addition (one can also verify that it is not closed under scalar multiplication, but this is unnecessary if one shows the failure of closure under addition): Not Closed under Addition: Let y1 , y2 ∈ S . (y1 + y2 ) + 2(y1 + y2 ) − (y1 + y2 ) = y1 + y2 + 2(y1 + y2 ) − y1 − y2 = (y1 + 2y1 − y1 ) + (y2 + 2y2 − y2 ) Thus, y1 + y2 ∈ S . / = 1 + 1 = 2 = 1. Or alternatively: Not Closed under Scalar Multiplication: Let k ∈ R and y1 ∈ S . ((ky1 ) + 2(ky1 ) − (ky1 ) = ky1 + 2ky1 − ky1 = k (y1 + 2y1 − y1 ) = k · 1 = k = 1, unless k = 1. Therefore, ky1 ∈ S unless k = 1. / 1 −2 1 20. A = 4 −7 −2 . nullspace(A) = {x ∈ R3 : Ax = 0}. The RREF of the augmented matrix of −1 3 4 239 1000 the system Ax = 0 is 0 1 0 0 . Consequently, the linear system Ax = 0 has only the trivial solution 0010 (0, 0, 0), so nullspace(A) = {(0, 0, 0)}. 1 3 −2 1 6 . nullspace(A) = {x ∈ R3 : Ax = 0}. The RREF of the augmented matrix of 21. A = 3 10 −4 2 5 −6 −1 1 0 −8 −8 0 2 3 0 , so that nullspace(A) = {(8r + 8s, −2r − 3s, r, s) : r, s ∈ R}. the system Ax = 0 is 0 1 00 0 00 1 i −2 4i −5 . nullspace(A) = {x ∈ C3 : Ax = 0}. The RREF of the augmented matrix of 22. A = 3 −1 −3i i 1000 the system Ax = 0 is 0 1 0 0 . Consequently, the linear system Ax = 0 has only the trivial solution 0010 (0, 0, 0), so nullspace(A) = {(0, 0, 0)}. 23. Since the zero function y (x) = 0 for all x ∈ I is not a solution to the differential equation, the set of all solutions does not contain the zero vector from C 2 (I ), hence it is not a vector space at all and cannot be a subspace. 24. (a) As an example, we can let V = R2 . If we take S1 to be the set of points lying on the x-axis and S2 to be the set of points lying on the y -axis, then it is readily seen that S1 and S2 are both subspaces of V . However, the union of these subspaces is not closed under addition. For instance, the points (1, 0) and (0, 1) lie in S1 ∪ S2 , but (1, 0) + (0, 1) = (1, 1) ∈ S1 ∪ S2 . Therefore, the union S1 ∪ S2 does not form a subspace of V . (b) Since S1 and S2 are both subspaces of V , they both contain the zero vector. It follows that the zero vector is an element of S1 ∩ S2 , hence this subset is nonempty. Now let u and v be vectors in S1 ∩ S2 , and let c be a scalar. Then u and v are both in S1 and both in S2 . Since S1 and S2 are each subspaces of V , u + v and cu are vectors in both S1 and S2 , hence they are in S1 ∩ S2 . This implies that S1 ∩ S2 is a nonempty subset of V , which is closed under both addition and scalar multiplication. Therefore, S1 ∩ S2 is a subspace of V . (c) Note that S1 is a subset of S1 + S2 (every vector v ∈ S1 can be written as v + 0 ∈ S1 + S2 ), so S1 + S2 is nonempty. Closure under Addition: Now let u and v belong to S1 + S2 . We may write u = u1 + u2 and v = v1 + v2 , where u1 , v1 ∈ S1 and u2 , v2 ∈ S2 . Then u + v = (u1 + u2 ) + (v1 + v2 ) = (u1 + v1 ) + (u2 + v2 ). Since S1 and S2 are closed under addition, we have that u1 + v1 ∈ S1 and u2 + v2 ∈ S2 . Therefore, u + v = (u1 + u2 ) + (v1 + v2 ) = (u1 + v1 ) + (u2 + v2 ) ∈ S1 + S2 . Hence, S1 + S2 is closed under addition. 240 Closure under Scalar Multiplication: Next, let u ∈ S1 + S2 and let c be a scalar. We may write u = u1 + u2 , where u1 ∈ S1 and u2 ∈ S2 . Thus, c · u = c · u1 + c · u2 , and since S1 and S2 are closed under scalar multiplication, c · u1 ∈ S1 and c · u2 ∈ S2 . Therefore, c · u = c · u1 + c · u2 ∈ S1 + S2 . Hence, S1 + S2 is closed under scalar multiplication. Solutions to Section 4.4 True-False Review: 1. TRUE. By its very definition, when a linear span of a set of vectors is formed, that span becomes closed under addition and under scalar multiplication. Therefore, it is a subspace of V . 2. FALSE. In order to say that S spans V , it must be true that all vectors in V can be expressed as a linear combination of the vectors in S , not simply “some” vector. 3. TRUE. Every vector in V can be expressed as a linear combination of the vectors in S , and therefore, it is also true that every vector in W can be expressed as a linear combination of the vectors in S . Therefore, S spans W , and S is a spanning set for W . 4. FALSE. To illustrate this, consider V = R2 , and consider the spanning set {(1, 0), (0, 1), (1, 1)}. Then the vector v = (2, 2) can be expressed as a linear combination of the vectors in S in more than one way: v = 2(1, 1) and v = 2(1, 0) + 2(0, 1). Many other illustrations, using a variety of different vector spaces, are also possible. 5. TRUE. To say that a set S of vectors in V spans V is to say that every vector in V belongs to span(S ). So V is a subset of span(S ). But of course, every vector in span(S ) belongs to the vector space V , and so span(S ) is a subset of V . Therefore, span(S ) = V . 6. FALSE. This is not necessarily the case. For example, the linear span of the vectors (1, 1, 1) and (2, 2, 2) is simply a line through the origin, not a plane. 7. FALSE. There are vector spaces that do not contain finite spanning sets. For instance, if V is the vector space consisting of all polynomials with coefficients in R, then since a finite spanning set could not contain polynomials of arbitrarily large degree, no finite spanning set is possible for V . 8. FALSE. To illustrate this, consider V = R2 , and consider the spanning set S = {(1, 0), (0, 1), (1, 1)}. The proper subset S = {(1, 0), (0, 1)} is still a spanning set for V . Therefore, it is possible for a proper subset of a spanning set for V to still be a spanning set for V . abc 9. TRUE. The general matrix 0 d e in this vector space can be written as aE11 + bE12 + cE13 + 00f dE22 + eE23 + f E33 , and therefore the matrices in the set {E11 , E12 , E13 , E22 , E23 , E33 } span the vector space. 10. FALSE. For instance, it is easily verified that {x2 , x2 + x, x2 + 1} is a spanning set for P2 , and yet, it contains only polynomials of degree 2. 241 11. FALSE. For instance, consider m = 2 and n = 3. Then one spanning set for R2 is {(1, 0), (0, 1), (1, 1), (2, 2)}, which consists of four vectors. On the other hand, one spanning set for R3 is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, which consists of only three vectors. 12. TRUE. This is explained in True-False Review Question 7 above. Problems: 1. {(1, −1), (2, −2), (2, 3)}. Since v1 = (1, −1), and v2 = (2, 3) are noncolinear, the given set of vectors does span R2 . (See the comment preceding Example 4.4.3 in the text.) 2. The given set of vectors does not span R2 since it does not contain two nonzero and non-colinear vectors. 3. The three vectors in the given set are all collinear. Consequently, the set of vectors does not span R2 . 4. Since 12 4 −1 5 −2 13 1 = −23 = 0, the given vectors are not coplanar, and therefore span R3 . 12 4 −2 3 −1 = −7 = 0, the given vectors are not coplanar, and therefore span R3 . Note that we 11 2 can simply ignore the zero vector (0, 0, 0). 5. Since 6. Since 2 31 −1 −3 1 4 53 = 0, the vectors are coplanar, and therefore the given set does not span R3 . 134 2 4 5 = 0, the vectors are coplanar, and therefore the given set does not span R3 . The linear 356 span of the vectors is those points (x, y, z ) for which the system 7. Since c1 (1, 2, 3) + c2 (3, 4, 5) + c3 (4, 5, 6) = (x, y, z ) 1 134x is consistent. Reducing the augmented matrix 2 4 5 y of this system yields 0 0 356z This system is consistent if and only if x − 2y + z = 0. Consequently, the linear span vectors consists of all points lying on the plane with the equation x − 2y + z = 0. 3 2 0 of 4 x 3 2x − y . 0 x − 2y + z the given set of 8. Let (x1 , x2 ) ∈ R2 and a, b ∈ R. (x1 , x2 ) = av1 + bv2 = a(2, −1) + b(3, 2) = (2a, −a) + (3b, 2b) = (2a + 3b, −a + 2b). It follows that 2x1 − 3x2 x1 + 2x2 2a + 3b = x1 and −a + 2b = x2 , which implies that a = and b = so (x1 , x2 ) = 7 7 2x1 − 3x2 x1 + 2x2 31 9 v1 + v2 . {v1 , v2 } spans R2 , and in particular, (5, −7) = v1 − v2 . 7 7 7 7 9. Let (x, y, z ) ∈ R3 and a, b, c ∈ R. (x, y, z ) = av1 + bv2 + cv3 = a(−1, 3, 2) + b(1, −2, 1) + c(2, 1, 1) = (−a, 3a, 2a) + (b, −2b, b) + (2c, c, c) = (−a + b + 2c, 3a − 2b + c, 2a + b + c). a + b + 2c = x These equalities result in the system: 3a − 2b + c = y 2a + b + c = z 242 Upon solving for a, b, and c we obtain a= −3x + y + 5z −x − 5y + 7z 7x + 3y − z ,b= , and c = . 16 16 16 Consequently, {v1 , v2 , v3 } spans R3 , and (x, y, z ) = −3x + y + 5z 16 v1 + −x − 5y + 7z 16 v2 + 7x + 3y − z 16 v3 . 10. Let (x, y ) ∈ R2 and a, b, c ∈ R. (x, y ) = av1 + bv2 + cv3 = a(1, 1) + b(−1, 2) + c(1, 4) = (a, a) + (−b, 2b) + (c, 4c) = (a − b + c, a + 2b + 4c). a−b+c=x These equalities result in the system: a + 2b + 4c = y Solving the system we find that a= y − x − 3c 2x + y − 6c , and b = 3 3 where c is a free real variable. It follows that (x, y ) = 2x + y − 6c 3 v1 + y − x − 3c 3 v2 + cv3 , which implies that the vectors v1 , v2 , and v3 span R2 . Moreover, if c = 0, then (x, y ) = 2x + y 3 v1 + y−x 3 v2 , so R2 = span{v1 , v2 } also. 11. x = (c1 , c2 , c2 − 2c1 ) = (c1 , 0, −2c1 ) + (0, c2 , c2 ) = c1 (1, 0, −2) + c2 (0, 1, 1) = c1 v1 + c2 v2 . Thus, {(1, 0, −2), (0, 1, 1)} spans S . 12. v = (c1 , c2 , c2 − 2c1 , c1 − 2c2 ) = (c1 , 0, −c1 , c1 ) + (0, c2 , c2 , −2c2 ) = c1 (1, 0, −1, 1) + c2 (0, 1, 1, −2). Thus, {(1, 0, −1, 1), (0, 1, 1, −2)} spans S . 13. x − 2y − z = 0 =⇒ x = 2y + z , so v ∈ R3 . =⇒ v = (2y + z, y, z ) = (2y, y, 0) + (z, 0, z ) = y (2, 1, 0) + z (1, 0, 1). Therefore S = {v ∈ R3 : v = a(2, 1, 0) + b(1, 0, 1), a, b ∈ R}, hence {(2, 1, 0), (1, 0, 1)} spans S . 123 x1 0 14. nullspace(A) = {x ∈ R3 : Ax = 0}. Ax = 0 =⇒ 1 3 4 x2 = 0 . 246 x3 0 Performing Gauss-Jordan elimination on the augmented matrix of the system we obtain: 1230 1230 1010 1 3 4 0 ∼ 0 1 1 0 ∼ 0 1 1 0 . 2460 0000 0000 From the last matrix we find that x1 = −r and x2 = −r where x3 = r, with r ∈ R. Consequently, x = (x1 , x2 , x3 ) = (−r, −r, r) = r(−1, −1, 1). 243 Thus, nullspace(A) = {x ∈ R3 : x = r(−1, −1, 1), r ∈ R} = span{(−1, −1, 1)}. 123 5 2 . nullspace(A) = {x ∈ R4 : Ax = 0}. The RREF of the augmented ma15. A = 1 3 4 2 4 6 1 − 10100 trix of this system is 0 1 1 0 0 . Consequently, nullspace(A) = {r(−1, −1, 1, 0) : r ∈ R} = 00010 span{(−1, −1, 1, 0)}. ab bc 16. A ∈ S =⇒ A = A= a0 00 where a, b, c ∈ R. Thus, 0b b0 + + 0 0 0 c 1 0 =a 0 0 +b 0 1 1 0 +c 00 01 = aA1 + bA2 + cA3 . Therefore S = span{A1 , A2 , A3 }. 17. S = 0α −α 0 A ∈ M2 (R) : A = ,α ∈ R = A ∈ M2 (R) : A = α 0 −1 1 0 = span 0 −1 1 0 18. (a) S = ∅ since 0 0 0 0 ∈ S. Closure under Addition: Let x, y ∈ S . Then, x11 0 x+y = x12 x22 + y11 0 y12 y22 = x11 + y11 0 x12 + y12 x22 + y22 , which implies that x + y ∈ S . Closure under Scalar Multiplication: Let r ∈ R and x ∈ S . Then rx = r x11 0 x12 x22 = rx11 0 rx12 rx22 , which implies thatrx ∈ S . Consequently, S is a subspace of M2 (R) by Theorem 4.3.2. (b) A = a11 0 a12 a22 Therefore, S = span = a11 1 0 10 + a12 00 0 01 , , 0 00 01 00 00 01 + a22 0 0 0 1 . . 19. Let v ∈ span{v1 , v2 } and a, b ∈ R. v = av1 + bv2 = a(1, −1, 2) + b(2, −1, 3) = (a, −a, 2a) + (2b, −b, 3b) = (a + 2b, −a − b, 2a + 3b). Thus, span{v1 , v2 } = {v ∈ R3 : v = (a + 2b, −a − b, 2a + 3b), a, b ∈ R}. Geometrically, span{v1 , v2 } is the plane through the origin determined by the two given vectors. The plane has parametric equations x = a + 2b, y = −a − b, and z = 2a + 3b. If a, b, and c are eliminated from the equations, then the resulting Cartesian equation is given by x − y − z = 0. . 244 20. Let v ∈ span{v1 , v2 } and a, b ∈ R. v = av1 + bv2 = a(1, 2, −1) + b(−2, −4, 2) = (a, 2a, −a) + (−2b, −4b, 2b) = (a − 2b, 2a − 4b, −a + 2b) = (a − 2b)(1, 2, −1) = k (1, 2, −1), where k = a − 2b. Thus, span{v1 , v2 } = {v ∈ R3 : v = k (1, 2, −1), k ∈ R}. Geometrically, span{v1 , v2 } is the line through the origin determined by the vector (1, 2, −1). 21. Let v ∈ span{v1 , v2 , v3 } and a, b, c ∈ R. v = av1 + bv2 + cv3 = a(1, 1, −1) + b(2, 1, 3) + c(−2, −2, 2) = (a, a, −a) + (2b, b, 3b) + (−2c, −2c, 2c) = (a + 2b − 2c, a + b − 2c, −a + 3b + 2c). a + 2b − 2c = x a + b − 2c = y Assuming that v = (x, y, z ) and using the last ordered triple, we obtain the system: −a + 3b + 2c = z Performing Gauss-Jordan elimination on the augmented matrix of the system, we obtain: 1 2 −2 x 1 2 −2 1 0 −2 x 2y − x 1 1 −2 y ∼ 0 −1 . 0 y−x ∼ 0 1 0 x−y −1 3 2z 0 5 0 x+z 00 0 5y − 4x + z It is clear from the last matrix that the subspace, S , of R3 is a plane through (0, 0, 0) with Cartesian equation 4x − 5y − z = 0. Moreover, {v1 , v2 } also spans the subspace S since v = av1 + bv2 + cv3 = a(1, 1, −1) + b(2, 1, 3) + c(−2, −2, 2) = a(1, 1, −1) + b(2, 1, 3) − 2c(1, 1, −1) = (a − 2c)(1, 1, −1) + b(2, 1, 3) = dv1 + bv2 where d = a − 2c ∈ R. 22. If v ∈ span{v1 , v2 } then there exist a, b ∈ R such that v = av1 + bv2 = a(1, −1, 2) + b(2, 1, 3) = (a, −a, 2a) + (2b, b, 3b) = (a + 2b, −a + b, 2a + 3b). a + 2b = 3 −a + b = 3 Hence, v = (3, 3, 4) is in span{v1 , v2 } provided there exists a, b ∈ R satisfying the system: 2a + 3b = 4 Solving this system we find that a = −1 and b = 2. Consequently, v = −v1 + 2v2 so that (3, 3, 4) ∈ span{v1 , v2 }. 23. If v ∈ span{v1 , v2 } then there exist a, b ∈ R such that v = av1 + bv2 = a(−1, 1, 2) + b(3, 1, −4) = (−a, a, 2a) + (3b, b, −4b) = (−a + 3b, a + b, 2a − 4b). −a + 3b = 5 a+b= 3 Hence v = (5, 3, −6) is in span{v1 , v2 } provided there exists a, b ∈ R satisfying the system: 2a − 4b = −6 Solving this system we find that a = 1 and b = 2. Consequently, v = v1 + 2v2 so that (5, 3, −6) ∈ span{v1 , v2 }. 24. If v ∈ span{v1 , v2 } then there exist a, b ∈ R such that v = av1 + bv2 = (3a, a, 2a) + (−2b, −b, b) = (3a − 2b, a − b, 2a + b). 3a − 2b = 1 a−b= 1 Hence v = (1, 1, −2) is in span{v1 , v2 } provided there exists a, b ∈ R satisfying the system: 2a + b = −2 245 1 1 1 −1 3 −2 1 reduces to 0 1 −2 , it follows that the system has no solution. Hence it Since 1 −1 0 0 2 2 1 −2 must be the case that (1, 1, −2) ∈ span{v1 , v2 }. / 25. If p ∈ span{p1 , p2 } then there exist a, b ∈ R such that p(x) = ap1 (x) + bp2 (x), so p(x) = 2x2 − x + 2 is in span{p1 , p2 } provided there exist a, b ∈ R such that 2x2 − x + 2 = a(x − 4) + b(x2 − x + 3) = ax − 4a + bx2 − bx + 3b = bx2 + (a − b)x + (3b − 4a). Equating like coefficients and solving, we find that a = 1 and b = 2. Thus, 2x2 − x + 2 = 1 · (x − 4) + 2 · (x2 − x + 3) = p1 (x) + 2p2 (x) so p ∈ span{p1 , p2 }. 26. Let A ∈ span{A1 , A2 , A3 } and c1 , c2 , c3 ∈ R. 1 −1 0 A = c1 A1 + c2 A2 + c3 A3 = c1 + c2 2 0 −2 = c1 2c1 −c1 0 + 0 −2c2 Therefore span{A1 , A2 , A3 } = c2 c2 + 3c3 c3 0 2c3 A ∈ M2 (R) : A = 1 1 + c3 30 12 c1 + 3c3 −c1 + c2 2c1 − 2c2 + c3 c2 + 2c3 c1 + 3c3 −c1 + c2 . 2c1 − 2c2 + c3 c2 + 2c3 = . 27. Let A ∈ span{A1 , A2 } and a, b ∈ R. 12 −2 1 a 2a −2b b a − 2b 2a + b A = aA1 + bA2 = a +b = + = . −1 3 1 −1 −a 3a b −b −a + b 3a − b a − 2b 2a + b So span{A1 , A2 } = A ∈ M2 (R) : A = . Now, to determine whether B ∈ span{A1 , A2 }, −a + b 3a − b a − 2b 2a + b 31 let = . This implies that a = 1 and b = −1, thus B ∈ span{A1 , A2 }. −a + b 3a − b −2 4 28. (a) The general vector in span{f, g } is of the form h(x) = c1 cosh x + c2 sinh x where c1 , c2 ∈ R. (b) Let h ∈ S and c1 , c2 ∈ R. Then h(x) = c1 f (x) + c2 g (x) = c1 cosh x + c2 sinh x ex − e−x c1 ex c1 e−x c2 ex c2 e−x ex + e−x + + − + c2 = 2 2 2 2 2 2 c1 + c2 x c1 − c2 −x x −x = e+ e = d1 e + d2 e 2 2 c1 + c2 c1 − c2 where d1 = and d2 = . Therefore S =span{ex , e−x }. 2 2 29. The origin in R3 . = c1 30. All points lying on the line through the origin with direction v1 . 31. All points lying on the plane through the origin containing v1 and v2 . 32. If v1 = v2 = 0, then the subspace is the origin in R3 . If at least one of the vectors is nonzero, then the subspace consists of all points lying on the line through the origin in the direction of the nonzero vector. 33. Suppose that S is a subset of S . We must show that every vector in span(S ) also belongs to span(S ). Every vector v that lies in span(S ) can be expressed as v = c1 v1 + c2 v2 + · · · + ck vk , where v1 , v2 , . . . , vk 246 belong to S . However, since S is a subset of S , v1 , v2 , . . . , vk also belong to S , and therefore, v belongs to span(S ). Thus, we have shown that every vector in span(S ) also lies in span(S ). 34. Proof of =⇒: We begin by supposing that span{v1 , v2 , v3 } = span{v1 , v2 }. Since v3 ∈ span{v1 , v2 , v3 }, our supposition implies that v3 ∈ span{v1 , v2 }, which means that v3 can be expressed as a linear combination of the vectors v1 and v2 . Proof of ⇐=: Now suppose that v3 can be expressed as a linear combination of the vectors v1 and v2 . We must show that span{v1 , v2 , v3 } = span{v1 , v2 }, and we do this by showing that each of these subsets is a subset of the other. Since it is clear that span{v1 , v2 } is a subset of span{v1 , v2 , v3 }, we focus our attention on proving that every vector in span{v1 , v2 , v3 } belongs to span{v1 , v2 }. To see this, suppose that v belongs to span{v1 , v2 , v3 }, so that we may write v = c1 v1 + c2 v2 + c3 v3 . By assumption, v3 can be expressed as a linear combination of v1 and v2 , so that we may write v3 = d1 v1 + d2 v2 . Hence, we obtain v = c1 v1 + c2 v2 + c3 v3 = c1 v1 + c2 v2 + c3 (d1 v1 + d2 v2 ) = (c1 + c3 d1 )v1 + (c2 + c3 d2 )v2 ∈ span{v1 , v2 }, as required. Solutions to Section 4.5 1. FALSE. For instance, consider the vector space V = R2 . Here are two different minimal spanning sets for V : {(1, 0), (0, 1)} and {(1, 0), (1, 1)}. Many other examples of this abound. 2. TRUE. We have seven column vectors, and each of them belongs to R5 . Therefore, the number of vectors present exceeds the number of components in those vectors, and hence they must be linearly dependent. 3. FALSE. For instance, the 7 × 5 zero matrix, 07×5 , does not have linearly independent columns. 4. TRUE. Any linear dependencies within the subset also represent linear dependencies within the original, larger set of vectors. Therefore, if the nonempty subset were linearly dependent, then this would require that the original set is also linearly dependent. In other words, if the original set is linearly independent, then so is the nonempty subset. 5. TRUE. This is stated in Theorem 4.5.21. 6. TRUE. If we can write v = c1 v1 + c2 v2 + · · · + ck vk , then {v, v1 , v2 , . . . , vk } is a linearly dependent set. 7. TRUE. This is a rephrasing of the statement in True-False Review Question 5 above. 8. FALSE. None of the vectors (1, 0), (0, 1), and (1, 1) in R2 are proportional to each other, and yet, they form a linearly dependent set of vectors. 9. FALSE. The illustration given in part (c) of Example 4.5.22 gives an excellent case-in-point here. Problems: 1. {(1, −1), (1, 1)}. These vectors are elements of R2 . Since there are two vectors, and the dimension of R2 11 = 0. Now is two, Corollary 4.5.15 states that the vectors will be linearly dependent if and only if −1 1 11 = 2 = 0. Consequently, the given vectors are linearly independent. −1 1 247 2. {(2, −1), (3, 2), (0, 1)}. These vectors are elements of R2 , but since there are three vectors, the vectors are linearly dependent by Corollary 4.5.15. Let v1 = (2, −1), v2 = (3, 2), v3 = (0, 1). We now determine a dependency relationship. The condition c1 v1 + c2 v2 + c3 v3 = 0 requires 2c1 + 3c2 = 0 and −c1 + 2c2 + c3 = 0. 1 0 −3 7 2 01 7 c1 = 3r, c2 = −2r, c3 = 7r. Consequently, 3v1 − 2v2 + 7v3 = 0. The RREF of the augmented matrix of this system is 0 0 . Hence the system has solution 3. {(1, −1, 0), (0, 1, −1), (1, 1, 1)}. These vectors are elements of R3 . 1 01 −1 1 1 = 3 = 0, so by Corollary 4.5.15, the vectors are linearly independent. 0 −1 1 4. {(1, 2, 3), (1, −1, 2), (1, −4, 1)}. These vectors are elements of R3 . 1 1 1 2 −1 −4 = 0, so by Corollary 4.5.15, the vectors are linearly dependent. Let v1 = (1, 2, 3), v2 = 3 2 1 (1, −1, 2), v3 = (1, −4, 1). We determine a dependency relation. The condition c1 v1 + c2 v2 + c3 v3 = 0 requires c1 + c2 + c3 = 0, 2c1 − c2 − 4c3 = 0, 3c1 + 2c2 + c3 = 0. 1 0 −1 0 2 0 . Hence c1 = r, c2 = −2r, c3 = r, The RREF of the augmented matrix of this system is 0 1 00 00 and so v1 − 2v2 + v3 = 0. 5. Given {(−2, 4, −6), (3, −6, 9)}. The vectors are linearly dependent because 3(−2, 4, −6) + 2(3, −6, 9) = (0, 0, 0), which gives a linear dependency relation. Alternatively, let a, b ∈ R and observe that a(−2, 4, −6) + b(3, −6, 9) = (0, 0, 0) =⇒ (−2a, 4a, −6a) + (3b, −6b, 9b) = (0, 0, 0). −2a + 3b = 0 4a − 6b = 0 The last equality results in the system: −6a + 9b = 0 1 −3 0 2 3 0 0 , which implies that a = b. Thus, the The RREF of the augmented matrix of this system is 0 2 0 00 given set of vectors is linearly dependent. 6. {(1, −1, 2), (2, 1, 0)}. Let a, b ∈ R. a(1, −1, 2) + b(2, 1, 0) = (0, 0, 0) =⇒ (a, −a, 2a) + (2b, b, 0) = (0, 0, 0) =⇒ (a + 2b, −a + b, 2a) = (0, 0, 0). The a + 2b = 0 −a + b = 0 . Since the only solution of the system is a = b = 0, it last equality results in the system: 2a = 0 follows that the vectors are linearly independent. 7. {(−1, 1, 2), (0, 2, −1), (3, 1, 2), (−1, −1, 1)}. These vectors are elements of R3 . Since there are four vectors, it follows from Corollary 4.5.15 that the vectors are linearly dependent. Let 248 v1 = (−1, 1, 2), v2 = (0, 2, −1), v3 = (3, 1, 2), v4 = (−1, −1, 1). Then c1 v1 + c2 v2 + c3 v3 + c4 v4 = 0 requires −c1 + 3c3 − c4 = 0, c1 + 2c2 + c3 − c4 = 0, 2c1 − c2 + 2c3 + c4 = 0. 2 100 0 5 The RREF of the augmented matrix of this system is 0 1 0 − 3 0 so that 5 0 0 1 −1 0 5 c1 = −2r, c2 = 3r, c3 = r, c4 = 5r. Hence, −2v1 + 3v2 + v3 + 5v4 = 0. 8. {(1, −1, 2, 3), (2, −1, 1, −1), (−1, 1, 1, 1)}. Let a, b, c ∈ R. a(1, −1, 2, 3) + b(2, −1, 1, −1) + c(−1, 1, 1, 1) = (0, 0, 0, 0) =⇒ (a, −a, 2a, 3a) + (2b, −b, b, −b) + (−c, c, c, c) = (0, 0, 0, 0) =⇒ (a + 2b − c, −a − b + c, 2a + b + c, 3a − b + c) = (0, 0, 0, 0). a + 2b − c = 0 −a − b + c = 0 The last equality results in the system: . The RREF of the augmented matrix of this system 2a + b + c = 0 3a − b + c = 0 1000 0 1 0 0 is 0 0 1 0 . Consequently, a = b = c = 0. Thus, the given set of vectors is linearly independent. 0000 9. {(2, −1, 0, 1), (1, 0, −1, 2), (0, 3, 1, 2), (−1, 1, 2, 1)}. These vectors are elements in R4 . By CET (row 1), we obtain: 2 1 0 −1 031 −1 3 1 −1 03 −1 03 1 012+ 0 −1 1 = 21 = 0. = 2 −1 1 2 − 0 −1 1 2 221 121 1 22 1 22 1 Thus, it follows from Corollary 4.5.15 that the vectors are linearly independent. 147 2 5 8 = 0, the given set of vectors is linearly dependent in R3 . Further, since the vectors 369 are not collinear, it follows that span{v1 , v2 , v3 } is a plane 3-space. 10. Since 11. (a) Since 2 1 −3 −1 3 −9 5 −4 12 = 0, the given set of vectors is linearly dependent. (b) By inspection, we see that v3 = −3v2 . Hence v2 and v3 are colinear and therefore span{v2 , v3 } is the line through the origin that has the direction of v2 . Further, since v1 is not proportional to either of these vectors, it does not lie along the same line, hence v1 is not in span{v2 , v3 }. 12. 1 1 k 0 2 k 1 k 6 = 10 02 0k 1 k−1 6−k = 2 k k−1 6−k = (3 − k )(k + 4). Now, the determinant is zero when 249 k = 3 or k = −4. Consequently, by Corollary 4.5.15, the vectors are linearly dependent if and only if k = 3 or k = −4. 13. {(1, 0, 1, k ), (−1, 0, k, 1), (2, 0, 1, 3)}. These vectors are elements in R4 . Let a, b, c ∈ R. a(1, 0, 1, k ) + b(−1, 0, k, 1) + c(2, 0, 1, 3) = (0, 0, 0, 0) =⇒ (a, 0, a, ka) + (−b, 0, kb, b) + (2c, 0, c, 3c) = (0, 0, 0, 0) =⇒ (a − b + 2c, 0, a + kb + c, ka + b + 3c) (0, 0, 0, 0). = a − b + 2c = 0 a + kb + c = 0 . Evaluating the determinant of the coefficient The last equality results in the system: ka + b + 3c = 0 matrix, we obtain 1 −1 2 1 0 0 1 k 1 = 1 k+1 −1 = 2(k + 1)(2 − k ). k k + 1 3 − 2k k 13 Consequently, the system has only the trivial solution, hence the given set of vectors are linearly independent if and only if k = 2, −1. 14. {(1, 1, 0, −1), (1, k, 1, 1), (2, 1, k, 1), (−1, 1, 1, k )}. These vectors are elements in R4 . Let a, b, c, d ∈ R. a(1, 1, 0, −1) + b(1, k, 1, 1) + c(2, 1, k, 1) + d(−1, 1, 1, k ) = (0, 0, 0, 0) =⇒ (a, a, 0, −a) + (b, kb, b, b) + (2c, c, kc, c) + (−d, d, d, kd) = (0, 0, 0, 0) =⇒ (a + b + 2c − d, a + kb + c + d, b + kc + d, −a + b + c + kd) = (0, 0, 0, 0). a + b + 2c − d = 0 a + kb + c + d = 0 The last equality results in the system: . By Corollary 3.2.5, this system has the b + kc + d = 0 −a + b + c + kd = 0 trivial solution if and only if the determinant of the coefficient matrix is zero. Evaluating the determinant of the coefficient matrix, we obtain: 1 1 2 −1 1 1 2 −1 k − 1 −1 2 0 −k 2 + k − 1 3 − k 0 k − 1 −1 2 1k1 1 1 k 1 k 1 = =1 = 01k 1 0 1 k 1 2 3 k−1 0 3 − 2k k−3 0 2 3 k−1 −1 1 1 k k2 − k + 1 1 k2 − k + 1 k − 3 = (k − 3)(k − 1)(k + 2). = (k − 3) 3 − 2k k−3 3 − 2k 1 For the original set of vectors to be linearly independent, we need a = b = c = 0. We see that this condition will be true provided that k ∈ R such that k = 3, 1, −2. = 15. Let a, b, c ∈ R. a 1 0 1 1 +b 2 −1 0 1 +c 3 0 6 4 = 0 0 0 0 a + 2b + 3c = 0 a + 2b + 3c a − b + 6c 00 a − b + 6c = 0 . =⇒ = . The last equation results in the system: 0 a + b + 4c 00 a + b + 4c = 0 10 50 The RREF of the augmented matrix of this system is 0 1 −1 0 , which implies that the system has 00 00 an infinite number of solutions. Consequently, the given matrices are linearly dependent. 250 16. Let a, b ∈ R. a 2 −1 3 4 −1 1 +b 2 3 = 0 0 0 0 2a − b = 0 3a + b = 0 2a − b −a + 2b 00 =⇒ = . The last equation results in the system: . This system 3a + b 4a + 3b 00 −a + 2b = 0 4a + 3b = 0 has only the trivial solution a = b = 0, thus the given matricies are linearly independent in M2 (R). 17. Let a, b, c ∈ R. a =⇒ 1 1 0 2 a − b + 2c b+c a + 2b + 5c 2a + b + 7c +b = −1 1 21 0 0 0 0 +c 21 57 = 0 0 0 0 . The last equation results in the system: 1030 0 1 1 0 The RREF of the augmented matrix of this system is 0 0 0 0 0000 an infinite number of solutions. Consequently, the given matrices are a − b + 2c = 0 a + 2b + 5c = 0 2a + b + 7c = 0 b+c=0 . , which implies that the system has linearly dependent. 18. Let a, b ∈ R. ap1 (x) + bp2 (x) = 0 =⇒ a(1 − x) + b(1 + x) = 0 =⇒ (a + b) + (−a + b)x = 0. a+b=0 Equating like coefficients, we obtain the system: . Since the only solution to this system is −a + b = 0 a = b = 0, it follows that the given vectors are linearly independent. 19. Let a, b ∈ R. ap1 (x) + bp2 (x) = 0 =⇒ a(2 + 3x) + b(4 + 6x) = 0 =⇒ (2a + 4b) + (3a + 6b)x = 0. 2a + 4b = 0 Equating like coefficients, we obtain the system: . The RREF of the augmented matrix of this 3a + 6b = 0 120 system is , which implies that the system has an infinite number of solutions. Thus, the given 000 vectors are linearly dependent. 20. Let c1 , c2 ∈ R. c1 p1 (x) + c2 p2 (x) = 0 =⇒ c1 (a + bx) + c2 (c + dx) = 0 =⇒ (ac1 + cc2 ) + (bc1 + dc2 )x = 0. ac1 + cc2 = 0 Equating like coefficients, we obtain the system: . The determinant of the matrix of coefbc1 + dc2 = 0 ac ficients is = ad − bc. Consequently, the system has just the trivial solution, and hence p1 (x) and bd p2 (x) are linearly independent if and only if ad − bc = 0. 21. Since cos 2x = cos2 x − sin2 x, f1 (x) = f3 (x) − f2 (x) so it follows that f1 , f2 , and f3 are linearly dependent in C ∞ (−∞, ∞). 4 22. Let v1 = (1, 2, 3), v2 = (−3, 4, 5), v3 = (1, − 3 , − 5 ). By inspection, we see that v2 = −3v3 . Further, 3 since v1 and v2 are not proportional they are linearly independent. Consequently, {v1 , v2 } is a linearly independent set of vectors and span{v1 , v2 } =span{v1 , v2 , v3 }. 23. Let v1 = (1, 2, −1), v2 = (3, 1, 5), v3 = (0, 0, 0), v4 = (−1, 2, 3). Since v3 = 0, it is certainly true that span{v1 , v2 v3 } =span{v1 , v2 , v3 v4 }. Further, since det[v1 , v2 , v3 ] = −42 = 0, {v1 , v2 , v3 } is a linearly independent set. 251 24. Since we have four vectors in R3 , the given set is linearly dependent. We could determine the specific linear dependency between the vectors to find a linearly independent subset, but in this case, if we just take any 1 13 three of the vectors, say (1, −1, 1), (1, −3, 1), (3, 1, 2), then −1 −3 1 = 2 = 0, so that these vectors are 1 12 linearly independent. Consequently, span{(1, −1, 1), (1, −3, 1), (3, 1, 2)} = span{(1, 1, 1), (1, −1, 1), (1, −3, 1), (3, 1, 2)}. 25. Let v1 = (1, 1, −1, 1), v2 = (2, −1, 3, 1), v3 = (1, 1, 2, 1), v4 = (2, −1, 2, 1). 1 21 2 1 −1 1 −1 Since = 0, the set {v1 , v2 , v3 , v4 } is linearly dependent. We now determine the linearly −1 32 2 1 11 1 dependent relationship. The RREF of the augmented matrix corresponding to the system 100 0 1 0 is 0 0 1 000 that a linearly c1 (1, 1, −1, 1) + c2 (2, −1, 3, 1) + c3 (1, 1, 2, 1) + c4 (2, −1, 2, 1) = (0, 0, 0, 0) 1 0 3 1 0 , so that c1 = −r, c2 = −3r, c3 = r, c4 = 3r, where r is a free variable. It follows −1 0 3 00 dependent relationship between the given set of vectors is −v1 − 3v2 + v3 + 3v4 = 0 so that v1 = −3v2 + v3 + 3v4 . Consequently, span{v2 , v3 , v4 } = span{v1 , v2 , v3 , v4 }, and {v2 , v3 , v4 } is a linearly independent set. 26. Let A1 = 1 3 2 4 , A2 = −1 5 2 7 , A3 = 3 1 2 1 . Then c1 A1 + c2 A2 + c3 A3 = 02 requires that c1 − c2 + 3c3 = 0, 2c1 + 2c2 + 2c3 = 0, 3c1 + 5c2 + c3 = 0, 4c1 + 7c2 + c3 = 0. 10 20 0 1 −1 0 . Consequently, the matrices are The RREF of the augmented matrix of this system is 0 0 0 0 00 00 linearly dependent. Solving the system gives c1 = −2r, c2 = c3 = r. Hence, a linearly dependent relationship is −2A1 + A2 + A3 = 02 . 27. We first determine whether the given set of polynomials is linearly dependent. Let p1 (x) = 2 − 5x, p2 (x) = 3 + 7x, p3 (x) = 4 − x. Then c1 (2 − 5x) + c2 (3 + 7x) + c3 (4 − x) = 0 requires 2c1 + 3c2 + 4c3 = 0 and − 5c1 + 7c2 − c3 = 0. 252 This system has solution (−31r, −18r, 29r), where r is a free variable. Consequently, the given set of polynomials is linearly dependent, and a linearly dependent relationship is −31p1 (x) − 18p2 (x) + 29p3 (x) = 0, or equivalently, p3 (x) = 1 [31p1 (x) + 18p2 (x)]. 29 Hence, the linearly independent set of vectors {2 − 5x, 3 + 7x} spans the same subspace of P1 as that spanned by {2 − 5x, 3 + 7x, 4 − x}. 28. We first determine whether the given set of polynomials is linearly dependent. Let p1 (x) = 2 + x2 , p2 (x) = 4 − 2x + 3x2 , p3 (x) = 1 + x. Then c1 (2 + x2 ) + c2 (4 − 2x + 3x2 ) + c3 (1 + x) = 0 leads to the system 2c1 + 4c2 + c3 = 0, −2c2 + c3 = 0, c1 + 3c2 = 0. This system has solution (−3r, r, 2r) where r is a free variable. Consequently, the given set of vectors is linearly dependent, and a specific linear relationship is −3p1 (x) + p2 (x) + 2p3 (x) = 0, or equivalently, p2 (x) = 3p1 (x) − 2p3 (x). Hence, the linearly independent set of vectors {2 + x2 , 1 + x} spans the same subspace of P2 as that spanned by the given set of vectors. 1 x x2 29. W [f1 , f2 , f3 ](x) = 0 1 2x 002 linearly independent on I . = 2. Since W [f1 , f2 , f3 ](x) = 0 on I , it follows that the functions are 30. W [f1 , f2 , f3 ](x) = sin x cos x tan x cos x − sin x sec2 x − sin x − cos x 2 tan x sec2 x = sin x cos x tan x cos x − sin x sec2 x 0 0 tan x(2 sec2 x + 1) = − tan x(2 sec2 x + 1). W [f1 , f2 , f3 ](x) is not always zero over I , so the vectors are linearly independent by Theorem 4.5.21 1 3x x2 − 1 2x 31. W [f1 , f2 , f3 ](x) = 0 3 = 00 2 independent set on I by Theorem 4.5.21. 3 0 2x 2 = 6 = 0 on I . Consequently, {f1 , f2 , f3 } is a linearly e2x e3x e−x 11 1 10 0 2e2x 3e3x −e−x = e4x 2 3 −1 = e4x 2 1 −3 4 5 −3 49 1 4e2x 9e3x e−x Wronskian is never zero, the functions are linearly indepenedent on (−∞, ∞). 32. W [f1 , f2 , f3 ](x) = = 12e4x . Since the 253 33. On [0, ∞), f2 = 7f1 , so that the functions are linearly dependent on this interval. Therefore W [f1 , f2 ](x) = 3x3 7x2 0 for x ∈ [0, ∞). However, on (−∞, 0), W [f1 , f2 ](x) = = −21x4 = 0. Since the Wronskian is 9x2 14x not zero for all x ∈ (−∞, ∞), the functions are linearly independent on that interval. 1 x 2x − 1 01 2 00 0 are linearly dependent on (−∞, ∞). 34. W [f1 , f2 , f3 ](x) = = 0. By inspection, we see that f3 = 2f2 − f1 , so that the functions 35. We show that the Wronskian (the determinant can be computed by cofactor expansion along the first row) is identically zero: ex ex ex e−x −e−x e−x cosh x sinh x cosh x = −(cosh x + sinh x) − (cosh x − sinh x) + 2 cosh x = 0. Thus, the Wronskian is identically zero on (−∞, ∞). Furthermore, {f1 , f2 , f3 } is a linearly dependent set because 1 1 1 1 1 1 ex + e−x − f1 (x) − f2 (x) + f3 (x) = − ex − e−x + cosh x = − ex − e−x + = 0 for all x ∈ I. 2 2 2 2 2 2 2 36. We show that the Wronskian is identically zero for f1 (x) = ax3 and f2 (x) = bx3 , which covers the functions in this problem as a special case: ax3 3ax2 bx3 3bx2 = 3abx5 − 3abx5 = 0. Next, let a, b ∈ R. If x ≥ 0, then af1 (x) + bf2 (x) = 0 =⇒ 2ax3 + 5bx3 = 0 =⇒ (2a + 5b)x3 = 0 =⇒ 2a + 5b = 0. If x ≤ 0, then af1 (x) + bf2 (x) = 0 =⇒ 2ax3 − 3bx3 = 0 =⇒ (2a − 3b)x3 = 0 =⇒ 2a − 3b = 0. Solving the resulting system, we obtain a = b = 0; therefore, {f1 , f2 } is a linearly independent set of vectors on (−∞, ∞). 37. (a) When x > 0, f2 (x) = 1 and when x < 0, f2 (x) = −1; thus f2 (0) does not exist, which implies that f2 ∈ C 1 (−∞, ∞). / (b) Let a, b ∈ R. On the interval (−∞, 0), ax + b(−x) = 0, which has more than the trivial solution for a and b. Thus, {f1 , f2 } is a linearly dependent set of vectors on (−∞, 0). On the interval [0, ∞), ax + bx = 0 =⇒ a + b = 0, which has more than the trivial solution for a and b. Therefore {f1 , f2 } is a linearly dependent set of vectors on [0, ∞). a−b=0 On the interval (−∞, ∞), a and b must satisfy: ax + b(−x) = 0 and ax + bx = 0, that is, . Since a+b=0 this system has only a = b = 0 as its solution, {f1 , f2 } is a linearly independent set of vectors on (−∞, ∞). 254 y y = f2(x) = -x f1(x) = - f2(x) on (-∞, 0) y = f1(x) = x = f2(x) f1(x) = f2(x) on (0, ∞) x y = f1(x) = x Figure 0.0.64: Figure for Exercise 37 38. Let a, b ∈ R. ax + bx = 0 if x = 0 (a + b)x = 0 =⇒ a(0) + b(1) = 0 if x = 0 b=0 =⇒ a = 0 and b = 0 so {f1 , f2 } is a linearly independent set on I . af1 (x) + bf2 (x) = 0 =⇒ 39. Let a, b, c ∈ R and x ∈ (−∞, ∞). af1 (x) + bf2 (x) + cf3 (x) = 0 =⇒ a(x − 1) + b(2x) + c(3) = 0 for x ≥ 1 2a(x − 1) + b(2x) + c(3) = 0 for x < 1 a + 2b = 0 and − a + 3c = 0 (a + 2b)x + (−a + 3c) = 0 =⇒ (2a + 2b)x + (−2a + 3c) = 0 2a + 2b = 0 and − 2a + 3c = 0. Since the only solution to this system of equations is a = b = c = 0, it follows that the given functions are linearly independent on (−∞, ∞). The domain space may be divided into three types of intervals: (1) interval subsets of (−∞, 1), (2) interval subset of [1, ∞), (3) intervals containing 1 where 1 is not an endpoint of the intervals. =⇒ For intervals of type (3): Intervals such as type (3) are treated as above [with domain space of (−∞, ∞)]: vectors are linearly independent. For intervals of type (1): a(2(x − 1)) + b(2x) + c(3) = 0 =⇒ (2a + 2b)x + (−2a + 3c) = 0 =⇒ 2a + 2b = 0, and −2a + 3c = 0. Since this system has three variables with only two equations, the solution to the system is not unique, hence intervals of type (1) result in linearly dependent vectors. For intervals of type (2): a(x − 1)+ b(2x)+ c(3) = 0 =⇒ a +2b = 0 and −a +3c = 0. As in the last case, this system has three variables with only two equations, so it must be the case that intervals of type (2) result in linearly dependent vectors. 40. (a) Let f0 (x) = 1, f1 (x) = x, f2 (x) = x2 , f3 (x) = x3 . 1 x x2 x3 0 1 2x 3x2 = 12 = 0. Hence, {f1 , f2 , f3 , f4 } is linearly independent on Then W [f1 , f2 , f3 , f4 ](x) = 002 6x 000 6 any interval. 255 1 x x2 · · · xn 0 1 2x · · · nxn−1 n−2 . (b) W [f0 , f1 , f2 , . . . , fn ](x) = 0 0 2 · · · n(n − 1)x .. . . .. . . .. . . 0 0 0 ··· n! The matrix corresponding to this determinant is upper triangular, so the value of the determinant is given by the product of all of the diagonal entries. W [f0 , f1 , f2 , . . . , fn ](x) = 1 · 1 · 2 · 6 · 24 · · · n!, which is not zero regardless of the actual domain of x. Consequently, the functions are linearly independent on any interval. 41. (a) Let f1 (x) = er1 x , f2 (x) = er2 x and f3 (x) = er3 x . Then 1 er1 x er2 x er3 x r1 er1 x r2 er2 x r3 er3 x = er1 x er2 x er3 x r1 W [f1 , f2 , f3 ](x) = 2 2 2 2 r1 r1 er1 x r2 er2 x r3 er3 x 1 r2 2 r2 1 r3 2 r3 = e(r1 +r2 +r3 )x (r3 − r1 )(r3 − r2 )(r2 − r1 ). If ri = rj for i = j , then W [f1 , f2 , f3 ](x) is never zero, and hence the functions are linearly independent on any interval. If, on the other hand, ri = rj with i = j , then fi − fj = 0, so that the functions are linearly dependent. Thus, r1 , r2 , r3 must all be different in order that f1 , f2 , f3 are linearly independent. (b) er1 x r1 er1 x 2 r1 er1 x . . . W [f1 , f2 , f3 , . . . , fn ](x) = er2 x · · · r2 er2 x · · · 2 r2 er2 x · · · . . . n r1 −1 er1 x n r2 −1 er2 x = er1 x er2 x · · · ern x ∗ r1 x r2 x =e e r1 x r2 x =e e rn x ···e ··· 1 r1 2 r1 . . . n r1 −1 ern x rn ern x 2 rn ern x . . . n rn−1 ern x 1 r2 2 r2 . . . n r2 −1 ··· ··· ··· ··· 1 rn 2 rn . . . n rn−1 V (r1 , r2 , . . . , rn ) rn x ···e (rm − ri ). 1≤i<m≤n Now from the last equality, we see that if ri = rj for i = j , where i, j ∈ {1, 2, 3, . . . , n}, then W [f1 , f2 , . . . , fn ](x) is never zero, and hence the functions are linearly independent on any interval. If, on the other hand, ri = rj with i = j , then f1 − fj = 0, so that the functions are linearly dependent. Thus {f1 , f2 , . . . , fn } is a linearly independent set if and only if all the ri are distinct for i ∈ {1, 2, 3, . . . , n}. (*Note: V (r1 , r2 , . . . , rn ) is the n × n Vandermonde determinant. See Section 3.3 Problem 21). 42. Let a, b ∈ R. Assume that av + bw = 0. Then a(αv1 + v2 ) + b(v1 + αv2 ) = 0 which implies that (αa + b)v1 + (a + bα)v2 = 0. Now since it is given that v1 and v2 are linearly independent, αa + b = 0 and α1 = 0. That is, if and a + bα = 0. This system has only the trivial solution for a and b if and only if 1α only if α2 − 1 = 0, or α = ±1. Therefore, the vectors are linearly independent provided that α = ±1. 43. It is given that v1 and v2 are linearly independent. Let u1 = a1 v1 + b1 v2 , u2 = a2 v1 + b2 v2 , and u3 = a3 v1 + b3 v2 where a1 , a2 , a3 , b1 , b2 , b3 ∈ R. Let c1 , c2 , c3 ∈ R. Then 256 c1 u1 + c2 u2 + c3 u3 = 0 =⇒ c1 (a1 v1 + b1 v2 ) + c2 (a2 v1 + b2 v2 ) + c3 (a3 v1 + b3 v2 ) = 0 =⇒ (c1 a1 + c2 a2 + c3 a3 )v1 + (c1 b1 + c2 b2 + c3 b3 )v2 = 0. c1 a1 + c2 a2 + c3 a3 = 0 Now since v1 and v2 are linearly independent: There are an infinite number c1 b1 + c2 b2 + c3 b3 = 0. of solutions to this homogeneous system since there are three unknowns but only two equations. Hence, {u1 , u2 , u3 } is a linearly dependent set of vectors. 44. Given that {v1 , v2 , . . . , vm } is a linearly independent set of vectors in a vector space V , and m aik vi , k ∈ {1, 2, . . . , n}. uk = (44.1) i=1 n (a) Let ck ∈ R, k ∈ {1, 2, . . . , n}. Using system (44.1) and ck uk = 0, we obtain: k=1 n m m k=1 aik vi i=1 n i=1 ck k=1 = 0 ⇐⇒ aik ck vi = 0. Since the vi for each i ∈ {1, 2, . . . , m} are linearly independent, n aik ck = 0, 1 ≤ i ≤ m. (44.2) k=1 But this is a system of m equations with n unknowns c1 , c2 , . . . , cn . Since n > m, the system has more unknowns than equations, and so has nontrivial solutions. Thus, {u1 , u2 , . . . , un } is a linearly dependent set. (b) If m = n, then the system (44.2) has a trivial solution ⇐⇒ the coefficient matrix of the system is invertible ⇐⇒ det[aij ] = 0. (c) If n < m, then the homogeneous system (44.2) has just the trivial solution if and only if rank(A) = n. Recall that for a homogeneous system, rank(A# ) = rank(A). (d) Corollary 4.5.15. 45. We assume that c1 (Av1 ) + c2 (Av2 ) + · · · + cn (Avn ) = 0. Our aim is to show that c1 = c2 = · · · = cn = 0. We manipulate the left side of the above equation as follows: c1 (Av1 ) + c2 (Av2 ) + · · · + cn (Avn ) = 0 A(c1 v1 ) + A(c2 v2 ) + · · · + A(cn vn ) = 0 A(c1 v1 + c2 v2 + · · · + cn vn ) = 0. Since A is invertible, we can left multiply the last equation by A−1 to obtain c1 v1 + c2 v2 + · · · + cn vn = 0. Since {v1 , v2 , . . . , vn } is linearly independent, we can now conclude directly that c1 = c2 = · · · = cn = 0, as required. 257 46. Assume that c1 v1 + c2 v2 + c3 v3 = 0. We must show that c1 = c2 = c3 = 0. Let us suppose for the moment that c3 = 0. In that case, we can solve the above equation for v3 : v3 = − c1 c2 v1 − v2 . c3 c3 However, this contradicts the assumption that v3 does not belong to span{v1 , v2 }. Therefore, we conclude that c3 = 0. Our starting equation therefore reduces to c1 v1 + c2 v2 = 0. Now the assumption that {v1 , v2 } is linearly independent shows that c1 = c2 = 0. Therefore, c1 = c2 = c3 = 0, as required. 47. Assume that c1 v1 + c2 v2 + · · · + ck vk + ck+1 vk+1 = 0. We must show that c1 = c2 = · · · = ck+1 = 0. Let us suppose for the moment that ck+1 = 0. In that case, we can solve the above equation for vk+1 : vk+1 = − c1 ck+1 v1 − c2 ck+1 v2 − · · · − ck ck+1 vk . However, this contradicts the assumption that vk+1 does not belong to span{v1 , v2 , . . . , vk }. Therefore, we conclude that ck+1 = 0. Our starting equation therefore reduces to c1 v1 + c2 v2 + · · · + ck vk = 0. Now the assumption that {v1 , v2 , . . . , vk } is linearly independent shows that c1 = c2 = · · · = ck = 0. Therefore, c1 = c2 = · · · = ck = ck+1 = 0, as required. 48. Let {v1 , v2 , . . . , vk } be a set of vectors with k ≥ 2. Suppose that vk can be expressed as a linear combination of {v1 , v2 , . . . , vk−1 }. We claim that span{v1 , v2 , . . . , vk } = span{v1 , v2 , . . . , vk−1 }. Since every vector belonging to span{v1 , v2 , . . . , vk−1 } evidently belongs to span{v1 , v2 , . . . , vk }, we focus on showing that every vector in span{v1 , v2 , . . . , vk } also belongs to span{v1 , v2 , . . . , vk−1 }: Let v ∈ span{v1 , v2 , . . . , vk }. We therefore may write v = c1 v1 + c2 v2 + · · · + ck vk . By assumption, we may write vk = d1 v1 + d2 v2 + · · · + dk−1 vk−1 . Therefore, we obtain v = c1 v1 + c2 v2 + · · · + ck vk = c1 v1 + c2 v2 + · · · + ck−1 vk−1 + ck (d1 v1 + d2 v2 + · · · + dk−1 vk−1 ) = (c1 + ck d1 )v1 + (c2 + ck d2 )v2 + · · · + (ck−1 + ck dk−1 )vk−1 ∈ span{v1 , v2 , . . . , vk−1 }. This shows that every vector belonging to span{v1 , v2 , . . . , vk } also belongs to span{v1 , v2 , . . . , vk−1 }, as needed. 49. We first prove part 1 of Proposition 4.5.7. Suppose that we have a set {u, v} of two vectors in a vector space V . If {u, v} is linearly dependent, then we have cu + dv = 0, where c and d are not both zero. Without loss of generality, suppose that c = 0. Then we have d u = − v, c so that u and v are proportional. Conversely, if u and v are proportional, then v = cu for some constant c. Thus, cu − v = 0, which shows that {u, v} is linearly dependent. 258 For part 2 of Proposition 4.5.7, suppose the zero vector 0 belongs to a set S of vectors in a vector space V . Then 1 · 0 is a linear dependency among the vectors in S , and therefore S is linearly dependent. 50. Suppose that {v1 , v2 , . . . , vk } spans V and let v be any vector in V . By assumption, we can write v = c1 v1 + c2 v2 + · · · + ck vk , for some constants c1 , c2 , . . . , ck . Therefore, c1 v1 + c2 v2 + · · · + ck vk − v = 0 is a linear dependency among the vectors in {v, v1 , v2 , . . . , vk }. Thus, {v, v1 , v2 , . . . , vk } is linearly dependent. 51. Let S = {p1 , p2 , . . . , pk } and assume without loss of generality that the polynomials are listed in decreasing order by degree: deg(p1 ) > deg(p2 ) > · · · > deg(pk ). To show that S is linearly independent, assume that c1 p1 + c2 p2 + · · · + ck pk = 0. We wish to show that c1 = c2 = · · · = ck = 0. We require that each coefficient on the left side of the above equation is zero, since we have 0 on the right-hand side. Since p1 has the highest degree, none of the terms c2 p2 ,c3 p3 , . . . , ck pk can cancel the leading coefficient of p1 . Therefore, we conclude that c1 = 0. Thus, we now have c2 p2 + c3 p3 + · · · + ck pk = 0, and we can repeat this argument again now to show successively that c2 = c3 = · · · = ck = 0. Solutions to Section 4.6 1. FALSE. It is not enough that S spans V . It must also be the case that S is linearly independent. 2. FALSE. For example, R2 is not a subspace of R3 , since R2 is not even a subset of R3 . 3. TRUE. Any set of two non-proportional vectors in R2 will form a basis for R2 . 4. FALSE. We have dim[Pn ] = n + 1 and dim[Rn ] = n. 5. FALSE. For example, if V = R2 , then the set S = {(1, 0), (2, 0), (3, 0)}, consisting of 3 > 2 vectors, fails to span V , a 2-dimensional vector space. 6. TRUE. We have dim[P3 ] = 4, and so any set of more than four vectors in P3 must be linearly dependent (a maximal linearly independent set in a 4-dimensional vector space consists of four vectors). 7. FALSE. For instance, the two vectors 1 + x and 2 + 2x in P3 are linearly dependent. 8. TRUE. Since M3 (R) is 9-dimensional, any set of 10 vectors in this vector space must be linearly dependent by Theorem 4.6.4. 9. FALSE. Only linearly independent sets with fewer than n vectors can be extended to a basis for V . 10. TRUE. We can build such a subset by choosing vectors from the set as follows. Choose v1 to be any vector in the set. Now choose v2 in the set such that v2 ∈ span{v1 }. Next, choose v3 in the set such that v3 ∈ span{v1 , v2 }. Proceed in this manner until it is no longer possible to find a vector in the set that is not spanned by the collection of previously chosen vectors. This point will occur eventually, since V is 259 finite-dimensional. Moreover, the chosen vectors will form a linearly independent set, since each vi is chosen from outside span{v1 , v2 , . . . , vi−1 }. Thus, the set we obtain in this way is a linearly independent set of vectors that spans V , hence forms a basis for V . 11. FALSE. The set of all 3 × 3 upper triangular matrices forms a 6-dimensional subspace of M3 (R), not a 3-dimensional subspace. One basis is given by {E11 , E12 , E13 , E22 , E23 , E33 }. Problems: 1. dim[R2 ] = 2. There are two vectors, so if they are to form a basis for R2 , they need to be linearly 1 −1 independent: = 2 = 0. This implies that the vectors are linearly independent, hence they form a 1 1 2 basis for R . 2. dim[R3 ] = 3. There are three vectors, so if they are to form a basis for R3 , they need to be linearly 1 3 1 1 = 13 = 0. This implies that the vectors are linearly independent, hence they independent: 2 −1 1 2 −1 form a basis for R3 . 3. dim[R3 ] = 3. There are three vectors, so if they are to form a basis for R3 , they need to be linearly 1 2 3 5 11 = 0. This implies that the vectors are linearly dependent, hence they do not independent: −1 1 −2 −5 form a basis for R3 . 4. dim[R4 ] = 4. We need 4 linearly independent vectors in order to span R4 . However, there are only 3 vectors in this set. Thus, the vectors cannot be a basis for R4 . 5. dim[R4 ] = 4. There are four vectors, so if they 1 2 −1 1 2 −1 2 0 −1 2 1 1 1 −1 = independent: 0 3 1 0 3 1 1 0 −5 0 2 −1 −2 2 Since this determinant is nonzero, the given vectors basis for R4 . are to form a basis for R3 , they need to be linearly 2 −7 0 −5 −1 2 −3 −3 31 1 = −11. 31 1= = 1 −5 0 −2 −5 0 −2 −2 are linearly independent. Consequently, they form a 6. dim[R4 ] = 4. There are four vectors, so if they are to form a basis for R3 , they need to be linearly 010k −1 1 0 00k −1 0 1 0 0 1 2 − −1 1 0 = −(−1 + 2k ) − (−k 2 ) = k 2 − 2k + 1 = =− independent: 0112 k01 k01 k001 2 (k − 1) = 0 when k = 1. Thus, the vectors will form a basis for R4 provided k = 1. 7. The general vector p(x) ∈ P3 can be represented as p(x) = a0 + a1 x + a2 x2 + a3 x3 . Thus P3 = span{1, x, x2 , x3 }. Further, {1, x, x2 , x3 } is a linearly independent set since W [1, x, x2 , x3 ] = 1 x x2 x3 0 1 2x 3x2 002 6x 000 6 = 12 = 0. Consequently, S = {1, x, x2 , x3 } is a basis for P3 and dim[P3 ] = 4. Of course, S is not the only basis for P3 . 260 8. Many acceptable bases are possible here. One example is S = {x3 , x3 + 1, x3 + x, x3 + x2 }. All of the polynomials in this set have degree 3. We verify that S is a basis: Assume that c1 (x3 ) + c2 (x3 + 1) + c3 (x3 + x) + c4 (x3 + x2 ) = 0. Thus, (c1 + c2 )x3 + c4 x2 + c3 x + c2 = 0, from which we quickly see that c1 = c2 = c3 = c4 = 0. Thus, S is linearly independent. Since P3 is 4-dimensional, we can now conclude from Corollary 4.6.13 that S is a basis for P3 . 9. Ax = 0 =⇒ 1 3 −2 −6 x1 x2 = 0 0 . The augmented matrix for this system is 1 30 −2 −6 0 ∼ 130 ; thus, x1 + 3x2 = 0, or x1 = 3x2 . Let x2 = r so that x1 = 3r where r ∈ R. Consequently, 000 S = {x ∈ R2 : x = r(3, 1), r ∈ R} = span{(3, 1)}. It follows that {(3, 1)} is a basis for S and dim[S ] = 1. 000 x1 0 10. Ax = 0 =⇒ 0 0 0 x2 = 0 . The RREF of the augmented matrix for this system 010 x3 0 0100 is 0 0 0 0 . We see that x2 = 0, and x1 and x3 are free variables: x1 = r and x2 = s. Hence, 0000 (x1 , x2 , x3 ) = (r, 0, s) = r(1, 0, 0) + s(0, 0, 1), so that the solution set of the system is S = {x ∈ R3 : x = r(1, 0, 0) + s(0, 0, 1), r, s ∈ R}. Therefore we see that {(1, 0, 0), (0, 0, 1)} is a basis for S and dim[S ] = 2. 1 −1 4 x1 0 3 −2 x2 = 0 . The RREF of the augmented matrix for this system is 11. Ax = 0 =⇒ 2 1 2 −2 x3 0 10 20 0 1 −2 0 . If we let x3 = r then (x1 , x2 , x3 ) = (−2r, 2r, r) = r(−2, 2, 1), so that the solution set of 00 00 the system is S = {x ∈ R3 : x = r(−2, 2, 1), r ∈ R}. Therefore we see that {(−2, 2, 1)} is a basis for S and dim[S ] = 1. 1 −1 2 3 x1 0 2 −1 3 4 x2 0 = . The RREF of the augmented matrix for this system 12. Ax = 0 =⇒ 1 0 1 1 x3 0 3 −1 4 5 x4 0 10 1 10 0 1 −1 −2 0 . If we let x3 = r, x4 = s then x2 = r + 2s and x1 = −r − s. Hence, the solution set is 0 0 0 0 0 00 0 00 of the system is S = {x ∈ R4 : x = r(−1, 1, 1, 0) + s(−1, 2, 0, 1), r, s ∈ R} = span{(−1, 1, 1, 0), (−1, 2, 0, 1)}. Further, the vectors v1 = (−1, 1, 1, 0), v2 = (−1, 2, 0, 1) are linearly independent since c1 v1 + c2 v2 = 0 =⇒ c1 (−1, 1, 1, 0) + c2 (−1, 2, 0, 1) = (0, 0, 0, 0) =⇒ c1 = c2 = 0. Consequently, {(−1, 1, 1, 0), (−1, 2, 0, 1)} is a basis for S and dim[S ] = 2. 13. If we let y = r and z = s where r, s ∈ R, then x = 3r − s. It follows that any ordered triple in S can be written in the form: 261 (x, y, z ) = (3r − s, r, s) = (3r, r, 0) + (−s, 0, s) = r(3, 1, 0) + s(−1, 0, 1), where r, s ∈ R. If we let v1 = (3, 1, 0) and v2 = (−1, 0, 1), then S = {v ∈ R3 : v = r(3, 1, 0) + s(−1, 0, 1), r, s ∈ R} = span{v1 , v2 }; moreover, v1 and v2 are linearly independent for if a, b ∈ R and av1 + bv2 = 0, then a(3, 1, 0) + b(−1, 0, 1) = (0, 0, 0), which implies that (3a, a, 0) + (−b, 0, b) = (0, 0, 0), or (3a − b, a, b) = (0, 0, 0). In other words, a = b = 0. Since {v1 , v2 } spans S and is linearly independent, it is a basis for S . Also, dim[S ] = 2. 14. S = {x ∈ R3 : x = (r, r − 2s, 3s − 5r), r, s ∈ R} = {x ∈ R3 : x = (r, r, −5r) + (0, −2s, 3s), r, s ∈ R} = {x ∈ R3 : x = r(1, 1, −5) + s(0, −2, 3), r, s ∈ R}. Thus, S = span{(1, 1, −5), (0, −2, 3)}. The vectors v1 = (1, 1, −5) and v2 = (0, −2, 3) are linearly independent for if a, b ∈ R and av1 + bv2 = 0, then a(1, 1, −5)+b(0, −2, 3) = (0, 0, 0) =⇒ (a, a, −5a)+(0, −2b, 3b) = (0, 0, 0) =⇒ (a, a−2b, 3b−5a) = (0, 0, 0) =⇒ a = b = 0. It follows that {v1 , v2 } is a basis for S and dim[S ] = 2. ab 0c 15. S = {A ∈ M2 (R) : A = , a, b, c ∈ R}. Each vector in S can be written as 1 0 A=a so that S = span 1 0 0 0 0 0 , 1 0 a b c −a A=a so that S = span 1 0 0 −1 it follows that a basis for S is , 0 0 1 0 0 0 +b 00 01 10 00 , pendent, it follows that a basis for S is 16. S = {A ∈ M2 (R) : A = 0 0 1 0 00 01 +c , . Since the vectors in this set are clearly linearly inde, 0 0 1 0 , 0 0 0 1 , and therefore dim[S ] = 3. , a, b, c ∈ R}. Each vector in S can be written as 1 0 0 −1 , 1 0 0 −1 +b 0 0 1 0 +c 0 1 0 0 , 00 . Since this set of vectors is clearly linearly independent, 10 01 00 , , , and therefore dim[S ] = 3. 00 10 17. We see directly that v3 = 2v1 . Let v be an arbitrary vector in S . Then v = c1 v1 + c2 v2 + c3 v3 = (c1 + 2c3 )v1 + c2 v2 = d1 v1 + d2 v2 , where d1 = (c1 + 2c3 ) and d2 = c2 . Hence S = span{v1 , v2 }. Further, v1 and v2 are linearly independent for if a, b ∈ R and av1 + bv2 = 0, then a(1, 0, 1) + b(0, 1, 1) = (0, 0, 0) =⇒ (a, 0, a) + (0, b, b) = (0, 0, 0) =⇒ (a, b, a + b) = (0, 0, 0) =⇒ a = b = 0. Consequently, {v1 , v2 } is a basis for S and dim[S ] = 2. 262 18. f3 depends on f1 and f2 since sinh x = f1 (x) − f2 (x) ex − e−x . Thus, f3 (x) = . 2 2 ex e−x = −2 = 0 for all x ∈ R, so {f1 , f2 } is a linearly independent set. Thus, {f1 , f2 } x e −e−x is a basis for S and dim[S ] = 2. W [f1 , f2 ](x) = 19. The given set of matrices is linearly dependent because it contains the zero vector. Consequently, the 13 −1 4 5 −6 matrices A1 = , A2 = , A3 = span the same subspace of M2 (R) as that −1 2 11 −5 1 spanned by the original set. We now determine whether {A1 , A2 , A3 } is linearly independent. The vector equation: c1 A1 + c2 A2 + c3 A3 = 02 leads to the linear system c1 − c2 + 5c3 = 0, 3c1 + 4c2 − 6c3 = 0, −c1 + c2 − 5c3 = 0, 2c1 + c2 + c3 = 0. This system has solution (−2r, 3r, r), where r is a free variable. Consequently, {A1 , A2 , A3 } is linearly dependent with linear dependency relation −2A1 + 3A2 + A3 = 0, or equivalently, A3 = 2A1 − 3A2 . It follows that the set of matrices {A1 , A2 } spans the same subspace of M2 (R) as that spanned by the original set of matrices. Further {A1 , A2 } is linearly independent by inspection, and therefore it is a basis for the subspace. 20. (a) We must show that every vector (x, y ) ∈ R2 can be expressed as a linear combination of v1 and v2 . Mathematically, we express this as c1 (1, 1) + c2 (−1, 1) = (x, y ) which implies that c1 − c2 = x and c1 + c2 = y. Adding the equations here, we obtain 2c1 = x + y or c1 = 1 (x + y ). 2 Now we can solve for c2 : 1 (y − x). 2 Therefore, since we were able to solve for c1 and c2 in terms of x and y , we see that the system of equations is consistent for all x and y . Therefore, {v1 , v2 } spans R2 . c2 = y − c1 = (b) 1 −1 1 1 = 2 = 0, so the vectors are linearly independent. (c) We can draw this conclusion from part (a) alone by using Theorem 4.6.12 or from part (b) alone by using Theorem 4.6.10. 21. (a) We must show that every vector (x, y ) ∈ R2 can be expressed as a linear combination of v1 and v2 . Mathematically, we express this as c1 (2, 1) + c2 (3, 1) = (x, y ) 263 which implies that 2c1 + 3c2 = x and c1 + c2 = y. From this, we can solve for c1 and c2 in terms of x and y : c1 = 3y − x and c2 = x − 2y. Therefore, since we were able to solve for c1 and c2 in terms of x and y , we see that the system of equations is consistent for all x and y . Therefore, {v1 , v2 } spans R2 . (b) 2 1 3 1 = −1 = 0, so the vectors are linearly independent. (c) We can draw this conclusion from part (a) alone by using Theorem 4.6.12 or from part (b) alone by using Theorem 4.6.10. 22. dim[R3 ] = 3. There 0 Theorem 4.6.1. Since 6 3 is a basis for R3 . are 3 vectors, so if they are linearly independent, then they are a basis of R3 by 3 6 0 −3 = 81 = 0, the given vectors are linearly independent, and so {v1 , v2 , v3 } 3 0 23. dim[P2 ] = 3. There are 3 vectors, so {p1 , p2 , p3 } may be a basis for P2 depending on α. To be a basis, the set of vectors must be linearly independent. Let a, b, c ∈ R. Then ap1 (x) + bp2 (x) + cp3 (x) = 0 =⇒ a(1 + αx2 ) + b(1 + x + x2 ) + c(2 + x) = 0 =⇒ (a + + 2c) + (b + c)x + (aα + b)x2 = 0. Equating like coefficients in the last equality, we obtain the b a + b + 2c = 0 b + c = 0 Reduce the augmented matrix of this system. system: aα + b = 0. 1 1 20 11 2 1120 0 0 1 1 0 ∼ 0 1 1 0 ∼ 0 1 1 0 . α100 0 α − 1 2α 0 0 0 α+1 0 For this system to have only the trivial solution, the last row of the matrix must not be a zero row. This means that α = −1. Therefore, for the given set of vectors to be linearly independent (and thus a basis), α can be any value except −1. 24. dim[P2 ] = 3. There are 3 vectors, so in order to form a basis for P2 , they must be linearly independent. Since we are dealing with functions, we will use the Wronskian. W [p1 , p2 , p3 ](x) = 1 + x x(x − 1) 1 2x − 1 0 2 1 + 2x2 4x 4 = 1 + x x(x − 1) 1 2x − 1 0 2 1 + 2x 2 0 = −2. Since the Wronskian is nonzero, {p1 , p2 , p3 } is linearly independent, and hence forms a basis for P2 . 2 1 x 3x 2−1 25. W [p0 , p1 , p2 ](x) = 0 1 = 3 = 0, so {p0 , p1 , p2 } is a linearly independent set. Since 3x 00 3 dim[P2 ] = 3, it follows that {p0 , p1 , p2 } is a basis for P2 . 26. (a) Suppose that a, b, c, d ∈ R. −1 1 13 1 a +b +c 01 −1 0 1 0 2 +d 0 −1 2 3 = 02 264 −a + b + c = 0 a + 3b − d = 0 −a + b + c a + 3b − c =⇒ = 02 . The last matrix results in the system: The −b + c + 2d a + 2c + 3d −b + c + 2d = 0 a + 2c + 3d = 0. 1 −1 −1 00 0 4 1 −1 0 , which implies that a = b = RREF of the augmented matrix of this system is 0 0 5 7 0 0 0 0 10 c = d = 0. Hence, the given set of vectors is linearly independent. Since dim[M2 (R)] = 4, it follows that {A1 , A2 , A3 , A4 } is a basis for M2 (R). (b) We wish to express the vector 5 7 6 8 =a 5 7 6 8 −1 0 1 1 in the form +b 1 −1 3 0 +c 1 1 0 2 +d 0 −1 2 3 . Matching entries on each side of this equation (upper left, upper right, lower left, and lower right), we obtain a linear system with augmented matrix −1 11 05 1 3 0 −1 6 . 0 −1 1 2 7 1 02 38 Solving this system by Gaussian elimination, we find that a = − 34 , b = 12, c = − 55 , and d = 3 3 have 34 −1 1 55 1 0 56 0 −1 56 13 =− + 12 − + . 78 01 −1 0 12 2 3 3 3 3 56 3. Thus, we 27. x 1 1 −1 11 x 5 −6 2 (a) Ax = 0 =⇒ 2 −3 x3 5 0 2 −3 x4 0 = 0 . The augmented matrix for this linear system is 0 1 1 −1 10 1 1 −1 10 1 1 −1 10 11 −1 1 2 3 2 −3 5 −6 0 ∼ 0 −5 7 −8 0 ∼ 0 −5 7 −8 0 ∼ 0 1 −1.4 5 0 2 −3 0 0 −5 7 −8 0 0 0 0 00 00 0 1. A12 (−2), A13 (−5) 2. A23 (−1) 10 1.6 0 . 00 3. M2 (− 1 ) 5 We see that x3 = r and x4 = s are free variables, and therefore nullspace(A) is 2-dimensional. Now we can check directly that Av1 = 0 and Av2 = 0, and since v1 and v2 are linearly independent (they are non-proportional), they therefore form a basis for nullspace(A) by Corollary 4.6.13. (b) An arbitrary vector (x, y, z, w) in nullspace(A) can be expressed as a linear combination of the basis vectors: (x, y, z, w) = c1 (−2, 7, 5, 0) + c2 (3, −8, 0, 5), 265 where c1 , c2 ∈ R. 28. (a) An arbitrary matrix in S takes the form 1 = a 0 −1 0 −1 0 1 0 0 + b 0 0 0 1 0 −1 a b −a − b c d −c − d −a − c −b − d a + b + c + d −1 00 0 0 0 0 0 + c 1 0 −1 + d 0 1 −1 . 1 −1 0 1 0 −1 1 Therefore, we have the following basis for S : 1 0 −1 0 1 −1 0 00 0 , 0 0 0 , 1 −1 0 1 0 −1 1 −1 0 0 0 0 0 0 −1 , 0 1 −1 . 0 1 0 −1 1 From this, we see that dim[S ] = 4. (b) We see that each of the matrices 100 01 0 0 0 , 0 0 000 00 0 00 0 , 0 0 0 00 1 00 0 , 1 0 0 00 0 0 0 , 0 0 1 00 0 0 00 has a different row or column that does not sum to zero, and thus none of these matrices belong to S , and they are linearly independent from one another. Therefore, supplementing the basis for S in part (a) with the five matrices here extends the basis in (a) to a basis for M3 (R). 29. (a) An arbitrary matrix in S takes 1 = a 0 0 0 0 1 0 0 1 + b 0 0 1 1 0 0 Therefore, we have the following 0 100 0 0 1 , 0 010 1 a b c d e a+b+c−d−e the form b+c−d a+c−e d+e−c 00 1 00 0 0 0 0 0 1 + c 0 0 1 + d 1 0 −1 + e 0 1 −1 . 0 1 1 −1 −1 0 1 0 −1 1 basis for S : 10 0 0 1 , 0 00 1 0 1 0 0 1 , 1 1 −1 −1 0 0 0 0 0 0 −1 , 0 1 −1 . 0 1 0 −1 1 From this, we see that dim[S ] = 5. (b) We must include four additional matrices that are linearly independent and outside of S . The matrices E11 , E12 , E11 + E13 , and E22 will suffice in this case. 30. Let A ∈ Sym2 (R) so A = AT . A = a linear combination of 1 0 0 0 , 0 1 ab ∈ Sym2 (R); a, b, c ∈ R. This vector can be represented as bc 1 00 : , 0 01 266 10 01 00 10 01 00 +b +c . Since , , is a linearly independent 00 10 01 00 10 01 set that also spans Sym2 (R), it is a basis for Sym2 (R). Thus, dim[Sym2 (R)] = 3. Let B ∈ Skew2 (R) so B = 0b 01 01 −B T . Then B = =b , where b ∈ R. The set is linearly independent −b 0 −1 0 −1 0 01 and spans Skew2 (R) so that a basis for Skew2 (R) is . Consequently, dim[Skew2 (R)] = 1. Now, −1 0 dim[Sym2 (R)] = 4, and hence dim[Sym2 (R)] + dim[Skew2 (R)] = 3 + 1 = 4 = dim[M2 (R)]. A=a 31. We know that dim[Mn (R)] = n2 . Let S ∈ Symn (R) and let [Sij ] be the matrix with ones in the (i, j ) and (j, i) positions and zeroes elsewhere. Then the general n × n symmetric matrix can be expressed as: S = a11 S11 + a12 S12 +a13 S13 + · · · + a1n S1n + a22 S22 +a23 S23 + · · · + a2n S2n + ··· + an−1 n−1 Sn−1 n−1 + an−1 n Sn−1 n + ann Snn . We see that S has been resolved into a linear combination of n(n + 1) n + (n − 1) + (n − 2) + · · · + 1 = linearly independent matrices, which therefore form a basis for 2 n(n + 1) Symn (R); hence dim[Sym2 (R)] = . 2 Now let T ∈ Skewn (R) and let [Tij ] be the matrix with one in the (i, j )-position, negative one in the (j, i)position, and zeroes elsewhere, including the main diagonal. Then the general n × n skew-symmetric matrix can be expressed as: T = a12 T12 + a13 T13 +a14 T14 + · · · + a1n T1n + a23 T23 +a24 T24 + · · · + a2n T2n + ··· + an−1 n Tn−1 n (n − 1)n We see that T has been resolved into a linear combination of (n − 1)+(n − 2)+(n − 3)+ · · · +2+1 = 2 (n − 1)n linearly independent vectors, which therefore form a basis for Skewn (R); hence dim[Skewn (R)] = . 2 Consequently, using these results, we have n(n + 1) (n − 1)n dim[Symn (R)] + dim[Skew2 (R)] = + = n2 = dim[Mn (R)]. 2 2 32. (a) S is a two-dimensional subspace of R3 . Consequently, any two linearly independent vectors lying in this subspace determine a basis for S . By inspection we see, for example, that v1 = (4, −1, 0) and v2 = (3, 0, 1) both lie in the plane. Further, since they are not proportional, these vectors are linearly independent. Consequently, a basis for S is {(4, −1, 0), (3, 0, 1)}. (b) To extend the basis obtained in Part (a) to obtain a basis for R3 , we require one more vector that does not lie in S . For example, since v3 = (1, 0, 0) does not lie on the plane it is an appropriate vector. Consequently, a basis for R3 is {(4, −1, 0), (3, 0, 1), (1, 0, 0)}. 33. Each vector in S can be written as ab ba =a 1 0 0 1 +b 01 10 Consequently, a basis for S is given by the linearly independent set . 1 0 0 1 , 01 10 . To extend this 267 basis to M2 (R), we can choose, for example, the two vectors independent set 1 0 0 1 , 0 1 1 0 10 00 , , 0 0 1 0 1 0 0 0 and 0 0 1 0 . Then the linearly is a basis for M2 (R). 34. The vectors in S can be expressed as (2a1 + a2 )x2 + (a1 + a2 )x + (3a1 − a2 ) = a1 (2x2 + x + 3) + a2 (x2 + x − 1), and since {2x2 + x + 3, x2 + x − 1} is linearly independent (these polynomials are non-proportional) and certainly span S , they form a basis for S . To extend this basis to V = P2 , we must include one additional vector (since P2 is 3-dimensional). Any polynomial that is not in S will suffice. For example, x ∈ S , since x cannot be expressed in the form (2a1 + a2 )x2 + (a1 + a2 )x + (3a1 − a2 ), since the equations 2a1 + a2 = 0, a1 + a2 = 1, 3a1 − a2 = 0 are inconsistent. Thus, the extension we use as a basis for V here is {2x2 + x + 3, x2 + x − 1, x}. Many other correct answers can also be given here. 35. Since S is a basis for Pn−1 , S contains n vectors. Therefore, S ∪ {xn } is a set of n + 1 vectors, which is precisely the dimension of Pn . Moreover, xn does not lie in Pn−1 = span(S ), and therefore, S ∪ {xn } is linearly independent by Problem 47 in Section 4.5. By Corollary 4.6.13, we conclude that S ∪ {xn } is a basis for Pn . 36. Since S is a basis for Pn−1 , S contains n vectors. Therefore, S ∪ {p} is a set of n + 1 vectors, which is precisely the dimension of Pn . Moreover, p does not lie in Pn−1 = span(S ), and therefore, S ∪ {p} is linearly independent by Problem 47 in Section 4.5. By Corollary 4.6.13, we conclude that S ∪ {p} is a basis for Pn . 37. (a) Let ek denote the k th standard basis vector. Then a basis for Cn with scalars in R is given by {e1 , e2 , . . . , en , ie1 , ie2 , . . . , ien }, and the dimension is 2n. (b) Using the notation in part (a), a basis for Cn with scalars in C is given by {e1 , e2 , . . . , en }, and the dimension is n. Solutions to Section 4.7 True-False Review: 1. TRUE. This is the content of Theorem 4.7.1. The existence of such a linear combination comes from the fact that a basis for V must span V , and the uniqueness of such a linear combination follows from the linear independence of the vectors comprising a basis. 2. TRUE. This follows from the Equation [v]B = PB ←C [v]C , which is just Equation (4.7.6) with the roles of B and C reversed. 3. TRUE. The number of columns in the change-of-basis matrix PC ←B is the number of vectors in B , while the number of rows of PC ←B is the number of vectors in C . Since all bases for the vector space V contain the same number of vectors, this implies that PC ←B contains the same number of rows and columns. 268 4. TRUE. If V is an n-dimensional vector space, then PC ←B PB ←C = In = PB ←C PC ←B , which implies that PC ←B is invertible. 5. TRUE. This follows from the linearity properties: [v − w]B = [v + (−1)w]B = [v]B + [(−1)w]B = [v]B + (−1)[w]B = [v]B − [w]B . 6. FALSE. It depends on the order in which the vectors in the bases B and C are listed. For instance, if we consider the bases B = {(1, 0), (0, 1)} and C = {(0, 1), (1, 0)} for R2 , then although B and C contain the 1 0 same vectors, if we let v = (1, 0), then [v]B = while [v]C = . 0 1 7. FALSE. For instance, if we consider the bases B = {(1, 0), (0, 1)} and C = {(0, 1), (1, 0)} for R2 , and if 1 we let v = (1, 0) and w = (0, 1), then v = w, but [v]B = = [w]C . 0 8. TRUE. If B = {v1 , v2 , . . . , vn }, then the column vector [vi ]B is the ith standard basis vector (1 in the ith position and zeroes elsewhere). Thus, for each i, the ith column of PB ←B consists of a 1 in the ith position and zeroes elsewhere. This describes precisely the identity matrix. Problems: 1. Write (5, −10) = c1 (2, −2) + c2 (1, 4). Then 2c1 + c2 = 5 and − 2c1 + 4c2 = −10. Solving this system of equations gives c1 = 3 and c2 = −1. Thus, [v]B = 3 −1 . 2. Write (8, −2) = c1 (−1, 3) + c2 (3, 2). Then −c1 + 3c2 = 8 3c1 + 2c2 = −2. and Solving this system of equations gives c1 = −2 and c2 = 2. Thus, [v]B = −2 2 . 3. Write (−9, 1, −8) = c1 (1, 0, 1) + c2 (1, 1, −1) + c3 (2, 0, 1). Then c1 + c2 + 2c3 = −9 and c2 = 1 and c1 − c2 + c3 = −8. 4. Write (1, 7, 7) = c1 (1, −6, 3) + c2 (0, 5, −1) + c3 (3, −1, −1). 269 Then c1 + 3c3 = 1 − 6c1 + 5c2 − c3 = 7 and 3c1 − c2 − c3 = 7. and Using Gaussian elimination to solve this system of equations gives c1 = 4, c2 = 6, and c3 = −1. Thus, 4 [v]B = 6 . −1 5. Write (1, 7, 7) = c1 (3, −1, −1) + c2 (1, −6, 3) + c3 (0, 5, −1). Then 3c1 + c2 = 1 and − c1 − 6c2 + 5c3 = 7 and − c1 + 3c2 − c3 = 7. Using Gaussian elimination to solve this system of equations gives c1 = −1, c2 = 4, and c3 = 6. Thus, −1 [v]B = 4 . 6 6. Write (5, 5, 5) = c1 (−1, 0, 0) + c2 (0, 0, −3) + c3 (0, −2, 0). Then −c1 = 5 − 2c3 = 5 and and − 3c2 = 5. 5 Therefore, c1 = −5, c2 = − 3 , and c3 = − 5 . Thus, 2 −5 [v]B = −5/3 . −5/2 7. Write −4x2 + 2x + 6 = c1 (x2 + x) + c2 (2 + 2x) + c3 (1). Equating the powers of x on each side, we have c1 = −4 and c1 + 2c2 = 2 and 2c2 + c3 = 6. Solving this system of equations, we find that c1 = −4, c2 = 3, and c3 = 0. Hence, −4 [p(x)]B = 3 . 0 8. Write 15 − 18x − 30x2 = c1 (5 − 3x) + c2 (1) + c3 (1 + 2x2 ). Equating the powers of x on each side, we have 5c1 + c2 + c3 = 15 and − 3c1 = −18 and 2c3 = −30. 270 Solving this system of equations, we find that c1 = 6, c2 = 0, and c3 = −15. Hence, 6 0 . [p(x)]B = −15 9. Write 4 − x + x2 − 2x3 = c1 (1) + c2 (1 + x) + c3 (1 + x + x2 ) + c4 (1 + x + x2 + x3 ). Equating the powers of x on each side, we have c1 + c2 + c3 + c4 = 4 c2 + c3 + c4 = −1 and and c3 + c4 = 1 and c4 = −2. Solving this system of equations, we find that c1 = 5, c2 = −2, c3 = 3, and c4 = −2. Hence, 5 −2 [p(x)]B = 3 . −2 10. Write 8 + x + 6x2 + 9x3 = c1 (x3 + x2 ) + c2 (x3 − 1) + c3 (x3 + 1) + c4 (x3 + x). Equating the powers of x on each side, we have −c2 + c3 = 8 and c4 = 1 and c1 = 6 and c1 + c2 + c3 + c4 = 9. Solving this system of equations, we find that c1 = 6, c2 = −3, c3 = 5, and c4 = 1. Hence 6 −3 [p(x)]B = 5 . 1 11. Write −3 −2 −1 2 = c1 1 1 1 1 + c2 11 10 + c3 1 0 1 0 + c4 1 0 0 0 . Equating the individual entries of the matrices on each side of this equation (upper left, upper right, lower left, and lower right, respectively) gives c1 + c2 + c3 + c4 = −3 and c1 + c2 + c3 = −2 and c1 + c2 = −1 and c1 = 2. Solving this system of equations, we find that c1 = 2, c2 = −3, c3 = −1, and c4 = −1. Thus, 2 −3 [A]B = −1 . −1 12. Write −10 −15 16 −14 = c1 2 −1 3 5 + c2 0 −1 4 1 + c3 1 1 1 1 + c4 3 −1 2 5 . 271 Equating the individual entries of the matrices on each side of this equation (upper left, upper right, lower left, and lower right, respectively) gives 2c1 + c3 + 3c4 = −10, −c1 + 4c2 + c3 − c4 = 16, 3c1 − c2 + c3 + 2c4 = −15, 5c1 + c2 + c3 + 5c4 = −14. Solving this system of equations, we find that c1 = −2, c2 = 4, c3 = −3, and c4 = −1. Thus, −2 4 [A]B = −3 . −1 13. Write 5 7 6 8 = c1 −1 0 1 1 + c2 1 −1 3 0 + c3 1 1 0 2 + c4 0 −1 2 3 . Equating the individual entries of the matrices on each side of this equation (upper left, upper right, lower left, and lower right, respectively) gives −c1 + c2 + c3 = 5 and c1 + 3c2 − c4 = 6 and − c2 + c3 + 2c4 = 7 and c1 + 2c3 + 3c4 = 8. Solving this system of equations, we find that c1 = −34/3, c2 = 12, c3 = −55/3, and c4 = 56/3. Thus, −34/3 12 [A]B = −55/3 . 56/3 14. Write (x, y, z ) = c1 (0, 6, 3) + c2 (3, 0, 3) + c3 (6, −3, 0). Then 6c1 − 3c3 = y and 3c1 + 3c2 = z. 33 0z The augmented matrix for this linear system is 6 0 −3 y . We can reduce this to row-echelon form 03 6z 110 z/3 . Solving this system by back-subsitution gives x/3 as 0 1 2 0 0 9 y + 2x − 2z 3c2 + 6c3 = x c1 = 1 2 1 x+ y− z 9 9 9 and and 1 2 4 c2 = − x − y + z 9 9 9 and c3 = 2 1 2 x + y − z. 9 9 9 Hence, denote the ordered basis {v1 , v2 , v3 } by B , we have 1 2 1 9x + 9y − 9z [v]B = − 1 x − 2 y + 4 z . 9 9 9 1 2 2 9x + 9y − 9z 15. Write a0 + a1 x + a2 x2 = c1 (1 + x) + c2 x(x − 1) + c3 (1 + 2x2 ). 272 Equating powers of x on both sides of this equation, we have c1 + c3 = a0 c1 − c2 = a1 c2 + 2c3 = a2 . 1 0 1 a0 The augmented matrix corresponding to this system of equations is 1 −1 0 a1 . We can reduce this 0 1 2 a2 101 a0 . Thus, solving by back-substitution, we have a2 to row-echelon form as 0 1 2 0 0 1 −a0 + a1 + a2 c1 = 2a0 − a1 − a2 and and and c2 = 2a0 − 2a1 − a2 and c3 = −a0 + a1 + a2 . Hence, relative to the ordered basis B = {p1 , p2 , p3 }, we have 2a0 − a1 − a2 [p(x)]B = 2a0 − 2a1 − a2 . −a0 + a1 + a2 16. Let v1 = (9, 2) and v2 = (4, −3). Setting (9, 2) = c1 (2, 1) + c2 (−3, 1) and solving, we find c1 = 3 and c2 = −1. Thus, [v1 ]C = 3 −1 . Next, setting (4, −3) = c1 (2, 1) + c2 (−3, 1) and solving, we find c1 = −1 and c2 = −2. Thus, [v2 ]C = PC ←B = −1 −2 3 −1 −1 −2 . Therefore, . 17. Let v1 = (−5, −3) and v2 = (4, 28). Setting (−5, −3) = c1 (6, 2) + c2 (1, −1) and solving, we find c1 = −1 and c2 = 1. Thus, [v1 ]C = −1 1 . Next, setting (4, 28) = c1 (6, 2) + c2 (1, −1) and solving, we find c1 = 4 and c2 = −20. Thus, [v2 ]C = PC ←B = 4 −20 −1 4 1 −20 . Therefore, . 18. Let v1 = (2, −5, 0), v2 = (3, 0, 5), and v3 = (8, −2, −9). Setting (2, −5, 0) = c1 (1, −1, 1) + c2 (2, 0, 1) + c3 (0, 1, 3) 273 4 and solving, we find c1 = 4, c2 = −1, and c3 = −1. Thus, [v1 ]C = −1 . Next, setting −1 (3, 0, 5) = c1 (1, −1, 1) + c2 (2, 0, 1) + c3 (0, 1, 3) 1 and solving, we find c1 = 1, c2 = 1, and c3 = 1. Thus, [v2 ]C = 1 . Finally, setting 1 (8, −2, −9) = c1 (1, −1, 1) + c2 (2, 0, 1) + c3 (0, 1, 3) −2 and solving, we find c1 = −2, c2 = 5, and c3 = −4. Thus, [v3 ]C = 5 . Therefore, −4 4 1 −2 5 . PC ←B = −1 1 −1 1 −4 19. Let v1 = (−7, 4, 4), v2 = (4, 2, −1), and v3 = (−7, 5, 0). Setting (−7, 4, 4) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1) 0 and solving, we find c1 = 0, c2 = 5/3 and c3 = −7/3. Thus, [v1 ]C = 5/3 . Next, setting −7/3 (4, 2, −1) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1) and solving, we find c1 = 0, c2 = 19/3 and c3 = −7/3. Setting (4, 2, −1) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1) 3 and solving, we find c1 = 3, c2 = −2/3, and c3 = 1/3. Thus, [v2 ]C = −2/3 . Finally, setting 1/3 (−7, 5, 0) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1) 5 and solving, we find c1 = 5, c2 = −4, and c3 = −4. Hence, [v3 ]C = −4 . Therefore, −4 0 3 5 PC ←B = 5 − 2 −4 . 3 3 1 −7 −4 3 3 20. Let v1 = 7 − 4x and v2 = 5x. Setting 7 − 4x = c1 (1 − 2x) + c2 (2 + x) 274 3 2 and solving, we find c1 = 3 and c2 = 2. Thus, [v1 ]C = . Next, setting 5x = c1 (1 − 2x) + c2 (2 + x) and solving, we find c1 = −2 and c2 = 1. Hence, [v2 ]C = PC ←B = −2 1 3 −2 2 1 . Therefore, . 21. Let v1 = −4 + x − 6x2 , v2 = 6 + 2x2 , and v3 = −6 − 2x + 4x2 . Setting −4 + x − 6x2 = c1 (1 − x + 3x2 ) + c2 (2) + c3 (3 + x2 ) −1 and solving, we find c1 = −1, c2 = 3, and c3 = −3. Thus, [v1 ]C = 3 . Next, setting −3 6 + 2x2 = c1 (1 − x + 3x2 ) + c2 (2) + c3 (3 + x2 ) 0 and solving, we find c1 = 0, c2 = 0, and c3 = 2. Thus, [v2 ]C = 0 . Finally, setting 2 −6 − 2x + 4x2 = c1 (1 − x + 3x2 ) + c2 (2) + c3 (3 + x2 ) 2 and solving, we find c1 = 2, c2 = −1, and c3 = −2. Thus, [v3 ]C = −1 . Therefore, −2 −1 0 2 PC ←B = 3 0 −1 . −3 2 −2 22. Let v1 = −2 + 3x + 4x2 − x3 , v2 = 3x + 5x2 + 2x3 , v3 = −5x2 − 5x3 , and v4 = 4 + 4x + 4x2 . Setting −2 + 3x + 4x2 − x3 = c1 (1 − x3 ) + c2 (1 + x) + c3 (x + x2 ) + c4 (x2 + x3 ) 0 −2 and solving, we find c1 = 0, c2 = −2, c3 = 5, and c4 = −1. Thus, [v1 ]C = 5 . Next, setting −1 3x + 5x2 + 2x3 = c1 (1 − x3 ) + c2 (1 + x) + c3 (x + x2 ) + c4 (x2 + x3 ) 0 0 and solving, we find c1 = 0, c2 = 0, c3 = 3, and c4 = 2. Thus, [v2 ]C = . Next, solving 3 2 −5x2 − 5x3 = c1 (1 − x3 ) + c2 (1 + x) + c3 (x + x2 ) + c4 (x2 + x3 ) 275 0 0 and solving, we find c1 = 0, c2 = 0, c3 = 0, and c4 = −5. Thus, [v3 ]C = 0 . Finally, setting −5 4 + 4x + 4x2 = c1 (1 − x3 ) + c2 (1 + x) + c3 (x + x2 ) + c4 (x2 + x3 ) 2 2 and solving, we find c1 = 2, c2 = 2, c3 = 2, and c4 = 2. Thus, [v4 ]C = . Therefore, 2 2 00 02 −2 0 0 2 . PC ←B = 53 0 2 −1 2 −5 2 23. Let v1 = 2 + x2 , v2 = −1 − 6x + 8x2 , and v3 = −7 − 3x − 9x2 . Setting 2 + x2 = c1 (1 + x) + c2 (−x + x2 ) + c3 (1 + 2x2 ) 3 and solving, we find c1 = 3, c2 = 3, and c3 = −1. Thus, [v1 ]C = 3 . Next, solving −1 −1 − 6x + 8x2 = c1 (1 + x) + c2 (−x + x2 ) + c3 (1 + 2x2 ) −4 and solving, we find c1 = −4, c2 = 2, and c3 = 3. Thus, [v2 ]C = 2 . Finally, solving 3 −7 − 3x − 9x2 = c1 (1 + x) + c2 (−x + x2 ) + c3 (1 + 2x2 ) −2 and solving, we find c1 = −2, c2 = 1, and c3 = −5. Thus, [v3 ]C = 1 . Therefore, −5 3 −4 −2 2 1 . PC ←B = 3 −1 3 −5 24. Let v1 = 1 0 −1 −2 , v2 = 1 0 −1 −2 0 −1 3 0 , v3 = 1 1 + c2 = c1 1 1 35 00 , and v4 = 1 1 + c3 1 0 1 0 1 0 −2 −4 0 0 . Setting 10 00 + c4 −2 1 and solving, we find c1 = −2, c2 = c3 = c4 = 1. Thus, [v1 ]C = 1 . Next, setting 1 0 −1 3 0 c1 1 1 1 1 + c2 1 1 1 0 + c3 1 0 1 0 + c4 1 0 0 0 276 0 3 and solving, we find c1 = 0, c2 = 3, c3 = −4, and c4 = 1. Thus, [v2 ]C = −4 . Next, setting 1 3 0 5 0 c1 1 1 1 1 1 1 + c2 1 0 1 0 + c3 1 0 + c4 1 0 0 0 0 0 and solving, we find c1 = 0, c2 = 0, c3 = 5, and c4 = −2. Thus, [v3 ]C = 5 . Finally, setting −2 −2 −4 0 0 c1 1 1 1 1 1 1 + c2 1 0 + c3 1 0 1 0 + c4 1 0 0 0 0 0 and solving, we find c1 = 0, c2 = 0, c3 = −4, and c4 = 2. Thus, [v4 ]C = −4 . Therefore, we have 2 −2 0 0 0 1 3 0 0 . = 1 −4 5 −4 1 1 −2 2 PC ←B 25. Let v1 = E12 , v2 = E22 , v3 = E21 , and v4 0 0 [v1 ]C = , [v2 ]C = 0 1 = E11 . We see by 1 0 , [v3 ]C = 0 0 inspection that 0 0 , [v4 ]C = 1 0 0 1 . 0 0 Therefore, PC ←B 0 0 = 0 1 1 0 0 0 0 0 1 0 0 1 . 0 0 26. We could simply compute the inverse of the matrix obtained in Problem 16. For instructive purposes, however, we proceed directly. Let w1 = (2, 1) and w2 = (−3, 1). Setting (2, 1) = c1 (9, 2) + c2 (4, −3) and solving, we obtain c1 = 2/7 and c2 = −1/7. Thus, [w1 ]B = 2/7 −1/7 (−3, 1) = c1 (9, 2) + c2 (4, −3) . Next, setting 277 −1/7 −3/7 and solving, we obtain c1 = −1/7 and c2 = −3/7. Thus, [w2 ]B = PB ←C = 2 7 −1 7 −1 7 −3 7 . Therefore, . 27. We could simply compute the inverse of the matrix obtained in Problem 16. For instructive purposes, however, we proceed directly. Let w1 = (6, 2) and w2 = (1, −1). Setting (6, 2) = c1 (−5, −3) + c2 (4, 28) and solving, we obtain c1 = −5/4 and c2 = −1/16. Thus, [w1 ]B = −5/4 −1/16 . Next, setting (1, −1) = c1 (−5, −3) + c2 (4, 28) 25/4 −1/16 and solving, we obtain c1 = 25/4 and c2 = −1/16. Thus, [w2 ]B = PB ←C = 5 −4 25 4 1 − 16 1 − 16 . Therefore, . 28. We could simply compute the inverse of the matrix obtained in Problem 16. For instructive purposes, however, we proceed directly. Let w1 = (1, −1, 1), w2 = (2, 0, 1), and w3 = (0, 1, 3). Setting (1, −1, 1) = c1 (2, −5, 0) + c2 (3, 0, 5) + c3 (8, −2, −9) 1/5 and solving, we find c1 = 1/5, c2 = 1/5, and c3 = 0. Thus, [w1 ]B = 1/5 . Next, setting 0 (2, 0, 1) = c1 (2, −5, 0) + c2 (3, 0, 5) + c3 (8, −2, −9) −2/45 and solving, we find c1 = −2/45, c2 = 2/5, and c3 = 1/9. Thus, [w2 ]B = 2/5 . Finally, setting 1/9 (0, 1, 3) = c1 (2, −5, 0) + c2 (3, 0, 5) + c3 (8, −2, −9) −7/45 and solving, we find c1 = −7/45, c2 = 2/5, and c3 = −1/9. Thus, [w3 ]B = 2/5 . Therefore, −1/9 PB ←C 1 5 2 − 45 7 − 45 = 1 5 2 5 2 5 0 1 9 −1 9 . 278 29. We could simply compute the inverse of the matrix obtained in Problem 16. For instructive purposes, however, we proceed directly. Let w1 = 1 − 2x and w2 = 2 + x. Setting 1 − 2x = c1 (7 − 4x) + c2 (5x) and solving, we find c1 = 1/7 and c2 = −2/7. Thus, [w1 ]B = 1/7 −2/7 . Setting 2 + x = c1 (7 − 4x) + c2 (5x) 2/7 3/7 and solving, we find c1 = 2/7 and c2 = 3/7. Thus, [w2 ]B = PB ←C = 1 7 2 7 −2 7 3 7 . Therefore, . 30. We could simply compute the inverse of the matrix obtained in Problem 16. For instructive purposes, however, we proceed directly. Let w1 = 1 − x3 , w2 = 1 + x, w3 = x + x2 , and w4 = x2 + x3 . Setting a0 + a1 x + a2 x2 + a3 x3 = c1 (−2 + 3x + 4x2 − x3 ) + c2 (3x + 5x2 + 2x3 ) + c3 (−5x2 − 5x3 ) + c4 (4 + 4x + 4x2 ), the corresponding augmented matrix for the resulting equations that arise by equating the various powers of x is −2 0 0 4 a0 33 04 0 4 5 −5 4 0 , −1 2 −5 0 −1 which can be formulated as the linear system Ax = b, where −2 0 0 33 0 A= 4 5 −5 −1 2 −5 4 4 . 4 0 ??????? NEED CALCULATOR ?????? 31. We could simply compute the inverse of the matrix obtained in Problem 25. For instructive purposes, however, we proceed directly. Let w1 = E22 , w2 = E11 , w3 = E21 , and w4 = E12 . We see by inspection that 0 0 0 1 1 0 0 0 [w1 ]B = , [w2 ]B = , [w3 ]B = , [w4 ]B = . 0 0 1 0 0 1 0 0 Therefore PB ←C 0 1 = 0 0 0 0 0 1 0 0 1 0 1 0 . 0 0 279 32. We first compute [v]B and [v]C directly. Setting (−5, 3) = c1 (9, 2) + c2 (4, −3) 3 − 35 and solving, we obtain c1 = −3/35 and c2 = −37/35. Thus, [v]B = (−5, 3) = c1 (2, 1) + c2 (−3, 1) and solving, we obtain c1 = 4/5 and c2 = 11/5. Thus, [v]C = PC ←B = 3 −1 −1 −2 − 37 35 4 5 11 5 . Setting . Now, according to Problem 16, , so PC ←B [v]B = 3 −1 −1 −2 3 − 35 − 37 35 = 4 5 11 5 = [v]C , which confirms Equation (4.7.6). 33. We first compute [v]B and [v]C directly. Setting (−1, 2, 0) = c1 (−7, 4, 4) + c2 (4, 2, −1) + c3 (−7, 5, 0) and solving, we obtain c1 = 3/43, c2 = 12/43, and c3 = 10/43. Thus, [v]B = 3 43 12 43 . Setting 10 43 (−1, 2, 0) = c1 (1, 1, 0) + c2 (0, 1, 1) + c3 (3, −1, −1) 2 and solving, we obtain c1 = 2, c2 = −1, and c3 = −1. Thus, [v]C = −1 . Now, according to Problem −1 3 0 3 5 0 3 5 2 43 5 2 , so PC ←B [v]B = 5 − 2 −4 12 = −1 = [v]C , which 19, PC ←B = 3 − 3 −4 3 3 43 7 1 7 1 10 −1 −3 −4 −3 −4 3 3 43 confirms Equation (4.7.6). 34. We first compute [v]B and [v]C directly. Setting 6 − 4x = c1 (7 − 4x) + c2 (5x) and solving, we obtain c1 = 6/7 and c2 = −4/35. Thus, [v]B = 6 7 4 − 35 6 − 4x = c1 (1 − 2x) + c2 (2 + x) . Next, setting 280 and solving, we obtain c1 = 14/5 and c2 = 8/5. Thus, [v]C = 3 −2 2 1 PC ←B = 14 5 8 5 . Now, according to Problem 20, , so 3 −2 PC ←B [v]B = 2 1 6 7 4 − 35 = 14 5 8 5 = [v]C , which confirms Equation (4.7.6). 35. We first compute [v]B and [v]C directly. Setting 5 − x + 3x2 = c1 (−4 + x − 6x2 ) + c2 (6 + 2x2 ) + c3 (−6 − 2x + 4x2 ) 1 5 and solving, we obtain c1 = 1, c2 = 5/2, and c3 = 1. Thus, [v]B = 2 . Next, setting 1 and solving, we PC ←B = −1 3 −3 5 − x + 3x2 = c1 (1 − x + 3x2 ) + c2 (2) + c3 (3 + x2 ) 1 obtain c1 = 1, c2 = 2, and c3 = 0. Thus, [v]C = 2 . Now, according to Problem 21, 0 0 2 0 −1 , so 2 −2 1 1 −1 0 2 PC ←B [v]B = 3 0 −1 5 = 2 = [v]C , 2 0 −3 2 −2 1 which confirms Equation (4.7.6). 36. We first compute [v]B and [v]C directly. Setting −1 −1 −4 5 = c1 1 0 −1 −2 + c2 0 −1 3 0 + c3 3 0 5 0 + c4 −2 −4 0 0 −5/2 −13/6 and solving, we obtain c1 = −5/2, c2 = −13/6, c3 = 37/6, and c4 = 17/2. Thus, [v]B = 37/6 . Next, 17/2 setting −1 −1 11 11 11 10 = c1 + c2 + c3 + c4 −4 5 11 10 00 00 5 −9 and solving, we find c1 = 5, c2 = −9, c3 = 3, and c4 = 0. Thus, [v]C = 3 . Now, according to Problem 0 281 24, PC ←B −2 0 0 0 1 3 0 0 , and so = 1 −4 5 −4 1 1 −2 2 −2 0 0 0 1 3 0 0 PC ←B [v]B = 1 −4 5 −4 1 1 −2 2 −5/2 5 −13/6 −9 37/6 = 3 = [v]C , 17/2 0 which confirms Equation (4.7.6). 37. Write x = a1 v1 + a2 v2 + · · · + an vn . We have cx = c(a1 v1 + a2 v2 + · · · + an vn ) = (ca1 )v1 + (ca2 )v2 + · · · + (can )vn . Hence, [cx]B = ca1 ca2 . . . can = c a1 a2 . . . = c[x]B . an 38. We must show that {v1 , v2 , . . . , vn } is linearly independent and spans V . Check linear independence: Assume that c1 v1 + c2 v2 + · · · + cn vn = 0. We wish to show that c1 = c2 = · · · = cn = 0. Now by assumption, the zero vector 0 can be uniquely written as a linear combination of the vectors in {v1 , v2 , . . . , vn }. Since 0 = 0 · v1 + 0 · v2 + · · · + 0 · vn , we therefore conclude that c1 = c2 = · · · = cn = 0, as needed. Check spanning property: Let v be an arbitrary vector in V . By assumption, it is possible to express v (uniquely) as a linear combination of the vectors in {v1 , v2 , . . . , vn }; say v = a1 v1 + a2 v2 + · · · + an vn . Therefore, v lies in span{v1 , v2 , . . . , vn }. Since v is an arbitrary member of V , we conclude that {v1 , v2 , . . . , vn } spans V . 39. Let B = {v1 , v2 , . . . , vn } and let C = {vσ(1) , vσ(2) , . . . , vσ(n) }, where σ is a permutation of the set {1, 2, . . . , n}. We will show that PC ←B contains exactly one 1 in each row and column, and zeroes elsewhere (the argument for PB ←C is essentially identical, or can be deduced from the fact that PB ←C is the inverse of PC ←B ). Let i be in {1, 2, . . . , n}. The ith column of PC ←B is [vi ]C . Suppose that ki ∈ {1, 2, . . . , n} is such that σ (ki ) = i. Then [vi ]C is a column vector with a 1 in the ki th position and zeroes elsewhere. Since the values k1 , k2 , . . . , kn are distinct, we see that each column of PC ←B contains a single 1 (with zeroes elsewhere) in a different position from any other column. Hence, when we consider all n columns as a whole, each position in the column vector must have a 1 occurring exactly once in one of the columns. Therefore, PC ←B contains exactly one 1 in each row and column, and zeroes elsewhere. 282 Solutions to Section 4.8 True-False Review: 1. TRUE. Note that rowspace(A) is a subspace of Rn and colspace(A) is a subspace of Rm , so certainly if rowspace(A) = colspace(A), then Rn and Rm must be the same. That is, m = n. 2. FALSE. A basis for the row space of A consists of the nonzero row vectors of any row-echelon form of A. 3. FALSE. The nonzero column vectors of the original matrix A that correspond to the nonzero column vectors of a row-echelon form of A form a basis for colspace(A). 4. TRUE. Both rowspace(A) and colspace(A) have dimension equal to rank(A), the number of nonzero rows in a row-echelon form of A. Equivalently, their dimensions are both equal to the number of pivots occurring in a row-echelon form of A. 5. TRUE. For an invertible n × n matrix, rank(A) = n. That means there are n nonzero rows in a rowechelon form of A, and so rowspace(A) is n-dimensional. Therefore, we conclude that rowspace(A) = Rn . 6. TRUE. This follows immediately from the (true) statements in True-False Review Questions 4-5 above. Problems: 1. A row-echelon form of A is 1 −2 0 0 . Consequently, a basis for rowspace(A) is {(1, −2)}, whereas a basis for colspace(A) is {(1, −3)}. 1 1 −3 2 . Consequently, a basis for rowspace(A) is 0 1 −2 1 {(1, 1, −3, 2), (0, 1, −2, 1)}, whereas a basis for colspace(A) is {(1, 3), (1, 4)}. 123 3. A row-echelon form of A is 0 1 2 . Consequently, a basis for rowspace(A) is {(1, 2, 3), (0, 1, 2)}, 000 whereas a basis for colspace(A) is {(1, 5, 9), (2, 6, 10)}. 1 013 4. A row-echelon form of A is 0 0 0 . Consequently, a basis for rowspace(A) is {(0, 1, 1 )}, whereas 3 000 a basis for colspace(A) is {(3, −6, 12)}. 1 2 −1 3 0 0 0 1 . Consequently, a basis for rowspace(A) is 5. A row-echelon form of A is 0 0 0 0 00 00 {(1, 2, −1, 3), (0, 0, 0, 1)}, whereas a basis for colspace(A) is {(1, 3, 1, 5), (3, 5, −1, 7)}. 1 −1 2 3 2 −4 3 (Note: This is not row-echelon form, but it is not nec6. We can reduce A to 0 0 0 6 −13 essary to bring the leading nonzero element in each row to 1.). Consequently, a basis for rowspace(A) is {(1, −1, 2, 3), (0, 2, −4, 3), (0, 0, 6, −13)}, whereas a basis for colspace(A) is {(1, 1, 3), (−1, 1, 1), (2, −2, 4)}. 1 −1 2 1 . A row-echelon form of this matrix is 7. We determine a basis for the rowspace of the matrix 5 −4 7 −5 −4 2. A row-echelon form of A is 283 1 −1 2 0 1 −9 . Consequently, a basis for the subspace spanned by the given vectors is {(1, −1, 2), (0, 1, −9)}. 0 0 0 13 3 1 5 −1 . A row-echelon form of this matrix is 8. We determine a basis for the rowspace of the matrix 2 7 4 14 1 13 3 0 1 −2 . Consequently, a basis for the subspace spanned by the given vectors is {(1, 3, 3), (0, 1, −2)}. 0 0 0 00 0 1 1 −1 2 3 −4 . A row-echelon form of 9. We determine a basis for the rowspace of the matrix 2 1 1 2 −6 10 1 1 −1 2 this matrix is 0 1 −5 8 . Consequently, a basis for the subspace spanned by the given vectors is 00 00 {(1, 1, −1, 2), (0, 1, −5, 8)}. 1413 2 8 3 5 10. We determine a basis for the rowspace of the matrix 1 4 0 4 . A row-echelon form of this 2826 141 3 0 0 1 −1 . Consequently, a basis for the subspace spanned by the given vectors is matrix is 000 0 000 0 {(1, 4, 1, 3), (0, 0, 1, −1)}. 1 −3 . Consequently, a basis for rowspace(A) is {(1, −3)}, whereas a 0 0 basis for colspace(A) is {(−3, 1)}. Both of these subspaces are lines in the xy -plane. 124 12. (a) A row-echelon form of A is 0 1 1 . Consequently, a basis for rowspace(A) is 000 {(1, 2, 4), (0, 1, 1)}, whereas a basis for colspace(A) is {(1, 5, 3), (2, 11, 7)}. 11. A row-echelon form of A is (b) We see that both of these subspaces are 2-dimensional, and therefore each corresponds geometrically to a plane. By inspection, we see that the two basis vectors for rowspace(A) satisfy the equations 2x + y − z = 0, and therefore rowspace(A) corresponds to the plane with this equation. Similarly, we see that the two basis vectors for colspace(A) satisfy the equations 2x − y + z = 0, and therefore colspace(A) corresponds to the plane with this equations. 11 , colspace(A) is spanned by (1, 2), but if we permute the two rows of A, we obtain a 22 new matrix whose column space is spanned by (2, 1). On the other hand, if we multiply the first row by 2, then we obtain a new matrix whose column space is spanned by (2, 2). And if we add −2 times the first row to the second row, we obtain a new matrix whose column space is spanned by (1, 0). Therefore, in all cases, colspace(A) is altered by the row operations performed. 13. If A = 284 1 1 . We have −1 −1 rowspace(A) = {(r, r) : r ∈ R}, while colspace(A) = {(r, −r) : r ∈ R}. Thus, rowspace(A) and colspace(A) have no nonzero vectors in common. 14. Many examples are possible here, but an easy 2 × 2 example is the matrix A = Solutions to Section 4.9 True-False Review: 1. FALSE. For example, consider the 7 × 3 zero matrix, 07×3 . We have rank(07×3 ) = 0, and therefore by the Rank-Nullity Theorem, nullity(07×3 ) = 3. But |m − n| = |7 − 3| = 4. Many other examples can be given. In particular, provided that m > 2n, the m × n zero matrix will show the falsity of the statement. 2. FALSE. In this case, rowspace(A) is a subspace of R9 , hence it cannot possibly be equal to R7 . The correct conclusion here should have been that rowspace(A) is a 7-dimensional subspace of R9 . 3. TRUE. By the Rank-Nullity Theorem, rank(A) = 7, and therefore, rowspace(A) is a 7-dimensional subspace of R7 . Hence, rowspace(A) = R7 . 01 is an upper triangular matrix with two zeros appearing 00 on the main diagonal. However, since rank(A) = 1, we also have nullity(A) = 1. 4. FALSE. For instance, the matrix A = 5. TRUE. An invertible matrix A must have nullspace(A) = {0}, but if colspace(A) is also {0}, then A would be the zero matrix, which is certainly not invertible. 6. FALSE. For instance, if we take A = 1 + 1 = 2, but A + B = 1 0 1 0 10 00 0 0 and B = 1 0 , then nullity(A)+ nullity(B ) = , and nullity(A + B ) = 1. 7. FALSE. For instance, if we take A = 1 0 0 0 and B = 0 0 0 1 , then nullity(A)·nullity(B ) = 1 · 1 = 1, but AB = 02 , and nullity(AB ) = 2. 8. TRUE. If x belongs to the nullspace of B , then B x = 0. Therefore, (AB )x = A(B x) = A0 = 0, so that x also belongs to the nullspace of AB . Thus, nullspace(B ) is a subspace of nullspace(AB ). Hence, nullity(B ) ≤ nullity(AB ), as claimed. 9. TRUE. If y belongs to nullspace(A), then Ay = 0. Hence, if Axp = b, then A(y + xp ) = Ay + Axp = 0 + b = b, which demonstrates that y + xp is also a solution to the linear system Ax = b. Problems: 1. The matrix is already in row-echelon form. A vector (x, y, z, w) in nullspace(A) must satisfy x − 6z − w = 0. We see that y ,z , and w are free variables, and 1 6 0 6z + w 100 y : y, z, w ∈ R = span , , . nullspace(A) = 0 1 z 0 1 0 0 w 285 Therefore, nullity(A) = 3. Moreover, since this row-echelon form contains one nonzero row, rank(A) = 1. Since the number of columns of A is 4 = 1 + 3, the Rank-Nullity Theorem is verified. 2. We bring A to row-echelon form: 1 −1 2 0 0 REF(A) = . 1 A vector (x, y ) in nullspace(A) must satisfy x − 2 y = 0. Setting y = 2t, we get x = t. Therefore, nullspace(A) = t 2t :t∈R = span 1 2 . Therefore, nullity(A) = 1. Since REF(A) contains one nonzero row, rank(A) = 1. Since the number of columns of A is 2 = 1 + 1, the Rank-Nullity Theorem is verified. 3. We bring A to row-echelon form: 1 REF(A) = 0 0 1 −1 1 7 . 0 1 Since there are no unpivoted columns, there are no free variables in the associated homogeneous linear system, and so nullspace(A) = {0}. Therefore, nullity(A) = 0. Since REF(A) contains three nonzero rows, rank(A) = 3. Since the number of columns of A is 3 = 3 + 0, the Rank-Nullity Theorem is verified. 4. We bring A to row-echelon form: 1 4 −1 3 1 1 . REF(A) = 0 1 00 00 A vector (x, y, z, w) in nullspace(A) must satisfy x +4y − z +3w = 0 and y + z + w = 0. We see that z and w are free variables. Set z = t and w = s. Then y = −z − w = −t − s and x = z − 4y − 3w = t − 4(−t − s) − 3s = 5t + s. Therefore, 1 5 5t + s −t − s : s, t ∈ R = span −1 , −1 . nullspace(A) = t 1 0 1 0 s Therefore, nullity(A) = 2. Moreover, since REF(A) contains two nonzero rows, rank(A) = 2. Since the number of columns of A is 4 = 2 + 2, the Rank-Nullity Theorem is verified. 5. Since all rows (or columns) of this matrix are proportional to the first one, rank(A) = 1. Since A has two columns, we conclude from the Rank-Nullity Theorem that nullity(A) = 2 − rank(A) = 2 − 1 = 1. 6. The first and last rows of A are not proportional, but the middle rows are proportional to the first row. Therefore, rank(A) = 2. Since A has five columns, we conclude from the Rank-Nullity Theorem that nullity(A) = 5 − rank(A) = 5 − 2 = 3. 286 7. Since the second and third columns are not proportional and the first column is all zeros, we have rank(A) = 2. Since A has three columns, we conclude from the Rank-Nullity Theorem that nullity(A) = 3 − rank(A) = 3 − 2 = 1. 8. This matrix (already in row-echelon form) has one nonzero row, so rank(A) = 1. Since it has four columns, we conclude from the Rank-Nullity Theorem that nullity(A) = 4 − rank(A) = 4 − 1 = 3. 1 3 −1 4 9 11 . We quickly reduce this augmented 9. The augmented matrix for this linear system is 2 7 1 5 21 10 matrix to row-echelon form: 1 3 −1 4 0 1 11 3 . 00 00 A solution (x, y, z ) to the system will have a free variable corresponding to the third column: z = t. Then y + 11t = 3, so y = 3 − 11t. Finally, x + 3y − z = 4, so x = 4 + t − 3(3 − 11t) = −5 + 34t. Thus, the solution set is 34 −5 −5 + 34t 3 − 11t : t ∈ R = t −11 + 3 : t ∈ R . t 1 0 −5 34 Observe that xp = 3 is a particular solution to Ax = b, and that −11 forms a basis for 0 1 nullspace(A). Therefore, the set of solution vectors obtained does indeed take the form (4.9.3). 1 −1 2 3 6 10. The augmented matrix for this linear system is 1 −2 5 5 13 . We quickly reduce this aug2 −1 1 4 5 mented matrix to row-echelon form: 1 −1 2 3 6 0 1 −3 −2 −7 . 0 0 0 0 0 A solution (x, y, z, w) to the system will have free variables corresponding to the third and fourth columns: z = t and w = s. Then y − 3z − 2w = −7 requires that y = −7 + 3t + 2s, and x − y + 2z + 3w = 6 requires that x = 6 + (−7 + 3t + 2s) − 2t − 3s = −1 + t − s. Thus, the solution set is −1 + t − s −7 + 3t + 2s t s 1 −1 −1 2 −7 3 : s, t ∈ R = t + : s, t ∈ R . + s 0 0 1 0 1 0 287 −1 −7 Observe that xp = 0 is a particular solution to Ax = b, and that 0 −1 1 3 , 2 1 0 0 1 is a basis for nullspace(A). Therefore, the set of solution vectors obtained does indeed take the form (4.9.3). 1 1 −2 −3 3 −1 −7 2 . We quickly reduce this augmented 11. The augmented matrix for this linear system is 1 0 1 1 2 2 −4 −6 matrix to row-echelon form: −3 1 1 −2 1 0 1 − 11 . 4 4 00 1 1 There are no free variables in the solution set (since none of the first three columns is unpivoted), and we find 2 the solution set by back-substitution: −3 . It is easy to see that this is indeed a particular solution: 1 2 xp = −3 . Since the row-echelon form of A has three nonzero rows, rank(A) = 3. Thus, nullity(A) = 0. 1 Hence, nullspace(A) = {0}. Thus, the only term in the expression (4.9.3) that appears in the solution is xp , and this is precisely the unique solution we obtained in the calculations above. 12. By inspection, we see that a particular solution to this (homogeneous) linear system is xp = 0. We quickly reduce this augmented matrix to row-echelon form: 1 1 −1 5 0 0 1 −1 7 0 . 2 2 00 000 A solution (x, y, z, w) to the system will have free variables corresponding to the third and fourth columns: z = t and w = s. Then y − 1 z + 7 w = 0 requires that y = 1 t − 7 s, and x + y − z + 5w = 0 requires that 2 2 2 2 1 x = t − ( 2 t − 7 s) − 5s = 1 t − 3 s. Thus, the solution set is 2 2 2 1 1 3 −2 2t − 3s 2 2 1 1 −7 t − 7s 2 2 : s, t ∈ R = t 2 + s 2 : s, t ∈ R . 0 t 1 s 0 1 3 1 −2 2 −7 1 Since the vectors 2 and 2 form a basis for nullspace(A), our solutions to take the proper form 1 0 1 0 given in (4.9.3). 288 13. By the Rank-Nullity Theorem, rank(A) = 7 − nullity(A) = 7 − 4 = 3, and hence, colspace(A) is 3-dimensional. But since A has three rows, colspace(A) is a subspace of R3 . Therefore, since the only 3-dimensional subspace of R3 is R3 itself, we conclude that colspace(A) = R3 . Now rowspace(A) is also 3-dimensional, but it is a subspace of R5 . Therefore, it is not accurate to say that rowspace(A) = R3 . 14. By the Rank-Nullity Theorem, rank(A) = 4 − nullity(A) = 4 − 0 = 4, so we conclude that rowspace(A) is 4-dimensional. Since rowspace(A) is a subspace of R4 (since A contains four columns), and it is 4-dimensional, we conclude that rowspace(A) = R4 . Although colspace(A) is 4-dimensional, colspace(A) is a subspace of R6 , and therefore it is not accurate to say that colspace(A) = R4 . 15. If rowspace(A) = nullspace(A), then we know that rank(A) = nullity(A). Therefore, rank(A)+ nullity(A) must be even. But rank(A)+ nullity(A) is the number of columns of A. Therefore, A contains an even number of columns. 16. We know that rank(A) + nullity(A) = 7. But since A only has five rows, rank(A) ≤ 5. Therefore, nullity(A) ≥ 2. However, since nullspace(A) is a subspace of R7 , nullity(A) ≤ 7. Therefore, 2 ≤ nullity(A) ≤ 7. There are many examples of a 5 × 7 matrix A with nullity(A) = 2; one example is 1000000 0 1 0 0 0 0 0 0 0 1 0 0 0 0 . The only 5 × 7 matrix with nullity(A) = 7 is 05×7 , the 5 × 7 zero matrix. 0 0 0 1 0 0 0 0000100 17. We know that rank(A) + nullity(A) = 8. But since A only has three rows, rank(A) ≤ 3. Therefore, nullity(A) ≥ 5. However, since nullspace(A) is a subspace of R8 , nullity(A) ≤ 8. Therefore, 5 ≤ nullspace(A) ≤ 8. There are many examples of a 3 × 8 matrix A with nullity(A) = 5; one example is 10000000 0 1 0 0 0 0 0 0 . The only 3 × 8 matrix with nullity(A) = 8 is 03×8 , the 3 × 8 zero matrix. 00100000 18. If B x = 0, then AB x = A0 = 0. This observation shows that nullspace(B ) is a subspace of nullspace(AB ). On the other hand, if AB x = 0, then B x = (A−1 A)B x = A−1 (AB )x = A−1 0 = 0, so B x = 0. Therefore, nullspace(AB ) is a subspace of nullspace(B ). As a result, since nullspace(B ) and nullspace(AB ) are subspaces of each other, they must be equal: nullspace(AB ) = nullspace(B ). Therefore, nullity(AB ) = nullity(B ). Solutions to Section 4.10 True-False Review: 1. TRUE. This follows from the equivalence of (a) and (m) in the Invertible Matrix Theorem. 2. FALSE. If the matrix has n linearly independent rows, then by the equivalence of (a) and (m) in the Invertible Matrix Theorem, such a matrix would be invertible. But if that were so, then by part (j) of the Invertible Matrix Theorem, such a matrix would have to have n linearly independent columns. 3. FALSE. If the matrix has n linearly independent columns, then by the equivalence of (a) and (j) in the Invertible Matrix Theorem, such a matrix would be invertible. But if that were so, then by part (m) of the Invertible Matrix Theorem, such a matrix would have to have n linearly independent rows. 4. FALSE. An n × n matrix A with det(A) = 0 is not invertible by part (g) of the Invertible Matrix Theorem. Therefore, by the equivalence of (a) and (l) in the Invertible Matrix Theorem, the columns of A do not form a basis for Rn . 289 5. TRUE. If rowspace(A) = Rn , then by the equivalence of (a) and (n) in the Invertible Matrix Theorem, A is not invertible. Therefore, A is not row-equivalent to the identity matrix. Since B is row-equivalent to A, then B is not row-equivalent to the identity matrix, and therefore, B is not invertible. Hence, by part (k) of the Invertible Matrix Theorem, we conclude that colspace(B ) = Rn . 6. FALSE. If nullspace(A) = {0}, then A is invertible by the equivalence of (a) and (c) in the Invertible Matrix Theorem. Since E is an elementary matrix, it is also invertible. Therefore, EA is invertible. By part (g) of the Invertible Matrix Theorem, det(EA) = 0, contrary to the statement given. 7. FALSE. The matrix [A|B ] has 2n columns, but only n rows, and therefore rank([A|B ]) ≤ n. Hence, by the Rank-Nullity Theorem, nullity([A|B ]) ≥ n > 0. 8. TRUE. The first and third rows are proportional, and hence statement (m) in the Invertible Matrix Theorem is false. Therefore, by the equivalence with (a), the matrix is not invertible. 0100 1 0 0 0 9. FALSE. For instance, the matrix 0 0 0 1 is of the form given, but satisfies any (and all) of the 0010 statements of the Invertible Matrix Theorem. 121 10. FALSE. For instance, the matrix 3 6 2 is of the form given, but has a nonzero determinant, 100 and so by part (g) of the Invertible Matrix Theorem, it is invertible. Solutions to Section 4.11 1. FALSE. The converse of this statement is true, but for the given statement, many counterexamples exist. For instance, the vectors v = (1, 1) and w = (1, 0) in R2 are linearly independent, but they are not orthogonal. 2. FALSE. We have k v, k w = k v, k w = k k w, v = k 2 w, v = k 2 v, w , where we have used the axioms of an inner product to carry out these steps. Therefore, the result conflicts with the given statement, which must therefore be a false statement. 3. TRUE. We have c1 v1 + c2 v2 , w = c1 v1 , w + c2 v2 , w = c1 v1 , w + c2 v2 , w = c1 · 0 + c2 · 0 = 0. 4. TRUE. We have x + y, x − y = x, x − x, y + y, x − y, y = x, x − y, y = | x| 2 − | y| 2 . This will be negative if and only if | x| 2 < | y| 2 , and since | x| and | y| are nonnegative real numbers, | x| 2 < | y| 2 if and only if | x| < | y| . 5. FALSE. For example, if V is the inner product space of integrable functions on (−∞, ∞), then the formula b f, g = f (t)g (t)dt a 290 is a valid any product for any choice of real numbers a < b. See also Problem 9 in this section, in which a “non-standard” inner product on R2 is given. 6. TRUE. The angle between the vectors −2v and −2w is cos θ = (−2v) · (−2w) (−2)2 (v · w) v·w = = , | − 2v| | − 2w| (−2)2 | v| | w| | v| | w| and this is the angle between the vectors v and w. 7. FALSE. This definition of p, q will not satisfy the requirements of an inner product. For instance, if we take p = x, then p, p = 0, but p = 0. Problems: √ √ 1. v, w = 8, ||v|| = 3 3, ||w|| = 7. Hence, cos θ = v, w 8 = √ =⇒ θ ≈ 0.95 radians. ||v||||w|| 3 21 π π sin2 xdx x sin xdx = π, ||f || = 2. f , g = 0 π , ||g || = 2 = 0 cos θ = x2 dx = 0 π3 . Hence, 3 √ π π 1 /2 2 π π3 1/2 = 6 ≈ 0.68 radians. π 3 3. v, w = (2 + i)(−1 − i) + (3 − 2i)(1 + 3i) + (4 + i)(3 + i) = 19 + 11i. ||v|| = √ w, w = 22. ||w|| = v, v = √ 35. 4. Let A, B, C ∈ M2 (R). (1): A, A = a2 + a2 + a2 + a2 ≥ 0, and A, A = 0 ⇐⇒ a11 = a12 = a21 = a22 = 0 ⇐⇒ A = 0. 11 12 21 22 (2): A, B = a11 b11 + a12 b12 + a21 b21 + a22 b22 = b11 a11 + b12 a12 + b21 a21 + b22 a22 = B , A . (3): Let k ∈ R. k A, B = ka11 b11 + ka12 b12 + ka21 b21 + ka22 b22 = k (a11 b11 + a12 b12 + a21 b21 + a22 b22 ) = k A, B . (4): (A + B ), C = (a11 + b11 )c11 + (a12 + b12 )c12 + (a21 + b21 )c21 + (a22 + b22 )c22 = (a11 c11 + b11 c11 ) + (a12 c12 + b12 c12 ) + (a21 c21 + b21 c21 ) + (a22 c22 + b22 c22 ) = (a11 c11 + a12 c12 + a21 c21 + a22 c22 ) + (b11 c11 + b12 c12 + b21 c21 + b22 c22 = A, C + B , C . 5. We need only demonstrate one example showing that some property of an inner product is violated by the 1 0 given formula. Set A = . Then according to the given formula, we have A, A = −2, violating 0 −1 the requirement that u, u ≥ 0 for all vectors u. 6. A, B = 2 · 3 + (−1)1 + 3(−1) + 5 · 2 = 12. √ ||A|| = A, A = 2 · 2 + (−1)(−1) + 3 · 3 + 5 · 5 = 39. √ ||B || = B , B = 3 · 3 + 1 · 1 + (−1)(−1) + 2 · 2 = 15. √ √ 7. A, B = 13, ||A|| = 33, ||B || = 7. 291 8. Let p1 , p2 , p3 ∈ P1 where p1 (x) = a + bx, p2 (x) = c + dx, and p3 (x) = e + f x. Define p1 , p2 = ac + bd. (8.1) The properties 1 through 4 of Definition 4.11.3 must be verified. (1): p1 , p1 = a2 + b2 ≥ 0 and p1 , p1 = 0 ⇐⇒ a = b = 0 ⇐⇒ p1 (x) = 0. (2): p1 , p2 = ac + bd = ca + db = p2 , p1 . (3): Let k ∈ R. k p1 , p2 = kac + kbd = k (ac + bd) = k p1 , p2 . (4): p1 + p2 , p3 = (a + c)e + (b + d)f = ae + ce + bf + df = (ae + bf ) + (ce + df ) = p1 , p 3 + p 2 , p 3 . Hence, the mapping defined by (8.1) is an inner product in P1 . 9. Property 1: u, u = 2u1 u1 + u1 u2 + u2 u1 + 2u2 u2 = 2u2 + 2u1 u2 + 2u2 = (u1 + u2 )2 + u2 + u2 ≥ 0. 2 1 1 2 u, u = 0 ⇐⇒ (u1 + u2 )2 + u2 + u2 = 0 ⇐⇒ u1 = 0 = u2 ⇐⇒ u = 0. 2 1 Property 2: u, v = 2u1 v1 + u1 v2 + u2 v1 + 2u2 v2 = 2u2 v2 + u2 v1 + u1 v2 + 2u1 v1 = v, u . Property 3: k u, v = k (2u1 v1 + u1 v2 + u2 v1 + 2u2 v2 ) = 2ku1 v1 + ku1 v2 + ku2 v1 + 2ku2 v2 = 2(ku1 )v1 + (ku1 )v2 + (ku2 )v1 + 2(ku2 )v2 = k u, v = 2ku1 v1 + ku1 v2 + ku2 v1 + 2ku2 v2 = 2u1 (kv1 ) + u1 (kv2 ) + u2 (kv1 ) + 2u2 (kv2 ) = u, k v . Property 4: (u + v), w = (u1 + v1 , u2 + v2 ), (w1 , w2 ) = 2(u1 + v1 )w1 + (u1 + v1 )w2 + (u2 + v2 )w1 + 2(u2 + v2 )w2 = 2u1 w1 + 2v1 w1 + u1 w2 + v1 w2 + u2 w1 + v2 w1 + 2u2 w2 + 2v2 w2 = 2u1 w1 + u1 w2 + u2 w1 + 2u2 w2 + 2v1 w1 + v1 w2 + v2 w1 + 2v2 w2 = u, w + v, w . Therefore u, v = 2u1 v1 + u1 v2 + u2 v1 + 2u2 v2 defines an inner product on R2 . 10. (a) Using the defined inner product: v, w = (1, 0), (−1, 2) = 2 · 1(−1) + 1 · 2 + 0(−1) + 2 · 0 · 2 = 0. (b) Using the standard inner product: v, w = (1, 0), (−1, 2) = 1(−1) + 0 · 2 = −1 = 0. 11. (a) Using the defined inner product: v, w = 2 · 2 · 3 + 2 · 6 + (−1)3 + 2(−1)6 = 9 = 0. (b) Using the standard inner product: v, w = 2 · 3 + (−1)6 = 0. 12. (a) Using the defined inner product: v, w = 2 · 1 · 2 + 1 · 1 + (−2) · 2 + 2 · (−2) · 1 = −3 = 0. (b) Using the standard inner product: v, w = 1 · 2 + (−2)(1) = 0. 292 13. (a) Show symmetry: v, w = w, v . v, w = (v1 , v2 ), (w1 , w2 ) = v1 w1 − v2 w2 = w1 v1 − w2 v2 = (w1 , w2 ), (v1 , v2 ) = w, v . (b) Show k v, w = k v, w = v, k w . Note that k v = k (v1 , v2 ) = (kv1 , kv2 ) and k w = k (w1 , w2 ) = (kw1 , kw2 ). k v, w = (kv1 , kv2 ), (w1 , w2 ) = (kv1 )w1 − (kv2 )w2 = k (v1 w1 − v2 w2 ) = k (v1 , v2 ), (w1 , w2 ) = k v, w . Also, v, k w = (v1 , v2 ), (kw1 , kw2 ) = v1 (kw1 ) − v2 (kw2 ) = k (v1 w1 − v2 w2 ) = k (v1 , v2 ), (w1 , w2 ) = k v, w . (c) Show (u + v), w = u, w + v, w . Let w = (w1 , w2 ) and note that u + v = (u1 + v1 , u2 + v2 ). (u + v), w = (u1 + v1 , u2 + v2 ), (w1 , w2 ) = (u1 + v1 )w1 − (u2 + v2 )w2 = u1 w1 + v1 w1 − u2 w2 − v2 w2 = u1 w1 − u2 w2 + v1 w1 − v2 w2 = (u1 , v2 ), (w1 , w2 ) + (v1 , v2 ), (w1 , w2 ) = u, w + v, w . Property 1 fails since, for example, u, u < 0 whenever |u2 | > |u1 |. 2 2 2 2 14. v, v = 0 =⇒ (v1 , v2 ), (v1 , v2 ) = 0 =⇒ v1 − v2 = 0 =⇒ v1 = v2 =⇒ |v1 | = |v2 |. Thus, in this space, 2 null vectors are given by {(v1 , v2 ) ∈ R : |v1 | = |v2 |} or equivalently, v = r(1, 1) or v = s(1, −1) where r, s ∈ R. 2 2 2 2 15. v, v < 0 =⇒ (v1 , v2 ), (v1 , v2 ) < 0 =⇒ v1 − v2 < 0 =⇒ v1 < v2 . In this space, timelike vectors are 2 2 2 given by {(v1 , v2 ) ∈ R : v1 < v2 }. 2 2 2 2 16. v, v > 0 =⇒ (v1 , v2 ), (v1 , v2 ) > 0 =⇒ v1 − v2 > 0 =⇒ v1 > v2 . In this space, spacelike vectors are 2 2 given by {(v1 , v2 ) ∈ R2 : v1 > v2 }. 17. y Null Vectors Null Vectors Timelike Vectors Spacelike Vectors Spacelike Vectors x Spacelike Vectors Spacelike Vectors Timelike Vectors Null Vectors Null Vectors Figure 0.0.65: Figure for Exercise 17 18. Suppose that some ki ≤ 0. If ei denotes the standard basis vector in Rn with 1 in the ith position and zeros elsewhere, then the given formula yields ei , ei = ki ≤ 0, violating the first axiom of an inner product. Therefore, if the given formula defines a valid inner product, then we must have that ki > 0 for all i. Now we prove the converse. Suppose that each ki > 0. We verify the axioms of an inner product for the given proposed inner product. We have 2 2 2 v, v = k1 v1 + k2 v2 + · · · + kn vn ≥ 0, 293 and we have equality if and only if v1 = v2 = · · · = vn = 0. For the second axiom, we have w, v = k1 w1 v1 + k2 w2 v2 + · · · + kn wn vn = k1 v1 w1 + k2 v2 w2 + · · · + kn vn wn = v, w . For the third axiom, we have k v, w = k1 (kv1 )w1 + k2 (kv2 )w2 + · · · + kn (kvn )wn = k [k1 v1 w1 + k2 v2 w2 + · · · + kn vn wn ] = k v, w . Finally, for the fourth axiom, we have u + v, w = k1 (u1 + v1 )w1 + k2 (u2 + v2 )w2 + · · · + kn (un + vn )wn = [k1 u1 w1 + k2 u2 w2 + · · · + kn un wn ] + [k1 v1 w1 + k2 v2 w2 + · · · + kn vn wn ] = u, w + v, w . 19. We have v, 0 = v, 0 + 0 = 0 + 0, v = 0, v + 0, v = v, 0 + v, 0 = 2 v, 0 , which implies that v, 0 = 0. 20. (a) For all v, w ∈ V . ||v + w||2 = (v + w), (v + w) = v, v + w + w, v + w = v + w, v + v + w, w = v, v + w, v + v, w = v, v + v, w + w, v by Property 4 by Property 2 + w, w by Property 4 + w, w by Property 2 = ||v||2 + 2 v, w + ||w||2 . (b) This follows immediately by substituting v, w = 0 in the formula given in part (a). (c) (i) From part (a), it follows that ||v + w||2 = ||v||2 + 2 v, w + ||w||2 and ||v − w||2 = ||v + (−w)||2 = ||v||2 + 2 v, −w + || − w||2 = ||v||2 − 2 v, w + ||w||2 . Thus, ||v + w||2 − ||v − w||2 = ||v||2 + 2 v, w + ||w||2 − ||v||2 − 2 v, w + ||w||2 = 4 v, w . (ii) ||v + w||2 + ||v − w||2 = ||v||2 + 2 v, w + ||w||2 + ||v||2 − 2 v, w + ||w||2 = 2||v||2 + 2||w||2 = 2 ||v||2 + ||w||2 . 21. For all v, w ∈ V and vi , wi ∈ C. ||v + w||2 = v + w, v + w = v, v + w + w, v + w by Property 4 = v + w, v + v + w, w by Property 2 = v, v + w, v + v, w + w, w by Property 4 = v, v + w, v + v, w + w, w = v, v + w, w + v, w + v, w = ||v||2 + ||w||2 + 2Re{ v, w } = ||v||2 + 2Re{ v, w } + ||v||2 . Solutions to Section 4.12 294 1. TRUE. An orthonormal basis is simply an orthogonal basis consisting of unit vectors. 2. FALSE. The converse of this statement is true, but for the given statement, many counterexamples exist. For instance, the set of vectors {(1, 1), (1, 0)} in R2 is linearly independent, but does not form an orthogonal set. 3. TRUE. We can verify easily that π cos t sin tdt = 0 sin2 t π | = 0, 20 which means that {cos x, sin x} is an orthogonal set. Moreover, since they are non-proportional functions, they are linearly independent. Therefore, they comprise an orthogonal basis for the 2-dimensional inner product space span{cos x, sin x}. 4. FALSE. For instance, in R3 we can take the vectors x1 = (1, 0, 0), x2 = (1, 1, 0), and x3 = (0, 0, 1). Applying the Gram-Schmidt process to the ordered set {x1 , x2 , x3 } yields the standard basis {e1 , e2 , e3 }. How1 ever, applying the Gram-Schmidt process to the ordered set {x3 , x2 , x1 } yields the basis {x3 , x2 , ( 2 , − 1 , 0)} 2 instead. 5. TRUE. This is the content of Theorem 4.12.7. 6. TRUE. The vector P(w, v) is a scalar multiple of the vector v, and since v is orthogonal to u, P(w, v) is also orthogonal to u, so that its projection onto u must be 0. 7. TRUE. We have P(w1 + w2 , v) = w1 , v + w2 , v w1 , v w2 , v w1 + w2 , v v= v= v+ v = P(w1 , v) + P(w2 , v). 2 2 2 v v v v2 Problems: 1. (2, −1, 1), (1, 1, −1) = 2 + (−1) + (−1) = 0; (2, −1, 1), (0, 1, 1) = 0 + (−1) + 1 = 0; (1, 1, −1), (0, 1, 1) = 0 + 1 + (−1) = 0. Since each vector in the set is orthogonal to every other vector in the set, the vectors form an orthogonal set. To generate an orthonormal set, we divide each vector by its norm: √ √ ||(2, −1, 1)|| = √4 + 1 + 1 = √6, ||(1, 1, −1)|| = 1 + 1 + 1 = 3, and √ √ ||(0, 1, 1)|| = 0 + 1 + 1 = 2. Thus, the corresponding orthonormal set is: √ √ √ 2 3 6 (0, 1, 1) . (1, 1, −1), (2, −1, 1), 2 3 6 2. (1, 3, −1, 1), (−1, 1, 1, −1) = −1 + 3 + (−1) + (−1) = 0; (1, 3, −1, 1), (1, 0, 2, 1) = 1 + 0 + (−2) + 1 = 0; (−1, 1, 1, −1), (1, 0, 2, 1) = −1 + 0 + 2 + (−1) = 0. Since all vectors in the set are orthogonal to each other, they form an orthogonal set. To generate an orthonormal set, √ divide each vector by its norm: we √ √ ||(1, 3, −1, 1)|| = √+ 9 + 1 + 1 = √ = 2 3, 1 12 ||(−1, 1, 1, −1)|| = 1 + 1 + 1 + 1 = 4 = 2, and √ √ ||(1, 0, 2, 1)|| = 1 + 0 + 4 + 1 = 6. Thus, an orthonormal set is: √ √ 6 3 1 (1, 0, 2, 1) . (1, 3, −1, 1), (−1, 1, 1, −1), 6 6 2 295 3. (1, 2, −1, 0), (1, 0, 1, 2) = 1 + 0 + (−1) + 0 = 0; (1, 2, −1, 0), (−1, 1, 1, 0) = −1 + 2 + (−1) + 0 = 0; (1, 2, −1, 0), (1, −1, −1, 0) = 1 + (−2) + 1 + 0 = 0; (1, 0, 1, 2), (−1, 1, 1, 0) = −1 + 0 + 1 + 0 = 0; (1, 0, 1, 2), (1, −1, −1, 0) = 1 + 0 + (−1) + 0 = 0; (−1, 1, 1, 0), (1, −1, −1, 0) = −1 + (−1) + (−1) = −3. Hence, this is not an orthogonal set of vectors. 4. (1, 2, −1, 0, 3), (1, 1, 0, 2, −1) = 1 + 2 + 0 + 0 + (−3) = 0; (1, 2, −1, 0, 3), (4, 2, −4, −5, −4) = 4 + 4 + 4 + 0 + (−12) = 0; (1, 1, 0, 2, −1), (4, 2, −4, −5, −4) = 4 + 2 + 0 + (−10) + 4 = 0. Since all vectors in the set are orthogonal to each other, they form an orthogonal set. To generate an orthonormal set, we divide each vector by its norm: √ √ ||(1, 2, −1, 0, 3)|| = √1 + 4 + 1 + 0 + 9 = √15, ||(1, 1, 0, 2, −1)|| = 1 + 1 + 0 + 4 + 1 = 7, and√ √ ||(4, 2, −4, −5, −4)|| = 16 + 4 + 16 + 25 + 16 = 77. Thus, an orthonormal set is: √ √ √ 15 7 77 (1, 2, −1, 0, 3), (1, 1, 0, 2, −1), (4, 2, −4, −5, −4) . 15 7 77 5. We require that v1 , v2 = v1 , w = v2 , w = 0. Let w = (a, b, c) where a, b, c ∈ R. v1 , v2 = (1, 2, 3), (1, 1, −1) = 0. v1 , w = (1, 2, 3), (a, b, c) =⇒ a + 2b + 3c = 0. v2 , w = (1, 1, −1), (a, b, c) =⇒ a + b − c = 0. Letting the free variable c = t ∈ R, the system has the solution a = 5t, b = −4t, and c = t. Consequently, {(1, 2, 3), (1, 1, −1), (5t, −4t, t)} will form an orthogonal set whenever t = 0. To determine the corresponding orthonormal set, we must divide each vector by its√ norm: √ √ √ √ √ √ √ ||v1 || = 1 + 4 + 9 = 14, ||v2 || = 1 + 1 + 1 = 3, ||w|| = 25t2 + 16t2 + t2 = 42t2 = |t| 42 = t 42 if t ≥ 0. Setting t = 1, an orthonormal set is: √ √ √ 14 3 42 (1, 2, 3), (1, 1, −1), (5, −4, 1) . 14 3 42 6. (1 − i, 3 + 2i), (2 + 3i, 1 − i) = (1 − i)(2 + 3i) + (3 + 2i)(1 − i) = (1 − i)(2 − 3i) + (3 + 2i)(1 + i) = (2 − 3i − 2i − 3) + (3 + 3i + 2i − 2) = (−1 − 5i) + (1 + √ ) = 0. The vectors are orthogonal. 5i √ ||(1 − i, 3 + 2i)|| = (1 − i)(1 + i) + (3 + 2i)(3 − 2i) = 1 + 1 + 9 + 4 = 15. √ √ ||(2 + 3i, 1 − i)|| = (2 + 3i)(2 − 3i) + (1 − i)(1 + i) = 4 + 9 + 1 + 1 = 15. Thus, the corresponding orthonormal set is: √ √ 15 15 (2 + 3i, 1 − i) . (1 − i, 3 + 2i), 15 15 7. (1 − i, 1 + i, i), (0, i, 1 − i) = (1 − i) · 0 + (1 + i)(−i) + i(1 + i) = 0. (1 − i, 1 + i, i), (−3 + 3i, 2 + 2i, 2i) = (1 − i)(−3 − 3i) + (1 + i)(2 − 2i) + i(−2i) = 0. (0, i, 1 − i), (−3 + 3i, 2 + 2i, 2i) = 0 + i(2 − 2i) + (1 − i)(−2i) = (2i + 2) + (−2i − 2) = 0. Hence, the vectors are orthogonal. To obtain a corresponding orthonormal set, we divide each vector by its norm. √ √ ||(1 − i, 1 + i, i)|| = (1 − i)(1 + i) + (1 + i)(1 − i) + i(−i) = 1 + 1 + 1 + 1 + 1 = 5. 296 √ √ ||(0, i, 1 − i)|| = 0 + i(−i) + (1 − i)(1 + i) = 1 + 1 + 1 = 3. √ √ ||(−3 + 3i, 2 + 2i, 2i)|| = (−3 + 3i)(−3 − 3i) + (2 + 2i)(2 − 2i) + 2i(−2i) = 9 + 9 + 4 + 4 + 4 = 30. Consequently, an orthonormal set is: √ √ √ 5 3 30 (1 − i, 1 + i, i), (0, i, 1 − i), (−3 + 3i, 2 + 2i, 2i) . 5 3 30 8. Let z = a + bi where a, b ∈ R. We require that v, w = 0. (1 − i, 1 + 2i), (2 + i, a + bi) = 0 =⇒ (1 − i)(2 − i) + (1 + 2i)(a − bi) = 0 =⇒ 1 − 3i + a + 2b + (2a − b)i = 0. a + 2b = −1 Equating real parts and imaginary parts from the last equality results in the system: 2a − b = 3. This system has the solution a = 1 and b = −1; hence z = 1 − i. Our desired orthogonal set is given by {(1 − i, 1 + 2i), (2 + i, 1 − i)}. ||(1 − i, 1 + 2i)|| = ||(2 + i, 1 − i)|| = (1 − i)(1 + i) + (1 + 2i)(1 − 2i) = (2 + i)(2 − i) + (1 − i)(1 + i) = √ √ 1+1+1+4= 4+1+1+1= √ √ 7. 7. The corresponding orthonormal set is given by: √ 7 7 (2 + i, 1 − i) . (1 − i, 1 + 2i), 7 7 √ 1 1 − cos πx = 0. π −1 −1 1 1 sin πx f1 , f3 = 1, cos πx = cos πxdx = = 0. π −1 −1 1 11 f2 , f3 = sin πx, cos πx = sin πx cos πxdx = sin 2πxdx = 2 −1 −1 vectors are orthogonal. 1 √ ||f1 || = 1dx = [x]1 1 = 2. − 9. f1 , f2 = 1, sin πx = sin πxdx = −1 cos 2πx 4π −1 1 1 sin2 πxdx = ||f2 || = −1 −1 1 1 cos2 πxdx = ||f3 || = −1 Consequently, −1 √ 1 − cos 2πx dx = 2 x 2 1 1 + cos 2πx dx = 2 x 2 1 2 , sin πx, cos πx 2 1 1 · xdx = 10. f1 , f2 = 1, x = −1 f1 , f3 = 1, 3x2 − 1 = 2 1 −1 − −1 + −1 1 sin 2πx 4π 1 sin 2πx 4π 1 = 1. −1 1 = 1. −1 is an orthonormal set of functions on [−1, 1]. x2 2 1 = 0. −1 3x2 − 1 x3 − x dx = 2 2 1 = 0. −1 1 = 0. Thus, the −1 297 f2 , f3 = x, 1 3x2 − 1 = 2 −1 1 ||f1 || = dx = −1 1 2 ||f2 || = x dx = −1 √ [x]1 1 = − x3 3 1 1 3x4 x2 3x2 − 1 dx = − 2 24 2 x· = 0. Thus, the vectors are orthogonal. −1 2. √ 1 6 . 3 = −1 2 1 3x2 − 1 11 1 9x5 ||f3 || = dx = (9x4 − 6x2 + 1)dx = − 2x3 + x 2 4 −1 45 −1 To obtain a set of orthonormal vectors, we divide each vector by its norm: √ 1 = −1 √ √ √ f3 10 f2 6 2 f1 f3 . f2 , = f1 , = = 2 2 ||f3 || 2 ||f2 || ||f1 || √ √ 10 6 2 (3x2 − 1) x, , 4 2 2 √ Thus, is an orthonormal set of vectors. 1 11. f1 , f2 = sin πx sin 2πxdx = −1 1 2 1 (cos 3πx − cos πx)dx = 0. −1 1 11 (cos 4πx − cos 2πx)dx = 0. 2 −1 −1 1 1 1 f2 , f3 = sin 2πx sin 3πxdx = (cos 5πx − cos πx)dx = 0. 2 −1 −1 Therefore, {f1 , f2 , f3 } is an orthogonal set. f1 , f3 = sin πx sin 3πxdx = 1 1 sin2 πxdx = ||f1 || = −1 −1 1 1 sin2 2πxdx = ||f2 || = −1 −1 1 sin 2πx 1 x− 2 2π 1 (1 − cos 2πx)dx = 2 sin 4πx 1 x− 2 4π 1 (1 − cos 4πx)dx = 2 1 1 = 1. −1 1 = 1. −1 1 1 sin 6πx 1 x− = 1. (1 − cos 6πx)dx = 2 6π −1 −1 2 −1 Thus, it follows that {f1 , f2 , f3 } is an orthonormal set of vectors on [−1, 1]. sin2 3πxdx = ||f3 || = 1 12. f1 , f2 = cos πx, cos 2πx = cos πx cos 2πxdx = −1 1 f1 , f3 = cos πx, cos 3πx = cos πx cos 3πxdx = −1 1 f2 , f3 = cos 2πx, cos 3πx = 1 2 cos 2πx cos 3πxdx = −1 1 1 2 (cos 3πx + cos πx)dx = 0. −1 1 (cos 4πx + cos 2πx)dx = 0. −1 1 2 1 (cos 5πx + cos πx)dx = 0. −1 Therefore, {f1 , f2 , f3 } is an orthogonal set. 1 cos2 πxdx = ||f1 || = −1 1 cos2 2πxdx = ||f2 || = −1 1 2 1 2 1 (1 + cos 2πx)dx = −1 1 (1 + cos 4πx)dx = −1 1 sin 2πx x+ 2 2π 1 sin 4πx x+ 2 4π 1 = 1. −1 1 = 1. −1 10 . 5 298 1 1 11 1 sin 6πx (1 + cos 6πx)dx = x+ = 1. 2 −1 2 6π −1 −1 Thus, it follows that{f1 , f2 , f3 } is an orthonormal set of vectors on [−1, 1]. ||f3 || = cos2 3πxdx = 13. It is easily verified that A1 , A2 = 0, A1 , A3 = 0, A2 , A3 = 0. Thus we require a, b, c, d such that A1 , A4 = 0 =⇒ a + b − c + 2d = 0, A2 , A4 = 0 =⇒ −a + b + 2c + d = 0, A3 , A4 = 0 =⇒ a − 3b + 2d = 0. 1 3 Solving this system for a, b, c, and d, we obtain: a = c, b = − c, d = 0. Thus, 2 2 A4 = −1c 2 c 0 3 2c = 2c 3 −1 2 0 =k 3 −1 2 0 where k is any nonzero real number. 14. Let v1 = (1, −1, −1) and v√= (2, 1, −1). √ 2 u1 = v1 = (1, −1, −1), ||u1 || = 1 + 1 + 1 = 3. v2 , u1 = (2, 1, −1), (1, −1, −1) = 2 − 1 + 1 = 2. v2 , u1 2 1 u2 = v2 − u1 = (2, 1, −1) − (1, −1, −1) = (4, 5, −1). ||u1 ||2 3 3 1√ ||u2 || = (16 + 25 + 1)/9 = 42. Hence, an orthonormal basis is: 3 √ √ 42 3 (4, 5, −1) . (1, −1, −1), 42 3 15. Let v1 = (2, 1, −2) and v√= (1, 3, −1). √ 2 u1 = v1 = (2, 1, −2), ||u1 || = 4 + 1 + 4 = 9 = 3. v2 , u1 = (1, 3, −1), (2, 1, −2) = 2 · 1 + 1 · 3 + (−2)(−1) = 7. v2 , u1 7 5 u2 = v2 − u1 = (1, 3, −1) − (2, 1, −2) = (−1, 4, 1). 2 ||u1 || 9 9 √ 5√ 52 ||u2 || = 1 + 16 + 1 = . Hence, an orthonormal basis is: 9 3 √ 2 1 (−1, 4, 1) . (2, 1, −2), 6 3 16. Let v1 = (−1, 1, 1, 1) and v√= (1, 2, 1, 2). 2 u1 = v1 = (−1, 1, 1, 1), ||u1 || = 1 + 1 + 1 + 1 = 2. v2 , u1 = (1, 2, 1, 2), (−1, 1, 1, 1) = 1(−1) + 2 · 1 + 1 · 1 + 2 · 1 = 4. v2 , u1 u2 = v2 − u1 = (1, 2, 1, 2) − (−1, 1, 1, 1) = (2, 1, 0, 1). ||u1 ||2 √ √ ||u2 || = 4 + 1 + 0 + 1 = 6. Hence, an orthonormal basis is: √ 6 1 (2, 1, 0, 1) . (−1, 1, 1, 1), 6 2 17. Let v1 = (1, 0, −1, 0), v2 = √ , 1, −1, 0) and v√= (−1, 1, 0, 1). (1 3 u1 = v1 = (1, 0, −1, 0), ||u1 || = 1 + 0 + 1 + 0 = 2. , 299 v2 , u1 = (1, 1, −1, 0), (1, 0, −1, 0) = 1 · 1 + 1 · 0 + (−1)(−1) + 0 · 0 = 2. v2 , u1 u2 = v2 − u1 = (1, 1, −1, 0) − (1, 0, −1, 0) = (0, 1, 0, 0). ||u1 ||2 √ ||u2 || = 0 + 1 + 0 + 0 = 1. v3 , u1 = (−1, 1, 0, 1), (1, 0, −1, 0) = (−1)1 + 1 · 0 + 0(−1) + 1 · 0 = −1. v3 , u2 = (−1, 1, 0, 1), (0, 1, 0, 0) = (−1)0 + 1 · 1 + 0 · 0 + 1 · 0 = 1. v3 , u2 1 v3 , u1 1 u3 = v3 − u1 − u2 = (−1, 1, 0, 1) + (1, 0, −1, 0) − (0, 1, 0, 0) = (−1, 0, −1, 2); ||u1 ||2 ||u√ 2 || 2 2 2 1√ 6 ||u3 || = 1+0+1+4= . Hence, an orthonormal basis is: 2 2 √ √ 2 6 (1, 0, −1, 0), (0, 1, 0, 0), (−1, 0, −1, 2) . 2 6 18. Let v1 = (1, 2, 0, 1), v2 = √ , 1, 1, 0) and v3 √ (1, 0, 2, 1). (2 = u1 = v1 = (1, 2, 0, 1), ||u1 || = 1 + 4 + 0 + 1 = 6. v2 , u1 = (2, 1, 1, 0), (1, 2, 0, 1) = 2 · 1 + 1 · 2 + 1 · 0 + 0 · 1 = 4. v2 , u1 2 1 u2 = v2 − u1 = (2, 1, 1, 0) − (1, 2, 0, 1) = (4, −1, 3, −2). ||u1 ||2 3 3 1√ 1√ ||u2 || = 16 + 1 + 9 + 4 = 30. 3 3 v3 , u1 = (1, 0, 2, 1), (1, 2, 0, 1) = 1 · 1 + 0 · 2 + 2 · 0 + 1 · 1 = 2. 8 . 3 v3 , u1 v3 , u2 1 4 2 u1 − u2 = (1, 0, 2, 1) − (1, 2, 0, 1) − (4, −1, 3, −2) = (−1, −1, 3, 3); u3 = v3 − 2 2 ||u1 || ||u2 √ || 3 15 5 2√ 45 ||u3 || = 1+1+9+9= . Hence, an orthonormal basis is: 5 5 √ √ √ 6 30 5 (1, 2, 0, 1), (4, −1, 3, −2), (−1, −1, 3, 3) . 6 30 10 v3 , u2 = (1, 0, 2, 1), (4/3, −1/3, 1, −2/3) = 1(4/3) + 0(−1/3) + 2 · 1 + 1(−2/3) = 19. Let v1 = (1, 1, −1, 0), v2 = √ 1, 0, 1, 1) and v√= (2, −1, 2, 1). (− 3 u1 = v1 = (1, 1, −1, 0), ||u1 || = 1 + 1 + 1 + 0 = 3. v2 , u1 = (−1, 0, 1, 1), (1, 1, −1, 0) = −1 · 1 + 0 · 1 + 1(−1) + 1 · 0 = −2. v2 , u1 −2 1 u2 = v2 − u1 = (−1, 0, 1, 1) − (1, 1, −1, 0) = (−1, 2, 1, 3). 2 ||u1 || 3 3 √ 1√ 15 ||u2 || = 1+4+1+9= . 3 3 v3 , u1 = (2, −1, 2, 1), (1, 1, −1, 0) = 2 · 1 + −1 · 1 + 2(−1) + 1 · 0 = −1. 1 . 3 v3 , u1 v3 , u2 1 1 4 u3 = v3 − u1 − u2 = (2, −1, 2, 1) + (1, 1, −1, 0) − (−1, 2, 1, 3) = (3, −1, 2, 1); 2 ||u2 ||2 3 15 5 √ ||u1 || 4 15 ||u3 || = . Hence, an orthonormal basis is: 5 √ √ √ 15 15 3 (3, −1, 2, 1) . (−1, 2, 1, 3), (1, 1, −1, 0), 15 15 3 v3 , u2 = (2, −1, 2, 1), (−1/3, 2/3, 1/3, 1) = 2(−1/3) + (−1)(2/3) + 2(1/3) + 1 · 1 = 300 1 −2 1 1 1 . Hence, a basis for rowspace(A) is {(1, −2, 1), (0, 7, 1))}. 20. A row-echelon form of A is 0 7 0 00 The vectors in this basis are not orthogonal. We therefore use the Gram-Schmidt process to determine an orthogonal basis. Let v1 = (1, −2, 1), v2 = (0, 7, 1). Then an orthogonal basis for rowspace(A) is {u1 , u2 }, where 1 13 u1 = (1, −2, 1), u2 = (0, 7, 1) + (1, −2, 1) = (13, 16, 19). 6 6 21. Let v1 = (1 − i, 0, i) and v2 = (1, 1 + i, 0). √ u1 = v1 = (1 − i, 0, i), ||u1 || = (1 − i)(1 + i) + 0 + i(−i) = 3. v2 , u1 = (1, 1 + i, 0), (1 − i, 0, i) = 1(1 + i) + (1 + i)0 + 0(−i) = 1 + i. v2 , u1 1+i 1 u2 = v2 − u1 = (1, 1 + i, 0) − (1 − i, 0, i) = (1, 3 + 3i, 1 − i). ||u1 ||2 3 3 √ 1 21 ||u2 || = 1 + (3 + 3i)(3 − 3i) + (1 − i)(1 + i) = . Hence, an orthonormal basis is: 3 3 √ √ 3 21 (1 − i, 0, i), (1, 3 + 3i, 1 − i) . 3 21 22. Let v1 = (1 + i, i, 2 − i) and v2 = (1 + 2i, 1 − i, i). √ u1 = v1 = (1 + i, i, 2 − i), ||u1 || = (1 + i)(1 − i) + i(−i) + (2 − i)(2 + i) = 2 2. v2 , u1 = (1 + 2i)(1 − i) + (1 − i)(−i) + i(2 + i) = 1 + 2i. v2 , u1 1 1 u2 = v2 − u1 = (1 + 2i, 1 − i, i) − (1 + 2i)(1 + i, i, 2 − i) = (9 + 13i, 10 − 9i, −4 + 5i). 2 ||u1 || 8 8 √ 1 118 . Hence, an orthonormal ||u2 || = (9 + 13i)(9 − 13i) + (10 − 9i)(10 + 9i) + (−4 + 5i)(−4 − 5i) = 4 8 basis is: √ √ 2 118 (1 + i, i, 2 − i), (9 + 13i, 10 − 9i, −4 + 5i) . 4 236 23. Let f1 = 1, f2 = x and f3 = x2 . g1 = f1 = 1; 1 ||g1 ||2 = 1 dx = 1; f2 , g1 = 0 xdx = 0 x2 2 1 = 0 1 . 2 f2 , g1 1 1 g2 = f2 − g1 = x − = (2x − 1). ||g1 ||2 2 2 1 2 1 1 1 1 x3 x2 x 1 x2 − x + dx = − + = . ||g2 ||2 = x− dx = 2 4 3 2 40 12 0 0 1 1 x3 1 f3 , g1 = x2 dx = =. 30 3 0 1 1 1 x4 x3 1 f3 , g2 = x2 x − dx = − = . 2 4 60 12 0 f3 , g1 f3 , g2 1 1 1 g3 = f3 − g1 − g2 = x2 − − x − = (6x2 − 6x + 1). Thus, an orthogonal basis is given ||g1 ||2 ||g2 ||2 3 2 6 by: 1 1 1, (2x − 1), (6x2 − 6x + 1) . 2 6 301 24. Let f1 = 1, f2 = x2 and f3 = x4 for all x in [−1, 1]. 1 1 2 2 f1 , f2 = x2 dx = , f1 , f3 = x4 dx = , and f2 , f3 = 3 5 −1 −1 1 1 2 ||f1 ||2 = dx = 2, ||f2 ||2 = x4 dx = . 5 −1 −1 Let g1 = f1 = 1. f2 , g1 x2 , 1 1 1 g2 = f2 − g = x2 − · 1 = x2 − = (3x − 1). 21 ||g1 || ||1||2 3 3 g3 = f3 − 1 x6 dx = −1 2 . 7 x4 , x2 − 1 f3 , g1 f3 , g2 x4 , 1 3 g− g = x4 − ·1− 21 22 2 ||g1 || ||g2 || ||1|| ||x2 − 1 ||2 3 = x4 − x2 − 1 3 6x2 3 1 + = (35x4 − 30x2 + 3). 7 35 35 Thus, an orthogonal basis is given by: 1, 1 1 (3x2 − 1), (35x4 − 30x2 + 3) . 3 35 ππ 25. Let f1 = 1, f2 = sin x and f3 = cos x for all x in − , . 22 π /2 f1 , f2 = −π/2 π /2 f2 , f3 = π/2 sin xdx = [− cos 2x]−π/2 = 0. Therefore, f1 and f2 are orthogonal. sin x cos xdx = −π/2 nal. 1 2 π /2 f1 , f3 = −π/2 π /2 1 sin 2xdx = − cos 2x 4 −π/2 π /2 = 0. Therefore, f2 and f3 are orthogo−π/2 π/2 cos xdx = [sin x]−π/2 = 2. π /2 Let g1 = f1 = 1 so that ||g1 ||2 = dx = π . −π/2 π /2 2 π /2 1 − cos 2x π dx = . 2 2 −π/2 −π/2 f3 , g1 f3 , g2 2 1 g3 = f3 − g1 − g2 = cos x − · 1 − 0 · sin x = (π cos x − 2). Thus, an orthogonal basis for ||g1 ||2 ||g2 ||2 π π the subspace of C 0 [−π/2, π/2] spanned by {1, sin x, cos x} is: g2 = f2 = sin x, and ||g2 ||2 = sin xdx = 1, sin x, 26. Given A1 = 1 −1 2 1 and A2 = 2 −3 4 1 1 (π cos x − 2) . π . Using the Gram-Schmidt procedure: 1 −1 , A2 , B1 = 10 + 6 + 24 + 5 = 45, and ||B1 ||2 = 5 + 2 + 12 + 5 = 24. 2 1 1 A2 , B1 15 1 −1 2 −3 −9 8 8 B2 = A2 − B1 = − = . Thus, an orthogonal basis for the 1 4 1 2 1 −7 ||B1 ||2 8 4 8 subspace of M2 (R) spanned by A1 and A2 is: B1 = 1 −1 2 1 , 1 8 1 4 −9 8 −7 8 . 302 27. Given A1 = 0 1 1 0 , A2 = 0 1 1 1 1 1 and A3 = 1 0 . Using the Gram-Schmidt procedure: 1 , A2 , B1 = 5, and ||B1 ||2 = 5. 0 A2 , B1 01 01 00 B2 = A2 − B1 = − = . 11 10 01 ||B1 ||2 2 Also, A3 , B1 = 5, A3 , B2 = 0, and ||B2 || = 5, so that A3 , B2 A3 , B1 11 01 B1 − B2 = − B3 = A3 − −0 10 10 ||B1 ||2 ||B2 ||2 basis for the subspace of M2 (R) spanned by A1 , A2 , and A3 is: B1 = 0 1 01 10 , 1 0 0 0 , 0 0 0 0 0 1 0 1 = 1 0 0 0 . Thus, an orthogonal , which is the subspace of all symmetric matrices in M2 (R). 28. Given p1 (x) = 1 − 2x + 2x2 and p2 (x) = 2 − x − x2 . Using the Gram-Schmidt procedure: q1 = 1 − 2x + 2x2 , p2 , q1 = 2 · 1 + (−1)(−2) + (−1)2 = 2, ||q1 ||2 = 12 + (−2)2 + 22 = 9. So, 2 1 p2 , q 1 q 2 = p2 − q1 = 2 − x − x2 − (1 − 2x + 2x2 ) = (16 − 5x − 13x2 ). Thus, an orthogonal basis for the ||q1 ||2 9 9 subspace spanned by p1 and p2 is {1 − 2x + 2x2 , 16 − 5x − 13x2 }. 29. Given p1 (x) = 1 + x2 , p2 (x) = 2 − x + x3 , and p3 (x) = −x + 2x2 . Using the Gram-Schmidt procedure: q1 = 1 + x2 , p2 , q1 = 2 · 1 + (−1)(0) + 0 · 1 + 1 · 2 = 2, and q1 2 = 12 + 12 = 2. So, p2 , q 1 q 2 = p2 − q1 = 2 − x + x3 − (1 + x2 ) = 1 − x − x2 + x3 . Also, ||q1 ||2 p3 , q1 = 0 · 1 + (−1)0 + 2 · 1 + 02 = 2 p3 , q2 = 0 · 1 + (−1)2 + 2(−1) + 0 · 1 = −1, and ||q2 ||2 = 12 + (−1)2 + (−1)2 + 12 = 4 so that p3 , q 1 p3 , q 2 1 1 q 3 = p3 − q− q = −x + 2x2 − (1 + x2 ) + (1 − x − x2 + x3 ) = (−3 − 5x + 3x2 + x3 ). Thus, an 21 22 ||q1 || ||q2 || 4 4 orthogonal basis for the subspace spanned by p1 , p2 , and p3 is {1 + x2 , 1 − x − x2 + x3 , −3 − 5x + 3x2 + x3 }. 30. {u1 , u2 , v} is a linearly independent set of vectors, and u1 , u2 = 0. If we let u3 = v + λu1 + µu2 , then it must be the case that u3 , u1 = 0 and u3 , u2 = 0. u3 , u1 = 0 =⇒ v + λu1 + µu2 , u1 = 0 =⇒ v, u1 + λ u1 , u1 + µ u2 , u1 = 0 =⇒ v, u1 + λ u1 , u1 + µ · 0 = 0 v, u1 . =⇒ λ = − ||u1 ||2 u3 , u2 = 0 =⇒ v + λu1 + µu2 , u2 = 0 =⇒ v, u2 + λ u1 , u2 + µ u2 , u2 = 0 =⇒ v, u2 + λ · 0 + µ u2 , u2 = 0 v, u2 =⇒ µ = − . ||u2 ||2 v, u1 v, u2 Hence, if u3 = v − u− u2 , then {u1 , u2 , u3 } is an orthogonal basis for the subspace spanned 21 ||u1 || ||u2 ||2 by {u1 , u2 , v}. 31. It was shown in Remark 3 following Definition 4.12.1 that each vector ui is a unit vector. Moreover, for i = j, 1 1 1 ui , uj = vi , vj = vi , vj = 0, vi vj vi vj since vi , vj = 0. Therefore {u1 , u2 , . . . , uk } is an orthonormal set of vectors. 303 32. Set z = x − P(x, v1 ) − P(x, v2 ) − · · · − P(x, vk ). To verify that z is orthogonal to vi , we compute the inner product of z with vi : z, vi = x − P(x, v1 ) − P(x, v2 ) − · · · − P(x, vk ), vi = x, vi − P(x, v1 ), vi − P(x, v2 ), vi − · · · − P(x, vk ), vi . Since P(x, vj ) is a multiple of vj and the set {v1 , v2 , . . . , vk } is orthogonal, all of the subtracted terms on the right-hand side of this expression are zero except P(x, vi ), vi = x, vi x, vi vi , vi = vi , vi = x, vi . vi 2 vi 2 Therefore, z, vi = x, vi − x, vi = 0, which implies that z is orthogonal to vi . 33. We must show that W ⊥ is closed under addition and closed under scalar multiplication. Closure under addition: Let v1 and v2 belong to W ⊥ . This means that v1 , w = v2 , w = 0 for all w ∈ W . Therefore, v1 + v2 , w = v1 , w + v2 , w = 0 + 0 = 0 for all w ∈ W . Therefore, v1 + v2 ∈ W ⊥ . Closure under scalar multiplication: Let v belong to W ⊥ and let c be a scalar. This means that v, w = 0 for all w ∈ W . Therefore, cv, w = c v, w = c · 0 = 0, which shows that cv ∈ W ⊥ . 34. In this case, W ⊥ consists of all (x, y, z ) ∈ R3 such that (x, y, z ), (r, r, −r) = 0 for all r ∈ R. That is, rx + ry − rz = 0. In particular, we must have x + y − z = 0. Therefore, W ⊥ is the plane x + y − z = 0, which can also be expressed as W ⊥ = span{(−1, 0, 1), (0, 1, 1)}. 35. In this case, W ⊥ consists of all (x, y, z, w) ∈ R4 such that (x, y, z, w), (0, 1, −1, 3) = (x, y, z, w), (1, 0, 0, 3) = 0. This requires that y − z + 3w = 0 and x + 3w = 0. We can associated an augmented matrix with this 0 1 −1 3 0 system of linear equations: . Note that z and w are free variables: z = s and w = t. 10 030 Then y = s − 3t and x = −3t. Thus, W ⊥ = {(−3t, s−3t, s, t) : s, t ∈ R} = {t(−3, −3, 0, 1)+s(0, 1, 1, 0) : s, t ∈ R} = span{(−3, −3, 0, 1), (0, 1, 1, 0)}. xy zw 36. In this case, W ⊥ consists of all 2 × 2 matrices The set of symmetric matrices is spanned by the matrices that are orthogonal to all symmetric matrices. 10 00 , 0 1 1 0 0 0 , 0 1 . Thus, we must have xy zw , 10 00 = xy zw , 0 1 1 0 = xy zw , 0 0 0 1 = 0. 304 Thus, x = 0, y + z = 0 and w = 0. Therefore W⊥ = −z 0 0 z :z∈R 0 −1 1 0 = span , which is precisely the set of 2 × 2 skew-symmetric matrices. 37. Suppose v belongs to both W and W ⊥ . Then v, v = 0 by definition of W ⊥ , which implies that v = 0 by the first axiom of an inner product. Therefore W and W ⊥ can contain no common elements aside from the zero vector. 38. Suppose that W1 is a subset of W2 and let v be a vector in (W2 )⊥ . This means that v, w2 = 0 for all w2 ∈ W2 . In particular, v is orthogonal to all vectors in W1 (since W1 is merely a subset of W2 ). Thus, v ∈ (W1 )⊥ . Hence, we have shown that every vector belonging to (W2 )⊥ also belongs to (W1 )⊥ . 39. (a) Using technology we find that π π sin nx dx = 0, π cos nx dx = 0, −π sin nx cos mx dx = 0. −π Further, for m = n, −π π π sin nx sin mx dx = 0 and cos nx cos mx dx = 0. −π −π Consequently the given set of vectors is orthogonal on [−π, π ]. (b) Multiplying (4.12.7) by cos mx and integrating over [−π, π ] yields π f (x) cos mx dx = −π 1 a0 2 π π ∞ cos mx dx + −π (an cos nx + bn sin nx) cos mx dx. −π n=1 Assuming that interchange of the integral and infinite summation is permissible, this can be written π f (x) cos mx dx = −π which reduces to 1 a0 2 ∞ π n=1 π −π cos mx dx + −π π f (x) cos mx dx = −π 1 a0 2 (an cos nx + bn sin nx) cos mx dx. π π cos2 mx dx cos mx dx + am −π −π where we have used the results from part (a). When m = 0, this gives π f (x) dx = −π 1 a0 2 π dx = πa0 =⇒ a0 = −π 1 π π f (x) dx, −π whereas for m = 0, π π cos2 mx dx = πam =⇒ am = f (x) cos mx dx = am −π −π 1 π π f (x) cos mx dx. −π (c) Multiplying (4.12.7) by sin(mx), integrating over [−π, π ], and interchanging the integration and summation yields π 1 f (x) sin mx dx = a0 2 −π ∞ π π n=1 −π sin mx dx + −π (an cos nx + bn sin nx) sin mx dx. 305 Using the results from (a), this reduces to π π sin2 mx dx = πbn =⇒ bn = f (x) sin mx dx = bn −π −π 1 π π f (x) sin mx dx. −π (d) The Fourier coefficients for f are a0 = bn = 1 π 1 π π xdx = 0, an = −π π x sin nx dx = − −π π 1 π x cos nx dx = 0, −π 2 2 cos nπ = (−1)n+1 . n n The Fourier series for f is ∞ 2 (−1)n+1 sin nx. n n=1 (e) The approximations using the first term, the first three terms, the first five terms, and the first ten terms in the Fourier series for f are shown in the accompanying figures. S3(x) 3 2 1 x 3 2 1 1 2 3 1 2 3 Figure 0.0.66: Figure for Exercise 39(e) - 3 terms included These figures suggest that the Fourier series is converging to the function f (x) at all points in the interval (π, π ). Solutions to Section 4.13 Problems: 306 S5(x) 3 2 1 3 2 1 1 2 3 x 1 2 3 Figure 0.0.67: Figure for Exercise 39(e) - 5 terms included S10(x) 3 2 1 3 2 1 1 2 3 x 1 2 3 Figure 0.0.68: Figure for Exercise 39(e) - 10 terms included 1. Write v = (a1 , a2 , a3 , a4 , a5 ) ∈ R5 . Then we have (r + s)v = (r + s)(a1 , a2 , a3 , a4 , a5 ) = ((r + s)a1 , (r + s)a2 , (r + s)a3 , (r + s)a4 , (r + s)a5 ) = (ra1 + sa1 , ra2 + sa2 , ra3 + sa3 , ra4 + sa4 , ra5 + sa5 ) = (ra1 , ra2 , ra3 , ra4 , ra5 ) + (sa1 , sa2 , sa3 , sa4 , sa5 ) = r(a1 , a2 , a3 , a4 , a5 ) + s(a1 , a2 , a3 , a4 , a5 ) = rv + sv. 307 2. Write v = (a1 , a2 , a3 , a4 , a5 ) and w = (b1 , b2 , b3 , b4 , b5 ) in R5 . Then we have r(v + w) = r((a1 , a2 , a3 , a4 , a5 ) + (b1 , b2 , b3 , b4 , b5 )) = r(a1 + b1 , a2 + b2 , a3 + b3 , a4 + b4 , a5 + b5 ) = (r(a1 + b1 ), r(a2 + b2 ), r(a3 + b3 ), r(a4 + b4 ), r(a5 + b5 )) = (ra1 + rb1 , ra2 + rb2 , ra3 + rb3 , ra4 + rb4 , ra5 + rb5 ) = (ra1 , ra2 , ra3 , ra4 , ra5 ) + (rb1 , rb2 , rb3 , rb4 , rb5 ) = r(a1 , a2 , a3 , a4 , a5 ) + r(b1 , b2 , b3 , b4 , b5 ) = r v + r w. 3. NO. This set of polynomials is not closed under scalar multiplication. For example, the polynomial 1 p(x) = 2x belongs to the set, but 3 p(x) = 2 x does not belong to the set (since 2 is not an even integer). 3 3 4. YES. This set of polynomials forms a subspace of the vector space P5 . To confirm this, we will check that this set is closed under addition and scalar multiplication: Closure under Addition: Let p(x) = a0 + a1 x + a4 x4 + a5 x5 and q (x) = b0 + b1 x + b4 x4 + b5 x5 be polynomials in the set under consideration (their x2 and x3 terms are zero). Then p(x) + q (x) = (a0 + b0 ) + (a1 + b1 )x + · · · + (a4 + b4 )x4 + (a5 + b5 )x5 is again in the set (since it still has no x2 or x3 terms). So closure under addition holds. Closure under Scalar Multiplication: Let p(x) = a0 + a1 x + a4 x4 + a5 x5 be in the set, and let k be a scalar. Then kp(x) = (ka0 ) + (ka1 )x + (ka4 )x4 + (ka5 )x5 , which is again in the set (since it still has no x2 or x3 terms). So closure under scalar multiplication holds. 5. NO. We can see immediately that the zero vector (0, 0, 0) is not a solution to this linear system (the first equation is not satisfied by the zero vector), and therefore, we know at once that this set cannot be a vector space. 6. YES. The set of solutions to this linear system forms a subspace of R3 . To confirm this, we will check that this set is closed under addition and scalar multiplication: Closure under Addition: Let (a1 , a2 , a3 ) and (b1 , b2 , b3 ) be solutions to the linear system. This means that 4a1 − 7a2 + 2a3 = 0, 5a1 − 2a2 + 9a3 = 0 4b1 − 7b2 + 2b3 = 0, 5b1 − 2b2 + 9b3 = 0. and Adding the equations on the left, we get 4(a1 + b1 ) − 7(a2 + b2 ) + 2(a3 + b3 ) = 0, so the vector (a1 + b1 , a2 + b2 , a3 + b3 ) satisfies the first equation in the linear system. Likewise, adding the equations on the right, we get 5(a1 + b1 ) − 2(a2 + b2 ) + 9(a3 + b3 ) = 0, so (a1 + b1 , a2 + b2 , a3 + b3 ) also satisfies the second equation in the linear system. Therefore, (a1 + b1 , a2 + b2 , a3 + b3 ) is in the solution set for the linear system, and closure under addition therefore holds. 308 Closure under Scalar Multiplication: Let (a1 , a2 , a3 ) be a solution to the linear system, and let k be a scalar. We have 4a1 − 7a2 + 2a3 = 0, 5a1 − 2a2 + 9a3 = 0, and so, multiplying both equations by k , we have k (4a1 − 7a2 + 2a3 ) = 0, k (5a1 − 2a2 + 9a3 ) = 0, or 4(ka1 ) − 7(ka2 ) + 2(ka3 ) = 0, 5(ka1 ) − 2(ka2 ) + 9(ka3 ) = 0. Thus, the vector (ka1 , ka2 , ka3 ) is a solution to the linear system, and closure under scalar multiplication therefore holds. 7. NO. This set is not closed under addition. For example, the vectors 1 1 1 1 and −1 1 1 1 both belong to the set (their entries are all nonzero), but 1 1 1 1 + −1 1 1 1 0 2 = 2 2 , which does not belong to the set (some entries are zero, and some are nonzero). So closure under addition fails, and therefore, this set does not form a vector space. 12 forms a subspace of M2 (R). To 22 confirm this, we will check that this set is closed under addition and scalar multiplication: 8. YES. The set of 2 × 2 real matrices that commute with C = Closure under Addition: Let A and B be 2 × 2 real matrices that commute with C . That is, AC = CA and BC = CB . Then (A + B )C = AC + BC = CA + CB = C (A + B ), so A + B commutes with C , and therefore, closure under addition holds. Closure under Scalar Multiplication: Let A be a 2 × 2 real matrix that commutes with C , and let k be a scalar. Then since AC = CA, we have (kA)C = k (AC ) = k (CA) = C (kA), so kA is still in the set. Thus, the set is also closed under scalar multiplication. 1 9. YES. The set of functions f : [0, 1] → [0, 1] such that f (0) = f 4 = f 1 = f 3 = f (1) = 0 is a 2 4 subspace of the vector space of all functions [0, 1] → [0, 1]. We confirm this by checking that this set is closed under addition and scalar multiplication: Closure under Addition: Let g and h be functions such that g (0) = g 1 4 =g 1 2 =g 3 4 = g (1) = 0 h(0) = h 1 4 =h 1 2 =h 3 4 = h(1) = 0. and 309 Now (g + h)(0) = g (0) + h(0) = 0 + 0 = 0, 1 4 1 2 3 4 (g + h) (g + h) (g + h) =g =g =g 1 4 1 2 3 4 +h +h +h 1 4 1 2 3 4 = 0 + 0 = 0, = 0 + 0 = 0, = 0 + 0 = 0, (g + h)(1) = g (1) + h(1) = 0 + 0 = 0. Thus, g + h belongs to the set, and so the set is closed under addition. Closure under Scalar Multiplication: Let g be a function with 1 4 g (0) = g and let k be a scalar. Then =g 1 2 =g 3 4 = g (1) = 0, (kg )(0) = k · g (0) = k · 0 = 0, (kg ) (kg ) (kg ) 1 4 1 2 3 4 1 4 1 2 3 4 =k·g =k·g =k·g = k · 0 = 0, = k · 0 = 0, = k · 0 = 0, (kg )(1) = k · g (1) = k · 0 = 0. Thus, kg belongs to the set, and so the set is closed under scalar multiplication. 10. NO. This set is not closed under addition. For example, the function f defined by f (x) = x belongs to the set, and the function g defined by g (x) = if x ≤ 1/2 if x > 1/2 x, 0, belongs to the set. But if x ≤ 1/2 if x > 1/2. 2x, x, (f + g )(x) = Then f + g : [0, 1] → [0, 1], but for 0 < x ≤ 1 , |(f + g )(x)| = 2x > x, so f + g is not in the set. Therefore, 2 the set is not closed under addition. 11. NO. This set is not closed under addition. For example, if we let A= 1 0 0 1 B= and 0 0 1 0 2 1 then A2 = A is symmetric, and B 2 = 02 is symmetric, but (A + B )2 = 1 0 1 1 2 = 1 0 , 310 is not symmetric, so A + B is not in the set. Thus, the set in question is not closed under addition. 12. YES. Let us describe geometrically the set of points equidistant from (−1, 2) and (1, −2). If (x, y ) is such a point, then using the distance formula and equating distances to (−1, 2) and (1, −2), we have (x + 1)2 + (y − 2)2 = (x − 1)2 + (y + 2)2 or (x + 1)2 + (y − 2)2 = (x − 1)2 + (y + 2)2 or x2 + 2x + 1 + y 2 − 4y + 4 = x2 − 2x + 1 + y 2 + 4y + 4. Cancelling like terms and rearranging this equation, we have 4x = 8y . So the points that are equidistant 1 from (−1, 2) and (1, −2) lie on the line through the origin with equation y = 2 x. Any line through the origin 2 2 of R is a subspace of R . Therefore, this line forms a vector space. 13. NO. This set is not closed under addition, nor under scalar multiplication. For instance, the point (5, −3, 4) is a distance 5 from (0, −3, 4), so the point (5, −3, 4) lies in the set. But the point (10, −6, 8) = 2(5, −3, 4) is a distance √ √ (10 − 0)2 + (−6 + 3)2 + (8 − 4)2 = 100 + 9 + 16 = 125 = 5 from (0, −3, 4), so (10, −6, 8) is not in the set. So the set is not closed under scalar multiplication, and hence does not form a subspace. 14. We must check each of the vector space axioms (A1)-(A10). Axiom (A1): Assume that (a1 , a2 ) and (b1 , b2 ) belong to V . Then a2 , b2 > 0. Hence, (a1 , a2 ) + (b1 , b2 ) = (a1 + b1 , a2 b2 ) ∈ V, since a2 b2 > 0. Thus, V is closed under addition. Axiom (A2): Assume that (a1 , a2 ) ∈ V , and let k be a scalar. Note that since a2 > 0, the expression ak > 0 for every k ∈ R. Hence, k (a1 , a2 ) = (ka1 , ak ) ∈ V , thereby showing that V is closed under scalar 2 2 multiplication. Axiom (A3): Let (a1 , a2 ), (b1 , b2 ) ∈ V . We have (a1 , a2 ) + (b1 , b2 ) = (a1 + b1 , a2 b2 ) = (b1 + a1 , b2 a2 ) = (b1 , b2 ) + (a1 , a2 ), as required. Axiom (A4): Let (a1 , a2 ), (b1 , b2 ), (c1 , c2 ) ∈ V . We have ((a1 , a2 ) + (b1 , b2 )) + (c1 , c2 ) = (a1 + b1 , a2 b2 ) + (c1 , c2 ) = ((a1 + b1 ) + c1 , (a2 b2 )c2 ) = (a1 + (b1 + c1 ), a2 (b2 c2 )) = (a1 , a2 ) + (b1 + c1 , b2 c2 ) = (a1 , a2 ) + ((b1 , b2 ) + (c1 , c2 )), 311 as required. Axiom (A5): We claim that (0, 1) is the zero vector in V . To see this, let (b1 , b2 ) ∈ V . Then (0, 1) + (b1 , b2 ) = (0 + b1 , 1 · b2 ) = (b1 , b2 ). Since this holds for every (b1 , b2 ) ∈ V , we conclude that (0, 1) is the zero vector. Axiom (A6): We claim that the additive inverse of a vector (a1 , a2 ) in V is the vector (−a1 , a−1 ) (Note that 2 a−1 > 0 since a2 > 0.) To check this, we compute as follows: 2 (a1 , a2 ) + (−a1 , a−1 ) = (a1 + (−a1 ), a2 a−1 ) = (0, 1). 2 2 Axiom (A7): We have 1 · (a1 , a2 ) = (a1 , a1 ) = (a1 , a2 ) 2 for all (a1 , a2 ) ∈ V . Axiom (A8): Let (a1 , a2 ) ∈ V , and let r and s be scalars. Then we have (rs)(a1 , a2 ) = ((rs)a1 , ars ) 2 = (r(sa1 ), (as )r ) 2 = r(sa1 , as ) 2 = r(s(a1 , a2 )), as required. Axiom (A9): Let (a1 , a2 ) and (b1 , b2 ) be members of V , and let r be a scalar. We have r((a1 , a2 ) + (b1 , b2 )) = r(a1 + b1 , a2 b2 ) = (r(a1 + b1 ), (a2 b2 )r ) = (ra1 + rb1 , ar br ) 22 = (ra1 , ar ) + (rb1 , br ) 2 2 = r(a1 , a2 ) + r(b1 , b2 ), as required. Axiom (A10): Let (a1 , a2 ) ∈ V , and let r and s be scalars. We have (r + s)(a1 , a2 ) = ((r + s)a1 , ar+s ) 2 = (ra1 + sa1 , ar as ) 22 r = (ra1 , a2 ) + (sa1 , as ) 2 = r(a1 , a2 ) + s(a1 , a2 ), as required. 15. We must show that W is closed under addition and closed under scalar multiplication: Closure under Addition: Let (a, 2a ) and (b, 2b ) be elements of W . Now consider the sum of these elements: (a, 2a ) + (b, 2b ) = (a + b, 2a 2b ) = (a + b, 2a+b ) ∈ W, which shows that W is closed under addition. 312 Closure under Scalar Multiplication: Let (a, 2a ) be an element of W , and let k be a scalar. Then k (a, 2a ) = (ka, (2a )k ) = (ka, 2ka ) ∈ W, which shows that W is closed under scalar multiplication. Thus, W is a subspace of V . 16. Note that 3(1, 2) = (3, 23 ) = (3, 8), so the second vector is a multiple of the first one under the vector space operations of V from Problem 14. Therefore, {(1, 2), (3, 8)} is linearly dependent. 17. We show that S = {(1, 4), (2, 1)} is linearly independent and spans V . S is linearly independent: Assume that c1 (1, 4) + c2 (2, 1) = (0, 1). This can be written (c1 , 4c1 ) + (2c2 , 1c2 ) = (0, 1) or (c1 + 2c2 , 4c1 ) = (0, 1). In order for 4c1 = 1, we must have c1 = 0. And then in order for c1 + 2c2 = 0, we must have c2 = 0. Therefore, S is linearly independent. S spans V : Consider an arbitrary vector (a1 , a2 ) ∈ V , where a2 > 0. We must find constants c1 and c2 such that c1 (1, 4) + c2 (2, 1) = (a1 , a2 ). Thus, (c1 , 4c1 ) + (2c2 , 1c2 ) = (a1 , a2 ) or (c1 + 2c2 , 4c1 ) = (a1 , a2 ). Hence, c1 + 2c2 = a1 and 4c1 = a2 . From the second equation, we conclude that c1 = log4 (a2 ). Thus, from the first equation, 1 (a1 − log4 (a2 )). 2 Hence, since we were able to find constants c1 and c2 in order that c2 = c1 (1, 4) + c2 (2, 1) = (a1 , a2 ), we conclude that {(1, 4), (2, 1)} spans V . 18. This really hinges on whether or not the given vectors are linearly dependent or linearly independent. If we assume that c1 (2 + x2 ) + c2 (4 − 2x + 3x2 ) + c3 (1 + x) = 0, 313 then (2c1 + 4c2 + c3 ) + (−2c2 + c3 )x + (c1 + 3c2 )x2 = 0. Thus, we have − 2c2 + c3 = 0 2c1 + 4c2 + c3 = 0 c1 + 3c2 = 0. Since the matrix of coefficients 2 41 0 −2 1 1 30 fails to be invertible, we conclude that there will be non-trivial solutions for c1 , c2 , and c3 . Thus, the polynomials are linearly dependent. We can therefore remove a vector from the set without decreasing the span. We remove 1 + x, leaving us with {2 + x2 , 4 − 2x + 3x2 }. Since these polynomials are not proportional, they are now linearly independent, and hence, they are a basis for their span. Hence, dim{2 + x2 , 4 − 2x + 3x2 , 1 + x} = 2. 19. NO. This set is not closed under scalar multiplication. For example, (1, 1) belongs to W , but 2 · (1, 1) = (2, 2) does not belong to W . 20. NO. This set is not closed under scalar multiplication. For example, (1, 1) belongs to W , but 2 · (1, 1) = (2, 2) does not belong to W . 21. NO. The zero vector (zero matrix) is not an orthogonal matrix. Any subspace must contain a zero vector. 22. YES. We show that W is closed under addition and closed under scalar multiplication. Closure under Addition: Assume that f and g belong to W . Thus, f (a) = 2f (b) and g (a) = 2g (b). We must show that f + g belongs to W . We have (f + g )(a) = f (a) + g (a) = 2f (b) + 2g (b) = 2[f (b) + g (b)] = 2(f + g )(b), so f + g ∈ W . So W is closed under addition. Closure under Scalar Multiplication: Assume that f belongs to W and k is a scalar. Thus, f (a) = 2f (b). Moreover, (kf )(a) = kf (a) = k (2f (b)) = 2(kf (b)) = 2(kf )(b), so kf ∈ W . Thus, W is closed under scalar multiplication. 23. YES. We show that W is closed under addition and closed under scalar multiplication. Closure under Addition: Assume that f and g belong to W . Thus, b b f (x)dx = 0 and g (x)dx = 0. a a Hence, so f + g ∈ W . g (x)dx = 0 + 0 = 0, f (x)dx + (f + g )(x)dx = a b b b a a 314 Closure under Scalar Multiplication: Assume that f belongs to W and k is a scalar. Thus, b f (x)dx = 0. a Therefore, b b (kf )(x)dx = a b f (x)dx = k · 0 = 0, kf (x)dx = k a a so kf ∈ W . 24. YES. We show that W is closed under addition and scalar multiplication. Closure under Addition: Assume that b1 a2 d1 , c2 f1 e2 a1 c1 e1 b2 d2 ∈ W. f2 Thus, a1 + b1 = c1 + f1 , a1 − c1 = e1 − f1 − d1 , a2 + b2 = c2 + f2 , a2 − c2 = e2 − f2 − d2 . Hence, (a1 + a2 ) + (b1 + b2 ) = (c1 + c2 ) + (f1 + f2 ) and (a1 + a2 ) − (c1 + c2 ) = (e1 + e2 ) − (f1 + f2 ) − (d1 + d2 ). Thus, a1 c1 e1 b1 a2 d1 + c2 f1 e2 b2 a1 + a2 d2 = c1 + c2 f2 e1 + e2 b1 + b2 d1 + d2 ∈ W, f1 + f2 which means W is closed under addition. Closure under Scalar Multiplication: Assume that ab c d ∈W ef and k is a scalar. Then we have a + b = ka − kc = ke − kf − kd. Therefore, a k c e c + f and a − c = e − f − d. Thus, ka + kb = kc + kf and b ka kb d = kc kd ∈ W, ke kf f so W is closed under scalar multiplication. 25. (a) NO, (b) YES. Since 3 vectors are required to span R3 , S cannot span V . However, since the vectors are not proportional, they are linearly independent. 6 −3 2 1 1 , 26. (a) YES, (b) YES. If we place the three vectors into the columns of a 3 × 3 matrix 1 1 −8 −1 we observe that the matrix is invertible. Hence, its columns are linearly independent. Since we have 3 linearly independent vectors in the 3-dimensional space R3 , we have a basis for R3 . 315 27. (a) NO, (b) YES. Since we have only 3 vectors in a 4-dimensional vector space, they cannot possibly span R4 . To check linear independence, we place the vectors into the columns of a matrix: 61 1 −3 1 −8 2 1 −1 . 00 0 The first three rows for the same invertible matrix as in the previous problem, so the reduced row-echelon form 100 0 1 0 is 0 0 1 , so there are no free variables, and hence, the column vectors form a linearly independent 000 set. 28. (a) YES, (b) NO. Since (0, 0, 0) is a member of S , S cannot be linearly independent. However, the set {(10, −6, 5), (3, −3, 2), (6, 4, −1)} is linearly independent (these vectors can be made to form a 3 × 3 matrix that is invertible), and thus, S must span at least a 3-dimensional space. Since dim[R3 ] = 3, we know that S spans R3 . 29. (a) YES, (b) YES. Consider the linear equation c1 (2x − x3 ) + c2 (1 + x + x2 ) + 3c3 + c4 x = 0. Then (c2 + 3c3 ) + (2c1 + c2 + c4 )x + c2 x2 − c1 x3 = 0. From the latter equation, we see that c1 = c2 = 0 (looking at the x2 and x3 coefficients) and thus, c3 = c4 = 0 (looking at the constant term and x coefficient). Thus, c1 = c2 = c3 = c4 = 0, and hence, S is linearly independent. Since we have four linearly independent vectors in the 4-dimensional vector space P3 , we conclude that these vectors also span P3 . 30. (a) NO, (b) NO. The set S only contains four vectors, although dim[P4 ] = 5, so it is impossible for S to span P4 . Alternatively, simply note that none of the polynomials in S contain an x3 term. To check linear independence, consider the equation c1 (x4 + x + 2 + 1) + c2 (x2 + x + 1) + c3 (x + 1) + c4 (x4 + 2x + 3) = 0. Rearranging this, we have (c1 + c4 )x4 + (c1 + c2 )x2 + (c2 + c3 + 2c4 )x + (c1 + c2 + 3c4 ) = 0, and so we look to solve the linear system with augmented matrix 11130 1 1 1 30 0 1 1 2 0 0 1 1 2 0 ∼ = 1 1 0 0 0 0 0 −1 −3 0 10010 0 −1 −1 −2 0 111 011 001 000 3 2 3 0 0 0 , 0 0 where we have added −1 times the first row to each of the third and fourth rows in the first step, and zeroed at the last row in the second step. We see that c4 is a free variable, which means that a nontrivial solution to the linear system exists, and therefore, the original vectors are linearly dependent. 316 31. (a) NO, (b) YES. The vector space M2×3 (R) is 6-dimensional, and since only four vectors belong to S , S cannot possibly span M2×3 (R). On the other hand, if we form the linear system with augmented matrix −1 3 −1 −11 0 0 2 −2 −6 0 0 1 −3 −5 0 , 01 3 1 0 12 2 −2 0 13 1 −5 0 row reduction shows that each column of a row-echelon form contains a pivot, and therefore, the vectors are linearly independent. 32. (a) NO, (b) NO. Since M2 (R) is only 4-dimensional and S contains 5 vectors, S cannot possibly be linearly independent. Moreover, each matrix in S is symmetric, and therefore, only symmetric matrices can be found in the span(S ). Thus, S fails to span M2 (R). 33. Assume that {v1 , v2 , v3 } is linearly independent, and that v4 does not lie in span{v1 , v2 , v3 }. We will show that {v1 , v2 , v3 , v4 } is linearly independent. To do this, assume that c1 v1 + c2 v2 + c3 v3 + c4 v4 = 0. We must prove that c1 = c2 = c3 = c4 = 0. If c4 = 0, then we rearrange the above equation to show that v4 = − c2 c3 c1 v1 − v2 − v3 , c4 c4 c4 which implies that v4 ∈ span{v1 , v2 , v3 }, contrary to our assumption. Therefore, we know that c4 = 0. Hence the equation above reduces to c1 v1 + c2 v2 + c3 v3 = 0, and the linear independence of {v1 , v2 , v3 } now implies that c1 = c2 = c3 = 0. Therefore, c1 = c2 = c3 = c4 = 0, as required. 34. Note that v and w are column vectors in Rm . We have v · w = vT w. Since w ∈ nullspace(AT ), we have AT w = 0, and since v ∈ colspace(A), we can write v = Av1 for some v1 ∈ Rm . Therefore, T T T v · w = vT w = (Av1 )T w = (v1 AT )w = v1 (AT w) = v1 (0) = 0, as desired. 35. (a): Our proof here actually shows that the set of n × n skew-symmetric matrices forms a subspace of Mn (R) for all positive integers n. We show that W is closed under addition and scalar multiplication: Closure under Addition: Suppose that A and B are in W . This means that AT = −A and B T = −B . Then (A + B )T = AT + B T = (−A) + (−B ) = −(A + B ), so A + B is skew-symmetric. Therefore, A + B belongs to W , and W is closed under addition. Closure under Scalar Multiplication: Suppose that A is in W and k is a scalar. We know that AT = −A. Then (kA)T = k (AT ) = k (−A) = −(kA), 317 so kA is skew-symmetric. Therefore, kA belongs to W , and W is closed under scalar multiplication. Therefore, W is a subspace. (b): An arbitrary 3 × 3 skew-symmetric matrix takes the form 0 ab 010 00 −a 0 c = a −1 0 0 + b 0 0 −b −c 0 000 −1 0 1 0 00 0 + c 0 0 1 . 0 0 −1 0 This results in 03 if and only if a = b = c = 0, so the three matrices appearing on the right-hand side are linearly independent. Moreover, the equation above also demonstrates that W is spanned by the three matrices appearing on the right-hand side. Therefore, these matrices are a basis for W : 010 001 0 00 0 1 . Basis = −1 0 0 , 0 0 0 , 0 000 −1 0 0 0 −1 0 Hence, dim[W ] = 3. (c): Since dim[M3 (R)] = 9, we must add an additional six (linearly independent) 3×3 matrices to form a basis for M3 (R). Using the notation prior to Example 4.6.3, we can use the matrices E11 , E22 , E33 , E12 , E13 , E23 to extend the basis in part (b) to a basis for M3 (R): 0 Basis for M3 (R) = −1 0 1 0 0 00 0 0 , 0 0 −1 0 0 1 0 00 0 , 0 0 1 , E11 , E22 , E33 , E12 , E13 , E23 0 0 −1 0 36. (a): We show that W is closed under addition and closed under scalar multiplication. Arbitrary elements of W have the form a b −a − b . c d −c − d −a − c −b − d a + b + c + d Closure under Addition: Let a1 b1 c1 d1 X1 = −a1 − c1 −b1 − d1 be elements of W . Then −a1 − b1 −c1 − d1 a1 + b1 + c1 + d1 a1 + a2 c1 + c2 X1 + X2 = −a1 − c1 − a2 − c2 a2 c2 X2 = −a2 − c2 and b 1 + b2 d1 + d2 −b1 − d1 − b2 − d2 b2 d2 −b2 − d2 −a2 − b2 −c2 − d2 a2 + b2 + c2 + d2 −a1 − b1 − a2 − b2 , −c1 − d1 − c2 − d2 a1 + b1 + c1 + d1 + a2 + b2 + c2 + d2 which is immediately seen to have row sums and column sums of zero. Thus, X1 + X2 belongs to W , and W is closed under addition. Closure under Scalar Multiplication: Let a1 c1 X1 = −a1 − c1 b1 d1 −b1 − d1 −a1 − b1 , −c1 − d1 a1 + b1 + c1 + d1 318 and let k be a scalar. Then ka1 kb1 k (−a1 − b1 ) , kc1 kd1 k (−c1 − d1 ) kX1 = k (−a1 − c1 ) k (−b1 − d1 ) k (a1 + b1 + c1 + d1 ) and we see by inspection that all row and column sums of kX1 are zero, so kX1 belongs to W . Therefore, W is closed under scalar multiplication. Therefore, W is a subspace of M3 (R). (b): We may write a b −a − b 1 = a 0 c d −c − d −a − c −b − d a + b + c + d −1 0 −1 0 1 −1 00 0 0 0 0 0 0 +b 0 0 0 +c 1 0 −1 +d 0 1 −1 . 0 1 0 −1 1 −1 0 1 0 −1 1 This results in 03 if and only if a = b = c = d = 0, so the matrices appearing on the right-hand side are linearly independent. Moreover, the equation above also demonstrates that W is spanned by the four matrices appearing on the right-hand side. Therefore, these matrices are a basis for W : 1 0 −1 0 1 −1 00 0 0 0 0 0 , 0 0 0 , 1 0 −1 , 0 1 −1 . Basis = 0 0 −1 0 1 0 −1 1 −1 0 1 0 −1 1 Hence, dim[W ] = 4. (c): Since dim[M3 (R)] = 9, we must add an additional five (linearly independent) 3 × 3 matrices to form a basis for M3 (R). Using the notation prior to Example 4.6.3, we can use the matrices E11 , E12 , E13 , E21 , and E31 to extend the basis from part (b) to a basis for M3 (R): Basis for M3 (R) = 1 0 −1 0 1 −1 0 0 −1 0 0 , 1 0 0 , 0 0 1 0 −1 1 −1 0 0 0 0 0 0 −1 , 0 1 −1 , E11 , E12 , E13 , E21 , E31 . 0 1 0 −1 1 37. (a): We must verify the axioms (A1)-(A10) for a vector space: Axiom (A1): Assume that (v1 , w1 ) and (v2 , w2 ) belong to V ⊕ W . Since v1 +V v2 ∈ V and w1 +W w2 ∈ W , then the sum (v1 , w1 ) + (v2 , w2 ) = (v1 +V v2 , w1 +W w2 ) lies in V ⊕ W . Axiom (A2): Assume that (v, w) belongs to V ⊕ W , and let k be a scalar. Since k ·V v ∈ V and k ·W w ∈ W , the scalar multiplication k · (v, w) = (k ·V v, k ·W w) lies in V ⊕ W . For the remainder of the axioms, we will omit the ·V and ·W notations. They are to be understood. 319 Axiom (A3): Assume that (v1 , w1 ), (v2 , w2 ) ∈ V ⊕ W . Then (v1 , w1 ) + (v2 , w2 ) = (v1 + v2 , w1 + w2 ) = (v2 + v1 , w2 + w1 ) = (v2 , w2 ) + (v1 , w1 ), as required. Axiom (A4): Assume that (v1 , w1 ), (v2 , w2 ), (v3 , w3 ) ∈ V ⊕ W . Then ((v1 , w1 ) + (v2 , w2 )) + (v3 , w3 ) = (v1 + v2 , w1 + w2 ) + (v3 , w3 ) = ((v1 + v2 ) + v3 , (w1 + w2 ) + w3 ) = (v1 + (v2 + v3 ), w1 + (w2 + w3 )) = (v1 , w1 ) + (v2 + v3 , w2 + w3 ) = (v1 , w1 ) + ((v2 , w2 ) + (v3 , w3 )), as required. Axiom (A5): We claim that the zero vector in V ⊕ W is (0V , 0W ), where 0V is the zero vector in the vector space V and 0W is the zero vector in the vector space W . To check this, let (v, w) ∈ V ⊕ W . Then (0V , 0W ) + (v, w) = (0V + v, 0W + w) = (v, w), which confirms that (0V , 0W ) is the zero vector for V ⊕ W . Axiom (A6): We claim that the additive inverse of the vector (v, w) ∈ V ⊕ W is the vector (−v, −w), where −v is the additive inverse of v in the vector space V and −w is the additive inverse of w in the vector space W . We check this: (v, w) + (−v, −w) = (v + (−v ), w + (−w)) = (0V , 0W ), as required. Axiom (A7): For every vector (v, w) ∈ V ⊕ W , we have 1 · (v, w) = (1 · v, 1 · w) = (v, w), where in the last step we have used the fact that Axiom (A7) holds in each of the vector spaces V and W . Axiom (A8): Let (v, w) be a vector in V ⊕ W , and let r and s be scalars. Using the fact that Axiom (A8) holds in V and W , we have (rs)(v, w) = ((rs)v, (rs)w) = (r(sv ), r(sw)) = r(sv, sw) = r(s(v, w)). Axiom (A9): Let (v1 , w1 ) and (v2 , w2 ) be vectors in V ⊕ W , and let r be a scalar. Then r((v1 , w1 ) + (v2 , w2 )) = r(v1 + v2 , w1 + w2 ) = (r(v1 + v2 ), r(w1 + w2 )) = (rv1 + rv2 , rw1 + rw2 ) = (rv1 , rw1 ) + (rv2 , rw2 ) = r(v1 , w1 ) + r(v2 , w2 )), as required. 320 Axiom (A10): Let (v, w) ∈ V ⊕ W , and let r and s be scalars. Then (r + s)(v, w) = ((r + s)v, (r + s)w) = (rv + sv, rw + sw) = (rv, rw) + (sv, sw) = r(v, w) + s(v, w), as required. (b): We show that {(v, 0) : v ∈ V } is a subspace of V ⊕ W , by checking closure under addition and closure under scalar multiplication: Closure under Addition: Suppose (v1 , 0) and (v2 , 0) belong to {(v, 0) : v ∈ V }, where v1 , v2 ∈ V . Then (v1 , 0) + (v2 , 0) = (v1 + v2 , 0) ∈ {(v, 0) : v ∈ V }, which shows that the set is closed under addition. Closure under Scalar Multiplication: Suppose (v, 0) ∈ {(v, 0) : v ∈ V } and k is a scalar. Then k (v, 0) = (kv, 0) is again in the set. Thus, {(v, 0) : v ∈ V } is closed under scalar multiplication. Therefore, {(v, 0) : v ∈ V } is a subspace of V ⊕ W . (c): Let {v1 , v2 , . . . , vn } be a basis for V , and let {w1 , w2 , . . . , wm } be a basis for W . We claim that S = {(vi , 0) : 1 ≤ i ≤ n} ∪ {(0, wj ) : 1 ≤ j ≤ m} is a basis for V ⊕ W . To show this, we will verify that S is a linearly independent set that spans V ⊕ W : Check that S is linearly independent: Assume that c1 (v1 , 0) + c2 (v2 , 0) + · · · + cn (vn , 0) + d1 (0, w1 ) + d2 (0, w2 ) + · · · + dm (0, wm ) = (0, 0). We must show that c1 = c2 = · · · = cn = d1 = d2 = · · · = dm = 0. Adding the vectors on the left-hand side, we have (c1 v1 + c2 v2 + · · · + cn vn , d1 w1 + d2 w2 + · · · + dm wm ) = (0, 0), so that c1 v1 + c2 v2 + · · · + cn vn = 0 and d1 w1 + d2 w2 + · · · + dm wm = 0. Since {v1 , v2 , . . . , vn } is linearly independent, c1 = c2 = · · · = cn = 0, and since {w1 , w2 , . . . , wm } is linearly independent, d1 = d2 = · · · = dm = 0. Thus, S is linearly independent. Check that S spans V ⊕ W : Let (v, w) ∈ V ⊕ W . We must express (v, w) as a linear combination of the vectors in S . Since {v1 , v2 , . . . , vn } spans V , there exist scalars c1 , c2 , . . . , cn such that v = c1 v1 + c2 v2 + · · · + cn vn , 321 and since {w1 , w2 , . . . , wm } spans W , there exist scalars d1 , d2 , . . . , dm such that w = d1 w1 + d2 w2 + · · · + dm wm . Then (v, w) = c1 (v1 , 0) + c2 (v2 , 0) + · · · + cn (vn , 0) + d1 (0, w1 ) + d2 (0, w2 ) + · · · + dm (0, wm ). Therefore, (v, w) is a linear combination of vectors in S , so S spans V ⊕ W . Therefore, S is a basis for V ⊕ W . Since S contains n + m vectors, dim[V ⊕ W ] = n + m. 38. There are many examples here. One such example is S = {x3 , x3 − x2 , x3 − x, x3 − 1}, a basis for P3 whose vectors all have degree 3. To see that S is a basis, note that S is linearly independent, since if c1 x3 + c2 (x3 − x2 ) + c3 (x3 − x) + c4 (x3 − 1) = 0, then (c1 + c2 + c3 + c4 )x3 − c2 x2 − c3 x − c4 = 0, and so c1 = c2 = c3 = c4 = 0. Since S is a linearly independent set of 4 vectors and dim[P3 ] = 4, S is a basis for P3 . 39. Let A be an m × n matrix. By the Rank-Nullity Theorem, dim[colspace(A)] + dim[nullspace(A)] = n. Since, by assumption, colspace(A) = nullspace(A) = r, n = 2r must be even. 40. The ith row of the matrix A is bi (c1 c2 . . . cn ). Therefore, each row of A is a multiple of the first row, and so rank(A) = 1. Thus, by the Rank-Nullity Theorem, nullity(A) = n − 1. 41. A row-echelon form of A is given by 12 00 a basis for the columnspace of A is given by −2 1 . Thus, a basis for the rowspace of A is given by {(1, 2)}, −3 −6 , and a basis for the nullspace of A is given by . All three subspaces are one-dimensional. −2 0 1/3 5/21 . Thus, a basis for the rowspace of A is 0 0 6 −1 15 {(1, −6, −2, 0), (0, 1, 3 , 21 )}, and a basis for the column space of A is given by 3 , 3 . For 7 21 the nullspace, we observe that the equations corresponding to the row-echelon form of A can be written as 1 −6 1 42. A row-echelon form of A is given by 0 0 0 x − 6y − 2z = 0 Set w = t and z = s. Then y = − 1 s − 3 nullspace(A) = − 5 21 t and 1 5 y + z + w = 0. 3 21 and x = − 10 t. Thus, 7 10 1 5 t, − s − t, s, t 7 3 21 : s, t ∈ R = t− 10 5 1 , − , 0, 1 + s 0, − , 1, 0 7 21 3 5 Hence, a basis for the nullspace of A is given by {(− 10 , − 21 , 0, 1), (0, − 1 , 1, 0)}. 7 3 . 322 1 0 43. A row-echelon form of A is given by 0 0 {(1, −2.5, −5), (0, 1, 1.3), (0, 0, 1)}, and a basis −4 0 6 −2 −2.5 −5 1 1.3 . Thus, a basis for the rowspace of A is given by 0 1 0 0 for the columnspace of A is given by 0 3 10 13 , , . 5 2 5 10 Moreover, we have that the nullspace of A is 0-dimensional, and so the basis is empty. 10 2 2 1 0 1 −1 −4 −3 . Thus, a basis for the rowspace of A is 44. A row-echelon form of A is given by 0 0 1 4 3 00 0 1 0 given by {(1, 0, 2, 2, 1), (0, 1, −1, −4, −3), (0, 0, 1, 4, 3), (0, 0, 0, 1, 0)}, a basis for the columnspace of A is given by 3 1 , 1 −2 5 5 0 2 , 1 1 −4 0 2 2 . , −2 −2 For the nullspace, if the variables corresponding the columns are (x, y, z, u, v ), then the row-echelon form tells us that v = t is a free variable, u = 0, z = −3t, y = 0, and x = 5t. Thus, nullspace(A) = {(5t, 0, −3t, 0, t) : t ∈ R} = {t(5, 0, −3, 0, 1) : t ∈ R}, and so a basis for the nullspace of A is {(5, 0, −3, 0, 1)}. 45. We will obtain bases for the rowspace, columnspace, and nullspace and orthonormalize them. A row 126 0 1 2 echelon form of A is given by 0 0 0 . We see that a basis for the rowspace of A is given by 000 {(1, 2, 6), (0, 1, 2)}. We apply Gram-Schmidt to this set, and thus we need to replace (0, 1, 2) by (0, 1, 2) − 14 (1, 2, 6) = 41 − 14 13 2 , ,− 41 41 41 So an orthogonal basis for the rowspace of A is given by (1, 2, 6) , − 14 13 2 , ,− 41 41 41 . . 323 To replace this with an orthonormal basis, we must normalize each vector. The first one has norm the second one has norm √3 . Hence, an orthonormal basis for the rowspace of A is 41 √ √ 41 41 (1, 2, 6), 41 3 − 14 13 2 , ,− 41 41 41 √ 41 and . Returning to the row-echelon form of A obtained above, we see that a basis for the columnspace of A is 2 1 21 , . 1 0 1 0 2 1 We apply Gram-Schmidt to this set, and thus we need to replace by 1 0 2 1 4 1 4 2 1 −1 − = 1 6 0 3 3 . 0 1 −2 So an orthogonal basis for the columnspace of A is given by 1 4 2 1 −1 , . 0 3 3 1 −2 √ √ The norms of these vectors are, respectively, 6 and 30/3. Hence, we normalize the above orthogonal basis to obtain the orthonormal basis for the columnspace: 1 4 √ √ 6 2 , 30 −1 . 6 0 30 3 1 −2 Returning once more to the row-echelon form of A obtained above, we see that, in order to find the nullspace of A, we must solve the equations x + 2y + 6z = 0 and y + 2z = 0. Setting z = t as a free variable, we find that y = −2t and x = −2t. Thus, a basis for the nullspace of A is {(−2, −2, 1)}, which can be normalized to 2 21 − ,− , . 3 33 13 5 0 1 3/2 46. A row-echelon form for A is 0 0 1 . Note that since rank(A) = 3, nullity(A) = 0, and so 0 0 0 00 0 there is no basis for nullspace(A). Moreover, rowspace(A) is a 3-dimensional subspace of R3 , and therefore, rowspace(A) = R3 . An orthonormal basis for this is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. 324 Finally, consider the columnspace of A. We must apply the Gram-Schmidt process to the three columns of A. Thus, we replace the second column vector by 3 1 −1 −3 −1 1 v2 = 2 − 4 0 = 2 . 5 1 1 5 1 1 Next, we replace the third column vector by 1 −1 5 3 1 −1 1 3 7 3 v3 = 3 − 0 − 2 = 0 2 2 1 1 −3 2 1 1 3 8 Hence, an orthogonal basis for the columnspace of A is 1 −1 −1 1 0 , 2 1 1 1 1 , Normalizing each vector yields the orthonormal basis −1 1 1 −1 1 1 0 , √ 2 2 8 1 1 1 1 3 3 0 −3 3 1 , 6 . . 3 3 0 −3 3 . 47. Let x1 = (5, −1, 2) and let x2 = (7, 1, 1). Using the Gram-Schmidt process, we have v1 = x1 = (5, −1, 2) and v2 = x2 − x2 , v1 36 6 12 11 7 v = (7, 1, 1) − (5, −1, 2) = (7, 1, 1) − (6, − , ) = (1, , − ). 21 | v1 | 30 55 5 5 Hence, an orthogonal basis is given by {(5, −1, 2), (1, 11 7 , − )}. 5 5 48. We already saw in Problem 26 that S spans R3 , so therefore an obvious orthogonal basis for span(S ) is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Alternatively, for practice with Gram-Schmidt, we would proceed as follows: Let x1 = (6, −3, 2), x2 = (1, 1, 1), and x3 = (1, −8, −1). Using the Gram-Schmidt process, we have v1 = x1 = (6, −3, 2), 325 x2 , v1 5 v = (1, 1, 1) − (6, −3, 2) = 21 | v1 | 49 v2 = x2 − 19 64 39 ,, 49 49 49 , and v3 = x3 − x3 , v2 4 38 x3 , v1 v1 − v2 = (1, −8, −1) − (6, −3, 2) + (19, 64, 39) = | v1 | 2 | v2 | 2 7 427 − 45 36 81 ,− , 61 61 61 . Hence, an orthogonal basis is given by 19 64 39 ,, 49 49 49 (6, −3, 2), ,− 45 36 81 ,− , 61 61 61 . 49. We already saw in Problem 29 that S spans P3 , so therefore we can apply Gram-Schmidt to the basis {1, x, x2 , x3 } for P3 , instead of the given set of polynomials. Let x1 = 1, x2 = x, x3 = x2 , and x4 = x3 . Using the Gram-Schmidt process, we have v1 = x1 = 1, v2 = x2 − v3 = x3 − x2 , v1 1 v =x− , 21 | v1 | 2 x3 , v1 x3 , v2 1 1 1 v1 − v2 = x2 − (x − ) − = x2 − x + , | v1 | 2 | v2 | 2 2 3 6 and v4 = x4 − x4 , v1 x4 , v2 x4 , v3 1 9 1 3 1 3 3 1 v− v− v = x3 − − (x − ) − (x2 − x + ) = x3 − x2 + x − . 21 22 23 | v1 | | v2 | | v3 | 4 10 2 2 6 2 5 20 Hence, an orthogonal basis is given by 1 1 3 3 1 1, x − , x2 − x + , x3 − x2 + x − 2 6 2 5 20 . 50. It is easy to see that the span of the set of vectors in Problem 32 is the set of all 2 × 2 symmetric matrices. Therefore, we can simply give the orthogonal basis 1 0 0 0 , 0 1 1 0 , 0 0 0 1 for the set of all 2 × 2 symmetric matrices. 51. We have u, v = (2, 3), (4, −1) = 2 · 4 + 3 · (−1) = 5, | u| = and so θ = cos−1 u, v | u| | v| = cos−1 √ 5 √ 13 17 √ 13, | v| = √ 17, ≈ 1.23 radians. 52. We have u, v = (−2, −1, 2, 4), (−3, 5, 1, 1) = (−2) · (−3) + (−1) · 5 + 2 · 1 + 4 · 1 = 7, | u| = 5, | v| = 6, and so θ = cos−1 u, v | u| | v| = cos−1 7 5·6 = cos−1 (7/30) ≈ 1.34 radians. 326 53. For Problem 51, we have u, v = (2, 3), (4, −1) = 2 · 2 · 4 + 3 · (−1) = 13, | u| = √ 17, | v| = √ 33, and so θ = cos−1 u, v | u| | v| = cos−1 √ 13 √ 17 33 ≈ 0.99 radians. For Problem 52, we have u, v = (−2, −1, 2, 4), (−3, 5, 1, 1) = 2 · (−2) · (−3) + (−1) · 5 + 2 · 1 + 4 · 1 = 13, | u| = √ 29, | v| = √ 45, and so θ = cos−1 u, v | u| | v| = cos−1 √ 13 √ 29 · 45 ≈ 1.20 radians. 54. (a): We must verify the four axioms for an inner product given in Definition 4.11.3. Axiom 1: We have p · p = p(t0 )p(t0 ) + p(t1 )p(t1 ) + · · · + p(tn )p(tn ) = p(t0 )2 + p(t1 )2 + · · · + p(tn )2 ≥ 0. Moreover, p(t0 )2 + p(t1 )2 + · · · + p(tn )2 = 0 ⇐⇒ p(t0 ) = p(t1 ) = · · · = p(tn ) = 0. But the only polynomial of degree ≤ n which has more than n roots is the zero polynomial. Thus, p · p = 0 ⇐⇒ p = 0. Axiom 2: We have p · q = p(t0 )q (t0 ) + p(t1 )q (t1 ) + · · · + p(tn )q (tn ) = q (t0 )p(t0 ) + q (t1 )p(t1 ) + · · · + q (tn )p(tn ) = q · p for all p, q ∈ Pn . Axiom 3: Let k be a scalar, and let p, q ∈ Pn . Then (kp) · q = (kp)(t0 )q (t0 ) + (kp)(t1 )q (t1 ) + · · · + (kp)(tn )q (tn ) = kp(t0 )q (t0 ) + kp(t1 )q (t1 ) + · · · + kp(tn )q (tn ) = k [p(t0 )q (t0 ) + p(t1 )q (t1 ) + · · · + p(tn )q (tn )] = k [p · q ], as required. Axiom 4: Let p1 , p2 , q ∈ Pn . Then we have (p1 + p2 ) · q = (p1 + p2 )(t0 )q (t0 ) + (p1 + p2 )(t1 )q (t1 ) + · · · + (p1 + p2 )(tn )q (tn ) = [p1 (t0 ) + p2 (t0 )]q (t0 ) + [p1 (t1 ) + p2 (t1 )]q (t1 ) + · · · + [p1 (tn ) + p2 (tn )]q (tn ) = [p1 (t0 )q (t0 ) + p1 (t1 )q (t1 ) + · · · + p1 (tn )q (tn )] + [p2 (t0 )q (t0 ) + p2 (t1 )q (t1 ) + · · · + p2 (tn )q (tn )] = (p1 · q ) + (p2 · q ), 327 as required. (b): The projection of p2 onto span{p0 , p1 } is given by p2 , p 0 p2 , p 1 20 0 p0 + p1 = p0 + p1 = 5. | p0 | 2 | p1 | 2 4 11 (c): We take q = t2 − 5. 55. Let x = (2, 3, 4) and let v = (6, −1, −4). We must find the length of x − P(x, v) = (2, 3, 4) − (2, 3, 4), (6, −1, −4) 7 (6, −1, −4) = (2, 3, 4) + (6, −1, −4) = | (6, −1, −4)| 53 148 152 184 , , 53 53 53 . 56. Note that x − y, vi = x, vi − y, vi = 0, by assumption. Let v be an arbitrary vector in V , and write v = a1 v1 + a2 v2 + · · · + an vn , for some scalars a1 , a2 , . . . , an . Observe that x − y, v = x − y, a1 v1 + a2 v2 + · · · + an vn = a1 x − y, v1 + a2 x − y, v2 + · · · + an x − y, vn = a1 · 0 + a2 · 0 + · · · + an · 0 = 0. Thus, x − y is orthogonal to every vector in V . In particular x − y, x − y = 0, and hence, x − y = 0. Therefore x = y. 57. Any of the conditions (a)-(p) appearing in the Invertible Matrix Theorem would be appropriate at this point in the text. Solutions to Section 5.1 True-False Review: ˙ 1. FALSE. The conditions T (u + v) = T (u) + T (v) and T (c · v) = ccT (v) must hold for all vectors u, v in V and for all scalars c, not just “for some”. 2. FALSE. The dimensions of the matrix A should be m × n, not n × m, as stated in the question. 3. FALSE. This will only necessarily hold for a linear transformation, not for more general mappings. 4. TRUE. This is precisely the definition of the matrix associated with a linear transformation, as given in the text. 5. TRUE. Since 0 = T (0) = T (v + (−v)) = T (v) + T (−v), we conclude that T (−v) = −T (v). 6. TRUE. Using the properties of a linear transformation, we have T ((c + d)v) = T (cv + dv) = T (cv) + T (dv) = cT (v) + dT (v), 328 as stated. Problems: 1. Let (x1 , x2 ), (y1 , y2 ) ∈ R2 and c ∈ R. T ((x1 , x2 ) + (y1 , y2 )) = T (x1 + y1 , x2 + y2 ) = (x1 + y1 + 2x2 + 2y2 , 2x1 + 2y1 − x2 − y2 ) = (x1 + 2x2 , 2x1 − x2 ) + (y1 + 2y2 , 2y1 − y2 ) = T (x1 , x2 ) + T (y1 , y2 ). T (c(x1 , x2 )) = T (cx1 , cx2 ) = (cx1 + 2cx2 , 2cx1 − cx2 ) = c(x1 + 2x2 , 2x1 − x2 ) = cT (x1 , x2 ). Thus, T is a linear transformation. 2. Let (x1 , x2 , x3 ), (y1 , y2 , y3 ) ∈ R3 and c ∈ R. T ((x1 , x2 , x3 ) + (y1 , y2 , y3 )) = T (x1 + y1 , x2 + y2 , x3 + y3 ) = (x1 + y1 + 3x2 + 3y2 + x3 + y3 , x1 + y1 − x2 − y2 ) = (x1 + 3x2 + x3 , x1 − x2 ) + (y1 + 3y2 + y3 , y1 − y2 ) = T (x1 , x2 , x3 ) + T (y1 , y2 , y3 ). T (c(x1 , x2 , x3 )) = T (cx1 , cx2 , cx3 ) = (cx1 +3cx2 + cx3 , cx1 − cx2 ) = c(x1 +3x2 + x3 , x1 − x2 ) = cT (x1 , x2 , x3 ). Thus, T is a linear transformation. 3. Let y1 , y2 ∈ C 2 (I ) and c ∈ R. Then, T (y1 + y2 ) = (y1 + y2 ) − 16(y1 + y2 ) = (y1 − 16y1 ) + (y2 − 16y2 ) = T (y1 ) + T (y2 ). T (cy1 ) = (cy1 ) − 16(cy1 ) = c(y1 − 16y1 ) = cT (y1 ). Consequently, T is a linear transformation. 4. Let y1 , y2 ∈ C 2 (I ) and c ∈ R. Then, T (y1 + y2 ) = (y1 + y2 ) + a1 (y1 + y2 ) + a2 (y1 + y2 ) = (y1 + a1 y1 + a2 y1 ) + (y2 + a1 y2 + a2 y2 ) = T (y1 ) + T (y2 ). T (cy1 ) = (cy1 ) + a1 (cy1 ) + a2 (cy1 ) = c(y1 + a1 y1 + a2 y1 ) = cT (y1 ). Consequently, T is a linear transformation. b 5. Let f, g ∈ V and c ∈ R. Then T (f + g ) = b (f + g )(x)dx = a b [f (x)+ g (x)]dx = a b f (x)dx + a T (f ) + T (g ). b T (cf ) = b [cf (x)]dx = c a f (x)dx = cT (f ). Therefore, T is a linear transformation. a 6. Let A1 , A2 , B ∈ Mn (R) and c ∈ R. T (A1 + A2 ) = (A1 + A2 )B − B (A1 + A2 ) = A1 B + A2 B − BA1 − BA2 = (A1 B − BA1 ) + (A2 B − BA2 ) = T (A1 ) + T (A2 ). T (cA1 ) = (cA1 B ) − B (cA1 ) = c(A1 B − BA1 ) = cT (A1 ). Consequently, T is a linear transformation. g (x)dx = a 329 7. Let A, B ∈ Mn (R) and c ∈ R. Then, S (A + B ) = (A + B ) + (A + B )T = A + AT + B + B T = S (A) + S (B ). S (cA) = (cA) + (cA)T = c(A + AT ) = cS (A). Consequently, S is a linear transformation. 8. Let A, B ∈ Mn (R) and c ∈ R. Then, n T (A + B ) = tr(A + B ) = k=1 n T (cA) = tr(cA) = n (akk + bkk ) = k=1 n cakk = c k=1 n akk + bkk = tr(A) + tr(B ) = T (A) + T (B ). k=1 akk = ctr(A) = cT (A). k=1 Consequently, T is a linear transformation. 9. Let x = (x1 , x2 ), y = (y1 , y2 ) be in R2 . Then T (x + y) = T (x1 + y1 , x2 + y2 ) = (x1 + x2 + y1 + y2 , 2), whereas T (x) + T (y) = (x1 + x2 , 2) + (y1 + y2 , 2) = (x1 + x2 + y1 + y2 , 4). We see that T (x + y) = T (x) + T (y), hence T is not a linear transformation. 10. Let A ∈ M2 (R) and c ∈ R. Then T (cA) = det(cA) = c2 det(A) = c2 T (A). Since T (cA) = cT (A) in general, it follows that T is not a linear transformation. 3 −2 . 1 5 1 3 12. If T (x1 , x2 ) = (x1 + 3x2 , 2x1 − 7x2 , x1 ), then A = [T (e1 ), T (e2 )] = 2 −7 . 1 0 11. If T (x1 , x2 ) = (3x1 − 2x2 , x1 + 5x2 ), then A = [T (e1 ), T (e2 )] = 13. If T (x1 , x2 , x3 ) = (x1 − x2 + x3 , x3 − x1 ), then A = [T (e1 ), T (e2 ), T (e3 )] = 14. If T (x1 , x2 , x3 ) = x1 + 5x2 − 3x3 , then A = [T (e1 ), T (e2 ), T (e3 )] = 1 1 −1 1 −1 01 5 −3 . −1 −1 15. If T (x1 , x2 , x3 ) = (x3 − x1 , −x1 , 3x1 + 2x3 , 0), then A = [T (e1 ), T (e2 ), T (e3 )] = 3 0 16. T (x) = Ax = 1 −4 3 7 x1 x2 = x1 + 3x2 −4x1 + 7x2 , which we write as T (x1 , x2 ) = (x1 + 3x2 , −4x1 + 7x2 ). 17. T (x) = Ax = 2 −1 5 3 1 −2 x1 x2 = x3 2x1 − x2 + 5x3 3x1 + x2 − 2x3 , which we write as T (x1 , x2 , x3 ) = (2x1 − x2 + 5x3 , 3x1 + x2 − 2x3 ). 0 0 0 0 . 1 0 . 2 0 330 2x1 + 2x2 − 3x3 2 2 −3 x1 2 x2 = 4x1 − x2 + 2x3 , which we write as 18. T (x) = Ax = 4 −1 5 7 −8 x3 5x1 + 7x2 − 8x3 T (x1 , x2 , x3 ) = (2x1 + 2x2 − 3x3 , 4x1 − x2 + 2x3 , 5x1 + 7x2 − 8x3 ). −3 −3x −2 −2x , which we write as 19. T (x) = Ax = 0 [x] = 0 1 x T (x) = (−3x, −2x, 0, x). 20. T (x) = Ax = 1 −4 −6 0 2 x1 x2 x3 x4 x5 = [x1 − 4x2 − 6x3 + 2x5 ], which we write as T (x1 , x2 , x3 , x4 , x5 ) = x1 − 4x2 − 6x3 + 2x5 . 21. Let u be a fixed vector in V , v1 , v2 ∈ V , and c ∈ R. Then T (v1 + v2 ) = u, v1 + v2 = u, v1 + u, v2 = T (v1 ) + T (v2 ). T (cv1 ) = u, cv1 = c u, v1 = cT (v1 ). Thus, T is a linear transformation. 22. We must show that the linear transformation T respects addition and scalar multiplication: T respects addition: Let v1 and v2 be vectors in V . Then we have T (v1 + v2 ) = ( u1 , v1 + v2 , u2 , v1 + v2 ) = ( v1 + v2 , u1 , v1 + v2 , u2 ) = ( v1 , u1 + v2 , u1 , v1 , u2 + v2 , u2 ) = ( v1 , u1 , v1 , u2 ) + ( v2 , u1 , v2 , u2 ) = ( u1 , v1 , u2 , v1 ) + ( u1 , v2 , u2 , v2 ) = T (v1 ) + T (v2 ). T respects scalar multiplication: Let v be a vector in V and let c be a scalar. Then we have T (cv) = ( u1 , cv = ( cv, u1 = (c v, u1 = c( v, u1 = c( u1 , v = cT (v). , u2 , cv ) , cv, u2 ) , c v, u2 ) , v, u2 ) , u2 , v ) 1 1 then det(D) = −2 = 0, so by Corollary 4.5.15 the vectors v1 = (1, 1) 1 −1 and v2 = (1, −1) are linearly independent. Since dim[R2 ] = 2, it follows from Theorem 4.6.10 that {v1 , v2 } is a basis for R2 . 23. (a) If D = [v1 , v2 ] = 331 (b) Let x = (x1 , x2 ) be an arbitrary vector in R2 . Since {v1 , v2 } forms a basis for R2 , there exist c1 and c2 such that (x1 , x2 ) = c1 (1, 1) + c2 (1, −1), that is, such that c1 + c2 = x1 , c1 − c2 = x2 . Solving this system yields c1 = 1 1 (x1 + x2 ), c2 = (x1 − x2 ). Thus, 2 2 (x1 , x2 ) = so that 1 1 (x1 + x2 )v1 + (x1 − x2 )v2 , 2 2 1 1 (x1 + x2 )v1 + (x1 − x2 )v2 2 2 1 1 (x1 + x2 )T (v1 ) + (x1 − x2 )T (v2 ) 2 2 1 1 (x1 + x2 )(2, 3) + (x1 − x2 )(−1, 1) 2 2 x1 3x2 + , 2x1 + x2 . 2 2 when (4, −2) is substituted for (x1 , x2 ), it follows that T (4, −2) = (−1, 6). T [(x1 , x2 )] = T = = = In particular, 24. The matrix of T is the 4 × 2 matrix [T (e1 ), T (e2 )]. Therefore, we must determine T (1, 0) and T (0, 1), which we can determine from the given information by using the linear transformation properties. A quick calculation shows that (1, 0) = − 2 (−1, 1) + 1 (1, 2), so 3 3 1 2 1 2 1 515 2 T (1, 0) = T (− (−1, 1)+ (1, 2)) = − T (−1, 1)+ T (1, 2) = − (1, 0, −2, 2)+ (−3, 1, 1, 1) = (− , , , −1). 3 3 3 3 3 3 333 Similarly, we have (0, 1) = 1 (−1, 1) + 1 (1, 2), so 3 3 1 1 1 1 1 1 21 1 T (0, 1) = T ( (−1, 1) + (1, 2)) = T (−1, 1) + T (1, 2) = (1, 0, −2, 2) + (−3, 1, 1, 1) = (− , , − , 1). 3 3 3 3 3 3 33 3 Therefore, we have the matrix of T : −5/3 −2/3 1/3 1/2 5/3 −1/3 . −1 1 25. The matrix of T is the 2 × 4 matrix [T (e1 ), T (e2 ), T (e3 ), T (e4 )]. Therefore, we must determine T (1, 0, 0, 0), T (0, 1, 0, 0), T (0, 0, 1, 0), and T (0, 0, 0, 1), which we can determine from the given information by using the linear transformation properties. We are given that T (1, 0, 0, 0) = (3, −2). Next, T (0, 1, 0, 0) = T (1, 1, 0, 0) − T (1, 0, 0, 0) = (5, 1) − (3, −2) = (2, 3), T (0, 0, 1, 0) = T (1, 1, 1, 0) − T (1, 1, 0, 0) = (−1, 0) − (5, 1) = (−6, −1), 332 and T (0, 0, 0, 1) = T (1, 1, 1, 1) − T (1, 1, 1, 0) = (2, 2) − (−1, 0) = (3, 2). Therefore, we have the matrix of T : 3 2 −6 −2 3 −1 3 2 . 26. The matrix of T is the 3 × 3 matrix [T (e1 ), T (e2 ), T (e3 )]. Therefore, we must determine T (1, 0, 0), T (0, 1, 0), and T (0, 0, 1), which we can determine from the given information by using the linear transformation properties. A quick calculation shows that (1, 0, 0) = (1, 2, 0) − 6(0, 1, 1) + 2(0, 2, 3), so T (1, 0, 0) = T (1, 2, 0) − 6T (0, 1, 1) + 2T (0, 2, 3) = (2, −1, 1) − 6(3, −1, −1) + 2(6, −5, 4) = (32, −5, 4). Similarly, (0, 1, 0) = 3(0, 1, 1) − (0, 2, 3), so T (0, 1, 0) = 3T (0, 1, 1) − T (0, 2, 3) = 3(3, −1, −1) − (6, −5, 4) = (3, 2, −7). Finally, (0, 0, 1) = −2(0, 1, 1) + (0, 2, 3), so T (0, 0, 1) = −2T (0, 1, 1) + T (0, 2, 3) = −2(3, −1, −1) + (6, −5, 4) = (0, −3, 6). Therefore, we have the matrix of T : 32 3 0 −5 2 −3 . 4 −7 6 27. The matrix of T is the 4 × 3 matrix [T (e1 ), T (e2 ), T (e3 )]. Therefore, we must determine T (1, 0, 0), T (0, 1, 0), and T (0, 0, 1), which we can determine from the given information by using the linear transformation properties. A quick calculation shows that (1, 0, 0) = 1 (0, −1, 4) − 1 (0, 3, 3) + 1 (4, 4, −1), so 4 4 4 T (1, 0, 0) = 1 1 1 1 1 1 3 1 T (0, −1, 4)− T (0, 3, 3)+ T (4, 4, −1) = (2, 5, −2, 1)− (−1, 0, 0, 5)+ (−3, 1, 1, 3) = (0, , 0, − ). 4 4 4 4 4 4 2 4 1 Similarly, (0, 1, 0) = − 5 (0, −1, 4) + 4 15 (0, 3, 3), so 1 4 2 2 17 T (0, 1, 0) = − (2, 5, −2, 1) + (−1, 0, 0, 5) = (− , −1, , ). 5 15 3 5 15 Finally, (0, 0, 1) = 1 (0, −1, 4) + 5 T (0, 0, 1) = 1 15 (0, 3, 3), so 1 1 1 1 1 28 T (0, −1, 4) + T (0, 3, 3) = (2, 5, −2, 1) + (−1, 0, 0, 5) = ( , 1, − , ). 5 15 5 15 3 5 15 Therefore, we have the matrix of T : 0 −2/3 1/3 3/2 −1 1 . 0 2/5 −2/5 −1/4 17/15 8/15 28. T (ax2 + bx + c) = aT (x2 )+ bT (x)+ cT (1) = a(3x +2)+ b(x2 − 1)+ c(x +1) = bx2 +(3a + c)x +(2a − b + c). 333 29. Using the linearity of T , we have T (2v1 + 3v2 ) = v1 + v2 and T (v1 + v2 ) = 3v1 − v2 . That is, 2T (v1 ) + 3T (v2 ) = v1 + v2 and T (v1 ) + T (v2 ) = 3v1 − v2 . Solving this system for the unknowns T (v1 ) and T (v2 ), we obtain T (v2 ) = 3v2 − 5v1 and T (v1 ) = 8v1 − 4v2 . 30. Since T is a linear transformation we obtain: T (x2 ) − T (1) = x2 + x − 3, 2T (x) = 4x, (30.1) (30.2) 3T (x) + 2T (1) = 2(x + 3) = 2x + 6. (30.3) From Equation (30.2) it follows that T (x) = 2x, so upon substitution into Equation (30.3) we have 3(2x) + 2T (1) = 2(x + 3) or T (1) = −2x + 3. Substituting this last result into Equation (30.1) yields T (x2 ) − (−2x + 3) = x2 + x − 3 so T (x2 ) = x2 − x. Now if a, b and c are arbitrary real numbers, then T (ax2 + bx + c) = aT (x2 ) + bT (x) + cT (1) = a(x2 − x) + b(2x) + c(−2x + 3) = ax2 − ax + 2bx − 2cx + 3c = ax2 + (−a + 2b − 2c)x + 3c. 31. Let v ∈ V . Since {v1 , v2 } is a basis for V , there exists a, b ∈ R such that v = av1 + bv2 . Hence T (v) = T (av1 + bv2 ) = aT (v1 ) + bT (v2 ) = a(3v1 − v2 ) + b(v1 + 2v2 ) = 3av1 − av2 + bv1 + 2bv2 = (3a + b)v1 + (2b − a)v2 . 32. Let v be any vector in V . Since {v1 , v2 , . . . , vk } spans V , we can write v = c1 v1 + c2 v2 + · · · + ck vk for suitable scalars c1 , c2 , . . . , ck . Then T (v) = T (c1 v1 + c2 v2 + · · · + ck vk ) = c1 T (v1 ) + c2 T (v2 ) + · · · + ck T (vk ) = c1 S (v1 ) + c2 S (v2 ) + · · · + ck S (vk ) = S (c1 v1 + c2 v2 + · · · + ck vk ) = S (v), as required. 33. Let v be any vector in V . Since {v1 , v2 , . . . , vk } is a basis for V , we can write v = c1 v1 + c2 v2 + · · · + ck vk for suitable scalars c1 , c2 , . . . , ck . Then T (v) = T (c1 v1 + c2 v2 + · · · + ck vk ) = c1 T (v1 ) + c2 T (v2 ) + · · · + ck T (vk ) = c1 0 + c2 0 + . . . ck 0 = 0, as required. 34. Let v1 and v2 be arbitrary vectors in V . Then, (T1 + T2 )(v1 + v2 ) = T1 (v1 + v2 ) + T2 (v1 + v2 ) = T1 (v1 ) + T1 (v2 ) + T2 (v1 ) + T2 (v2 ) = T1 (v1 ) + T2 (v1 ) + T1 (v2 ) + T2 (v2 ) = (T1 + T2 )(v1 ) + (T1 + T2 )(v2 ). Further, if k is any scalar, then (T1 + T2 )(k v) = T1 (k v) + T2 (k v) = kT1 (v) + kT2 (v) = k [T1 (v) + T2 (v)] = k (T1 + T2 )(v). It follows that T1 + T2 is a linear transformation. Now consider the transformation cT , where c is an arbitrary scalar. (cT )(v1 + v2 ) = cT (v1 + v2 ) = c[T (v1 ) + T (v2 )] = cT (v1 ) + cT (v2 ) = (cT )(v1 ) + (cT )(v2 ). 334 (cT )(k v1 ) = cT (k v1 ) = c[kT (v1 )] = (ck )T (v1 ) = (kc)T (v1 ) = k [cT (v1 )]. Thus, cT is a linear transformation. 35. (T1 + T2 )(x) = T1 (x) + T2 (x) = Ax + B x = (A + B )x = 5 6 2 −2 x1 x2 = 5x1 + 6x2 2x1 − 2x2 . Hence, (T1 + T2 )(x1 , x2 ) = (5x1 + 6x2 , 2x1 − 2x2 ). (cT1 )(x) = cT1 (x) = c(Ax) = (cA)x = 3c c −c 2c x1 x2 = 3cx1 + cx2 −cx1 + 2cx2 . Hence, (cT1 )(x1 , x2 ) = (3cx1 + cx2 , −cx1 + 2cx2 ). 36. (T1 + T2 )(x) = T1 (x) + T2 (x) = Ax + B x = (A + B )x. (cT1 )(x) = cT1 (x) = c(Ax) = (cA)x. 37. Problem 34 establishes that if T1 and T2 are in L(V, W ) and c is any scalar, then T1 + T2 and cT1 are in L(V, W ). Consequently, Axioms (A1) and (A2) are satisfied. A3: Let v be any vector in L(V, W ). Then (T1 + T2 )(v) = T1 (v) + T2 (v) = T2 (v) + T1 (v) = (T2 + T1 )(v). Hence T1 + T2 = T2 + T1 , therefore the addition operation is commutative. A4: Let T3 ∈ L(V, W ). Then [(T1 + T2 ) + T3 ] (v) = (T1 + T2 )(v) + T3 (v) = [T1 (v) + T2 (v)] + T3 (v) = T1 (v) + [T2 (v) + T3 (v)] = T1 (v) + (T2 + T3 )(v) = [T1 + (T2 + T3 )](v). Hence (T1 + T2 ) + T3 = T1 + (T2 + T3 ), therefore the addition operation is associative. A5: The zero vector in L(V, W ) is the zero transformation, O : V → W , defined by O(v) = 0, for all v in V, where 0 denotes the zero vector in V . To show that O is indeed the zero vector in L(V, W ), let T be any transformation in L(V, W ). Then (T + O)(v) = T (v) + O(v) = T (v) + 0 = T (v) for all v ∈ V, so that T + O = T . A6: The additive inverse of the transformation T ∈ L(V, W ) is the linear transformation −T defined by −T = (−1)T , since [T + (−T )](v) = T (v) + (−T )(v) = T (v) + (−1)T (v) = T (v) − T (v) = 0, for all v ∈ V , so that T + (−T ) = O. A7-A10 are all straightforward verifications. Solutions to Section 5.2 True-False Review: 1. FALSE. For example, T (x1 , x2 ) = (0, 0) is a linear transformation that maps every line to the origin. 2. TRUE. All of the matrices Rx , Ry , Rxy , LSx , LSy , Sx , and Sy discussed in this section are elementary matrices. 335 3. FALSE. A shear parallel to the x-axis composed with a shear parallel to the y -axis is given by matrix 1k 10 1 + kl k = , which is not a shear. 01 l1 l 1 4. TRUE. This is explained prior to Example 5.2.1. 5. FALSE. For example, 01 10 Rxy · Rx = 1 0 0 −1 0 l k 0 0 −1 1 0 = , and this matrix is not in the form of a stretch. 6. FALSE. For example, k 0 0 1 1 0 = 0 l is not a stretch. Problems: 1. T (1, 1) = (1, −1), T (2, 1) = (1, −2), T (2, 2) = (2, −2), T (1, 2) = (2, −1). y 2 1 x 1 2 -1 -2 Figure 0.0.69: Figure for Exercise 1 2. T (1, 1) = (0, 3), T (2, 1) = (1, 4), T (2, 2) = (0, 6), T (1, 2) = (−1, 5). 3. T (1, 1) = (2, 0), T (2, 1) = (3, −1), T (2, 2) = (4, 0), T (1, 2) = (3, 1). 4. T (1, 1) = (−4, −2), T (2, 1) = (−6, −4), T (2, 2) = (−8, −4), T (1, 2) = (−6, −2). 1 0 5. A = 6. 0 2 2 0 1. P12 2 1 1 ∼ =⇒ T (x) = Ax corresponds to a shear parallel to the y -axis. 2 0 0 0 2. M1 (1/2) 2 ∼ 1 0 0 2 3 ∼ 3. M2 (1/2). 1 0 0 1 So, 336 y 6 5 4 3 2 1 x -2 -1 1 2 Figure 0.0.70: Figure for Exercise 2 y 2 1 x 1 2 3 4 -1 Figure 0.0.71: Figure for Exercise 3 T (x) = Ax = P12 M1 (2)M2 (2)x which corresponds to a stretch in the y -direction, followed by a stretch in the x-direction, followed by a reflection in y = x. 7. A = 8. 1 3 −1 0 0 −1 0 1 =⇒ T (x) = Ax corresponds to a shear parallel to the y -axis. 1 ∼ 1 0 0 −1 2 ∼ 1 0 0 1 1. M1 (−1) 2. M2 (−1) So, 337 y 2 1 x -8 -6 -4 -2 1 2 -2 -4 Figure 0.0.72: Figure for Exercise 4 T (x) = Ax = M1 (−1)M2 (−1)x which corresponds to a reflection in the x-axis, followed by a reflection in the y -axis. 9. 1 −3 −2 8 1 −3 0 2 1 ∼ 2 ∼ 1 −3 0 1 1 0 3 ∼ 0 1 1. A12 (2) 2. M2 (1/2) 3. A21 (3). So, T (x) = Ax = A12 (−2)M2 (2)A21 (−3)x which corresponds to a shear parallel to the x-axis, followed by a stretch in the y -direction, followed by a shear parallel to the y -axis. 10. 1 3 2 4 1 ∼ 1 2 0 −2 1 0 0 −2 2 ∼ 3 ∼ 10 01 1. A12 (−3) 2. A21 (1) 3. M2 (−1/2). So, T (x) = Ax = A12 (3)A21 (−1)M2 (−2)x which corresponds to a reflection in the x-axis followed by a stretch in they y -direction, followed by a shear parallel to the x-axis, followed by a shear parallel to the y -axis. 1 0 10 = 0 −2 02 a stretch in the y -direction. 11. 12. −1 −1 −1 0 1 ∼ 1 0 0 −1 −1 −1 0 1 2 ∼ . So T (x) = Ax corresponds to a reflection in the x-axis followed by 1 0 1 1 3 ∼ 1 0 0 1 1. A12 (−1) 2. M1 (−1) 3. A21 (−1). So, T (x) = Ax = A12 (1)M1 (−1)A21 (1)x which corresponds to a shear parallel to the x-axis, followed by a reflection in the y -axis, followed by a shear parallel to the y -axis. 13. 338 R(θ) = cos θ 0 0 1 1 sin θ 0 1 1 0 = cos θ 0 0 1 1 sin θ 0 1 1 − tan θ 0 1 0 sec θ 1 − tan θ 0 sec θ cos θ 0 = 0 1 − tan θ cos θ 1 sin θ cos θ − sin θ sin θ cos θ which coincides with the matrix of the transformation of R2 corresponding to a rotation through an angle θ in the counter-clockwise direction. π 0 −1 14. The matrix for a counter-clockwise rotation through an angle θ = is . Now, 1 0 2 = 0 −1 1 0 1 ∼ 1 0 0 −1 2 ∼ 1 0 0 1 1. P12 2. M2 (−1) So, T (x) = Ax = P12 M2 (−1)x which corresponds to a reflection in the x-axis followed by a reflection in y = x. Solutions to Section 5.3 True-False Review: 1. FALSE. The statement should read dim[Ker(T )] + dim[Rng(T )] = dim[V ], not dim[W ] on the right-hand side. 2. FALSE. As a specific illustration, we could take T : P4 → R7 defined by T (a0 + a1 x + a2 x2 + a3 x3 + a4 x4 ) = (a0 , a1 , a2 , a3 , a4 , 0, 0), and it is easy to see that T is a linear transformation with Ker(T ) = {0}. Therefore, Ker(T ) is 0-dimensional. 3. FALSE. The solution set to the homogeneous linear system Ax = 0 is Ker(T ), not Rng(T ). 4. FALSE. Rng(T ) is a subspace of W , not V , since it consists of vectors of the form T (v), and these belong to W . 5. TRUE. From the given information, we see that Ker(T ) is at least 2-dimensional, and therefore, since M23 is 6-dimensional, the Rank-Nullity Theorem requires that Rng(T ) have dimension at most 6 − 2 = 4. 6. TRUE. Any vector of the form T (v) where v belongs to Rn can be written as Av, and this in turn can be expressed as a linear combination of the columns of A. Therefore, T (v) belongs to colspace(A). Problems: 1. T (7, 5, −1) = T (−21, −15, 2) = 7 5 = 0 =⇒ (7, 5, −1) ∈ Ker(T ). 0 −1 −21 1 −1 2 −2 −15 = =⇒ (−21, −15, 2) ∈ Ker(T ). / 1 −2 −3 3 2 1 −1 2 1 −2 −3 339 T (35, 25, −5) = 1 −1 2 1 −2 −3 35 25 = −5 0 0 =⇒ (35, 25, −5) ∈ Ker(T ). 2. Ker(T ) = {x ∈ R2 : T (x) = 0} = {x ∈ R2 : Ax = 0}. The augmented matrix of the system Ax = 0 is: 360 120 . It follows that , with reduced row-echelon form of 120 000 Ker(T ) = {x ∈ R2 : x = (−2t, t), t ∈ R} = {x ∈ R2 : x = t(−2, 1), t ∈ R}. Geometrically, this is a line in R2 . It is the subspace of R2 spanned by the vector (−2, 1). dim[Ker(T )] = 1. For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A, we see that colspace(A) is generated by the first column vector of A. Consequently, Rng(T ) = {y ∈ R2 : y = r(3, 1), r ∈ R}. Geometrically, this is a line in R2 . It is the subspace of R2 spanned by the vector (3, 1). dim[Rng(T )] = 1. Since dim[Ker(T )]+ dim[Rng(T )] = 2 = dim[R2 ], Theorem 5.3.8 is satisfied. 3. Ker(T ) = {x ∈ R3 : T (x) = 0} = {x ∈ R3 : Ax = 0}. The augmented matrix of the system Ax = 0 1 −1 0 0 1000 1 2 0 , with reduced row-echelon form 0 1 0 0 . Thus x1 = x2 = x3 = 0, so is: 0 0010 2 −1 1 0 Ker(T ) = {0}. Geometrically, this describes a point (the origin) in R3 . dim[Ker(T )] = 0. For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A, we see that colspace(A) is generated by the first three column vectors of A. Consequently, Rng(T ) = R3 , dim[Rng(T )] = dim[R3 ] = 3, and Theorem 5.3.8 is satisfied since dim[Ker(T )]+ dim[Rng(T )] = 0 + 3 = dim[R3 ]. 3 4. R3 Ker(T ) = {x ∈ : T (x) = 0} = {x ∈ R : Ax = 0}. 1 −2 10 2 −3 −1 0 , with reduced row-echelon form of 5 −8 −1 0 The augmented matrix of the system Ax = 0 is: 1 0 −5 0 0 1 −3 0 . Thus 00 0 0 Ker(T ) = {x ∈ R3 : x = t(5, 3, 1), t ∈ R}. Geometrically, this describes the line in R3 through the origin, spanned by (5, 3, 1). dim[Ker(T )] = 1. For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A, we see that a basis for colspace(A) is given by the first two column vectors of A. Consequently, Rng(T ) = {y ∈ R3 : y = r(1, 2, 5) + s(−2, −3, −8), r, s ∈ R}. Geometrically, this is a plane through the origin in R3 . dim[Rng(T )] = 2 and Theorem 5.3.8 is satisfied since dim[Ker(T )]+ dim[Rng(T )] = 1 + 2 = 3 = dim[R3 ]. 5. Ker(T ) = {x ∈ R3 : T (x) = 0} = {x ∈ R3 : Ax = 0}. The augmented matrix of the system Ax = 0 is: 1 −1 2 0 1 −1 20 , with reduced row-echelon form of . Thus −3 3 −6 0 0 000 Ker(T ) = {x ∈ R3 : x = r(1, 1, 0) + s(−2, 0, 1), r, s ∈ R}. 340 Geometrically, this describes the plane through the origin in R3 , which is spanned by the linearly independent set {(1, 1, 0), (−2, 0, 1)}. dim[Ker(T )] = 2. For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A, we see that a basis for colspace(A) is given by the first column vector of A. Consequently, Rng(T ) = {y ∈ R2 : y = t(1, −3), t ∈ R}. Geometrically, this is the line through the origin in R2 spanned by (1, −3). dim[Rng(T )] = 1 and Theorem 5.3.8 is satisfied since dim[Ker(T )]+ dim[Rng(T )] = 2 + 1 = 3 = dim[R3 ]. 6. Ker(T ) = {x ∈ R3 : T (x) = 0} = {x ∈ R3 : Ax = 0}. The augmented matrix of the system Ax = 0 is: 1320 1300 , with reduced row-echelon form of . Thus 2650 0010 Ker(T ) = {x ∈ R3 : x = r(−3, 1, 0), r ∈ R}. Geometrically, this describes the line through the origin in R3 , which is spanned by (−3, 1, 0). dim[Ker(T )] = 1. For the given transformation, Rng(T ) = colspace(A). From the preceding reduced row-echelon form of A, we see that a basis for colspace(A) is given by the first and third column vectors of A. Consequently, Rng(T ) = span{(1, 2), (2, 5)} = R2 , so that dim[Rng(T )] = 2. Geometrically, Rng(T ) is the xy -plane, and Theorem 5.3.8 is satisfied since dim[Ker(T )]+ dim[Rng(T )] = 1 + 2 = 3 = dim[R3 ]. −5/3 −2/3 1/3 1/2 7. The matrix of T in Problem 24 of Section 5.1 is A = 5/3 −1/3 . Thus, −1 1 Ker(T ) = nullspace(A) = {0} and −2/3 −5/3 1/3 , 1/2 . Rng(T ) = colspace(A) = span 5/3 −1/3 −1 1 8. The matrix of T in Problem 25 of Section 5.1 is A = 3 −2 2 −6 3 −1 3 2 . Thus, Ker(T ) = nullspace(A) = span{(16/13, 15/13, 1, 0), (−5/13, −12/13, 0, 1)} and Rng(T ) = colspace(A) = R2 . 32 3 0 2 −3 . Thus, 9. The matrix of T in Problem 26 of Section 5.1 is A = −5 4 −7 6 Ker(T ) = nullspace(A) = {0} and Rng(T ) = colspace(A) = R3 . 341 0 −2/3 1/3 3/2 −1 1 . Thus, 10. The matrix of T in Problem 27 of Section 5.1 is A = 0 2/5 −2/5 −1/4 17/15 8/15 Ker(T ) = nullspace(A) = {0} and Rng(T ) = colspace(A) = span 0 −2/3 3/2 −1 , 0 2/5 −1/4 17/15 1/3 1 , . −2/5 8/15 11. (a) Ker(T ) = {v ∈ R3 : u, v = 0}. For v to be in the kernel of T , u and v must be orthogonal. Since u is any fixed vector in R3 , then v must lie in the plane orthogonal to u. Hence dim[Ker(T )] = 2. (b) Rng(T ) = {y ∈ R : y = u, v , v ∈ R3 }, and dim[Rng(T )] = 1. 12. (a) Ker(S ) = {A ∈ Mn (R) : A − AT = 0} = {A ∈ Mn (R) : A = AT }. Hence, any matrix in Ker(S ) is symmetric by definition. (b) Since any matrix in Ker(S ) is symmetric, it has been shown that {A1 , A2 , A3 } is a spanning set for the set of all symmetric matrices in M2 (R), where A1 = 1 0 0 0 , A2 = 0 1 1 0 , A3 = 0 0 0 1 . Thus, dim[Ker(S )] = 3. 13. Ker(T ) = {A ∈ Mn (R) : AB − BA = 0} = {A ∈ M2 (R) : AB = BA}. This is the set of matrices that commute with B . 14. (a) Ker(T ) = {p ∈ P2 : T (p) = 0} = {ax2 + bx + c ∈ P2 : ax2 + (a + 2b + c)x + (3a − 2b − c) = 0, for all x}. Thus, for p(x) = ax2 + bx + c to be in Ker(T ), a, b, and c must satisfy the system: a =0 a + 2b + c = 0 3a − 2b − c = 0 Solving this system, we obtain that a = 0 and c = −2b. Consequently, all polynomials of the form 0x2 + bx + (−2b) are in Ker(T ), so Ker(T ) = {b(x − 2) : b ∈ R}. Since Ker(T ) is spanned by the nonzero vector x − 2, it follows that dim[Rng(T )] = 1. (b) In this case, Rng(T ) = {T (ax2 + bx + c) : a, b, c ∈ R} = {ax2 + (a + 2b + c)x + (3a − 2b − c) : a, b, c ∈ R} = {a(x2 + x + 3) + b(2x − 2) + c(x − 1) : a, b, c ∈ R} = {a(x2 + x + 3) + (2b + c)(x − 1) : a, b, c ∈ R} = span{x2 + x + 3, x − 1}. 342 Since the vectors in this spanning set are linearly independent on any interval, it follows that the spanning set is a basis for Rng(T ). Hence, dim[Rng(T )] = 2. 15. Ker(T ) = {p ∈ P2 : T (p) = 0} = {ax2 + bx + c ∈ P2 : (a + b) + (b − c)x = 0, for all x}. Thus, a, b, and c must satisfy: a + b = 0 and b − c = 0 =⇒ a = −b and b = c. Letting c = r ∈ R, we have ax2 + bx + c = r(−x2 + x + 1). Thus, Ker(T ) = {r(−x2 + x + 1) : r ∈ R} and dim[Ker(T )] = 1. Rng(T ) = {T (ax2 + bx + c) : a, b, c ∈ R} = {(a + b) + (b − c)x : a, b, c ∈ R} = {c1 + c2 x : c1 , c2 ∈ R}. Consequently, a basis for Rng(T ) is {1, x}, so that Rng(T ) = P1 , and dim[Rng(T )] = 2. 16. Ker(T ) = {p ∈ P1 : T (p) = 0} = {ax + b ∈ P1 : (b − a) + (2b − 3a)x + bx2 = 0, for all x}. Thus, a and b must satisfy: b−a = 0, 2b−3a = 0, and b = 0 =⇒ a = b = 0. Thus, Ker(T ) = {0} and dim[Ker(T )] = 0. Rng(T ) = {T (ax + b) : a, b ∈ R} = {(b − a) + (2b − 3a)x + bx2 : a, b ∈ R} = {−a(1 + 3x) + b(1 + 2x + x2 ) : a, b ∈ R} = span{1 + 3x, 1 + 2x + x2 }. Since the vectors in this spanning set are linearly independent on any interval, it follows that the spanning set is a basis for Rng(T ), and dim[Rng(T )] = 2. 17. T (v) = 0 ⇐⇒ T (av1 + bv2 + cv3 ) = 0 ⇐⇒ aT (v1 ) + bT (v2 ) + cT (v3 ) = 0 ⇐⇒ a(2w1 − w2 ) + b(w1 − w2 ) + c(w1 + 2w2 ) = 0 ⇐⇒ (2a + b + c)w1 + (−a − b + 2c)w2 = 0 ⇐⇒ 2a + b + c = 0 and a − b + 2c = 0. Reducing the augmented matrix of the system yields: 2 110 1 −1 2 0 ∼ 1 −1 2 0 2 110 ∼ 1 0 0 10 1 −1 0 . Setting c = r =⇒ b = r, a = −r. Thus, Ker(T ) = {v ∈ V : v = r(−v1 + v2 + v3 ), r ∈ R} and dim[Ker(T )] = 1. Rng(T ) = {T (v) : v ∈ V } = {(2a + b + c)w1 + (−a − b + 2c)w2 : a, b, c ∈ V } = span{w1 , w2 } = W . Consequently, dim[Rng(T )] = 2. 18. (a) If w ∈ Rng(T ), then T (v) = w for some v ∈ V , and since {v1 , v2 , . . . , vn } is a basis for V , there exist c1 , c2 , . . . , cn ∈ R for which v = c1 v1 + c2 v2 + · · · + cn vn . Accordingly, w = T (v) = a1 T (v1 ) + a2 T (v2 ) + · · · + an T (vn ). Thus, Rng(T ) = span{T (v1 ), T (v2 ), . . . , T (vn )}. We must show that {T (v1 ), T (v2 ), . . . , T (vn )} is also linearly independent. Suppose that b1 T (v1 ) + b2 T (v2 ) + · · · + bn T (vn ) = 0. Then T (b1 v1 + b2 v2 + · · · + bn vn ) = 0, so since Ker(T ) = {0}, b1 v1 + b2 v2 + · · · + bn vn = 0. Since {v1 , v2 , . . . , vn } is a linearly independent set, the preceding equation implies that b1 = b2 = · · · = bn = 0. 343 Consequently, {T (v1 ), T (v2 ), . . . , T (vn )} is a linearly independent set in W . Therefore, since we have already shown that it is a spanning set for Rng(T ), {T (v1 ), T (v2 ), . . . , T (vn )} is a basis for Rng(T ). (b) As an example, let T : R3 → R2 be defined by T ((a, b, c)) = (a, b) for all (a, b, c) in R3 . Then for the basis {e1 , e2 , e3 } of R3 , we have {T (e1 ), T (e2 ), T (e3 )} = {(1, 0), (0, 1), (0, 0)}, which is clearly not a basis for R2 . Solutions to Section 5.4 True-False Review: 1. FALSE. Many one-to-one linear transformations T : P3 → M32 can be constructed. One possible example would be to define a0 a1 T (a0 + a1 x + a2 x2 + a3 x3 ) = a2 a3 . 00 It is easy to check that with T so defined, T is a one-to-one linear transformation. 2. TRUE. We can define an isomorphism T a T 0 0 : V → M32 via bc ab d e = c d . 0f ef With T so defined, it is easy to check that T is an isomorphism. 3. TRUE. Both Ker(T1 ) and Ker(T2 T1 ) are subspaces of V1 , and since if T1 (v1 ) = 0, then (T2 T1 )v1 = T2 (T1 (v1 )) = T2 (0) = 0, we see that every vector in Ker(T1 ) belongs to Ker(T2 T1 ). Therefore, Ker(T1 ) is a subspace of Ker(T2 T1 ). 4. TRUE. Observe that T is not one-to-one since Ker(T ) = {0} (because Ker(T ) is 1-dimensional). Moreover, since M22 is 4-dimensional, then by the Rank-Nullity Theorem, Rng(T ) is 3-dimensional. Since P2 is 3-dimensional, we conclude that Rng(T ) = P2 ; that is, T is onto. 5. TRUE. Since M2 (R) is 4-dimensional, Rng(T ) can be at most 4-dimensional. However, P4 is 5dimensional. Therefore, any such linear transformation T cannot be onto. 6. TRUE. If we assume that (T2 T1 )v = (T2 T1 )w, then T2 (T1 (v)) = T2 (T1 (w)). Since T2 is one-to-one, then T1 (v) = T1 (w). Next, since T1 is one-to-one, we conclude that v = w. Therefore, T2 T1 is one-to-one. 7. FALSE. This linear transformation is onto one-to-one. The reason is essentially because the derivative of any constant is zero. Therefore, Ker(T ) consists of all constant functions, and therefore, Ker(T ) = {0}. 8. TRUE. Since M23 is 6-dimensional and Rng(T ) is only 4-dimensional, T is not onto. Moreover, since P3 is 4-dimensional, the Rank-Nullity Theorem implies that Ker(T ) is 0-dimensional. Therefore, Ker(T ) = {0}, and this means that T is one-to-one. 9. TRUE. Recall that dim[Rn ] = n and dim[Rm ] = m. In order for such an isomorphism to exist, Rn and Rm must have the same dimension; that is, m = n. 10. FALSE. For example, the vector space of all polynomials with real coefficients is an infinite-dimensional real vector space, and since Rn is finite-dimensional for all positive integers n, this statement is false. 11. FALSE. In order for this to be true, it would also have to be assumed that T1 is onto. For example, suppose V1 = V2 = V3 = R2 . If we define T2 (x, y ) = (x, y ) for all (x, y ) in R2 , then T2 is onto. However, 344 if we define T1 (x, y ) = (0, 0) for all (x, y ) in R2 , then (T2 T1 )(x, y ) = T2 (T1 (x, y )) = T2 (0, 0) = (0, 0) for all (x, y ) in R2 . Therefore T2 T1 is not onto, since Rng(T2 T1 ) = {(0, 0)}, even though T2 itself is onto. 12. TRUE. This is a direct application of the Rank-Nullity Theorem. Since T is assumed to be onto, Rng(T ) = R3 , which is 3-dimensional. Therefore the dimension of Ker(T ) is 8 − 3 = 5. Problems: 1. T1 T2 (x) = T1 (T2 (x)) = T1 (B x) = (AB )x, −1 2 15 −5 −5 AB = = 31 −2 0 1 15 −5 −5 x1 T1 T2 (x) = (AB )x = = 1 15 x2 14 7 x1 T2 T1 (x) = (BA)x = = 2 −4 x2 Clearly, T1 T2 = T2 T1 . 2. T2 (T1 (x)) = B (A(x)) = −1 1 so T1 T2 = AB . Similarly, T2 T1 = BA. 15 −1 2 14 7 . BA = = . −2 0 31 2 −4 −5x1 − 5x2 = (−5(x1 + x2 ), x1 + 15x2 ) and x1 + 15x2 14x1 + 7x2 = (7(2x1 + x2 ), 2(x1 − 2x2 )). 2x1 − 4x2 1 −1 3 2 x1 x2 = −1 1 x1 − x2 3x1 + 2x2 = 2x1 + 3x2 = 2 3 x. T1 T2 does not exist because T1 must have a domain of R2 , yet the range of T2 , which must be the domain of T1 , is R. 1 −1 x1 0 = . This matrix equation results in 2 −2 x2 0 the system: x1 − x2 = 0 and 2x1 − 2x2 = 0, or equivalently, x1 = x2 . Thus, 3. Ker(T1 ) = {x ∈ R2 : Ax = 0}. Ax = 0 =⇒ Ker(T1 ) = {x ∈ R2 : x = r(1, 1) where r ∈ R}. Geometrically, this is the line through the origin in R2 spanned by (1, 1). 2 1 x1 0 Ker(T2 ) = {x ∈ R2 : B x = 0}. B x = 0 =⇒ = . This matrix equation results in the 3 −1 x2 0 system: 2x1 + x2 = 0 and 3x1 − x2 = 0, or equivalently, x1 = x2 = 0. Thus, Ker(T2 ) = {0}. Geometrically, this is a point (the origin). 1 −1 2 1 x1 0 Ker(T1 T2 ) = {x ∈ R2 : (AB )x = 0}. (AB )x = 0 =⇒ = 2 −2 3 −1 x2 0 −1 2 x1 0 =⇒ = . This matrix equation results in the system: −2 4 x2 0 −x1 + 2x2 = 0 and − 2x1 + 4x2 = 0, or equivalently, x1 = 2x2 . Thus, Ker(T1 T2 ) = {x ∈ R2 : x = s(2, 1) where s ∈ R}. Geometrically, this is the line through the origin in R2 spanned by the vector (2, 1). 2 1 1 −1 x1 0 Ker(T2 T1 ) = {x ∈ R2 : (BA)x = 0}. (BA)x = 0 =⇒ = 3 −1 2 −2 x2 0 4 −4 x1 0 =⇒ = . This matrix equation results in the system: 1 −1 x2 0 4x1 − 4x2 = 0 and x1 − x2 = 0, or equivalently, x1 = x2 . Thus, Ker(T2 T1 ) = {x ∈ R2 : x = t(1, 1) where t ∈ R}. Geometrically, this is the line through the origin in R2 spanned by the vector (1, 1). 345 4. (T2 T1 )(A) = T2 (T1 (A)) = T2 (A − AT ) = (A − AT ) + (A − AT )T = (A − AT ) + (AT − A) = 0n . 5. (a) (T1 (f ))(x) = x (T2 (f ))(x) = a (T1 T2 )(f )(x) = (T2 T1 )(f )(x) = (T1 T2 )(f ) = d [sin(x − a)] = cos(x − a). dx x sin(t − a)dt = [− cos(t − a)]a = 1 − cos(x − a). d [1 − cos(x − a)] = sin(x − a) = f (x). dx x x cos(t − a)dt = [sin(t − a)]a = sin(x − a) = f (x). Consequently, a (T2 T1 )(f ) = f. x (b) (T1 T2 )(f ) = T1 (T2 (f )) = T1 f (x)dx a (T2 T1 )(g ) = T2 (T1 (g )) = T2 dg (x) dx x = a = d dx x f (x)dx = f (x). a dg (x) dx = g (x) − g (a). dx T2 T1 (v) = T2 [aT1 (v1 ) + bT1 (v2 )] = T2 [a(v1 − v2 ) + b(2v1 = T2 [(a + 2b)v1 + (b − a)v2 ] = (a + 2b)T2 (v1 ) + ( 6. Let v ∈ V so there exists a, b ∈ R such that v = av1 +bv2 . Then = (a + 2b)(v1 + 2v2 ) + (b − a)(3v1 − v2 ) = (5b − 7. Let v ∈ V so there exists a, b ∈ R such that v = av1 +bv2 . Then T2 T1 (v) = T2 [aT1 (v1 ) + bT1 (v2 )] = T2 [a(3v1 + v2 )] = 3a(−5v2 ) + a(−v1 + 6v2 ) = −av1 − 9av2 . 8. Ker(T ) = {x ∈ R2 : T (x) = 0} = {x ∈ R2 : Ax = 0}. The augmented matrix of the system Ax = 0 420 100 . It follows that Ker(T ) = {0}. Hence T is is: , with reduced row-echelon form of 010 130 one-to-one by Theorem 5.4.7. For the given transformation, Rng(T ) = {y ∈ R2 : Ax = y is consistent}. The augmented matrix of the 4 2 y1 1 0 (3y1 − y2 )/5 system Ax = y is with reduced row-echelon form of . The system is, 1 3 y2 0 1 (2y2 − y1 )/5 2 −1 therefore, consistent for all (y1 , y2 ), so that Rng(T ) = R . Consequently, T is onto. T exists since T has been shown to be both one-to-one and onto. Using the Gauss-Jordan method for computing A−1 , we have 421 130 Thus, A−1 = 3 10 1 − 10 1 −5 2 5 0 1 ∼ 1 0 1 2 5 2 1 4 −1 4 0 1 ∼ 3 10 10 1 0 1 − 10 −1 5 2 5 . , so T −1 (y) = A−1 y. 9. Ker(T ) = {x ∈ R2 : T (x) = 0} = {x ∈ R2 : Ax = 0}. The augmented matrix of the system Ax = 0 is: 1 20 120 , with reduced row-echelon form of . It follows that −2 −4 0 000 Ker(T ) = {x ∈ R2 : x = t(−2, 1), t ∈ R}. By Theorem 5.4.7, since Ker(T ) = {0}, T is not one-to-one. This also implies that T −1 does not exist. For the given transformation, Rng(T ) = {y ∈ R2 : Ax = y is consistent}. The augmented matrix of the 346 y1 12 1 2 y1 . The last row of with reduced row-echelon form of 0 0 2y1 + y2 −2 −4 y2 this matrix implies that 2y1 + y2 = 0 is required for consistency. Therefore, it follows that system Ax = y is Rng(T ) = {(y1 , y2 ) ∈ R2 : 2y1 + y2 = 0} = {y ∈ R2 : y = s(1, −2), s ∈ R}. T is not onto because dim[Rng(T )] = 1 = 2 = dim[R2 ]. 10. Ker(T ) = {x ∈ R3 : T (x) = 0} = {x ∈ R3 : Ax = 0}. The augmented matrix of the system Ax = 0 is: 1 2 −1 0 1 0 −7 0 , with reduced row-echelon form . It follows that 25 10 01 30 Ker(T ) = {(x1 , x2 , x3 ) ∈ R3 : x1 = 7t, x2 = −3t, x3 = t, t ∈ R} = {x ∈ R3 : x = t(7, −3, 1), t ∈ R}. By Theorem 5.4.7, since Ker(T ) = {0}, T is not one-to-one. This also implies that T −1 does not exist. For the given transformation, Rng(T ) = {y ∈ R2 : Ax = y is consistent}. The augmented matrix of the system Ax = y is 1 2 −1 y1 1 0 −7 5y1 − 2y2 ∼ . 25 1 y2 01 3 y2 − 2y1 We clearly have a consistent system for all y = (y1 , y2 ) ∈ R2 , thus Rng(T ) = R2 . Therefore, T is onto by Definition 5.3.3. 11. Reducing A to row-echelon form, we obtain 1 0 REF(A) = 0 0 3 1 0 0 5 2 . 0 0 We quickly find that Ker(T ) = nullspace(A) = span{(1, −2, 1)}. Moreover, 1 0 34 Rng(T ) = colspace(A) = span , 4 5 2 1 . Based on these calculations, we see that T is neither one-to-one nor onto. 12. We have and so 1 2 1 (0, 0, 1) = − (2, 1, −3) + (1, 0, 0) + (0, 1, 0), 3 3 3 1 2 1 T (0, 0, 1) = − T (2, 1, −3) + T (1, 0, 0) + T (0, 1, 0) 3 3 3 1 2 1 = − (7, −1) + (4, 5) + (−1, 1) 3 3 3 = (0, 4). (a) From the given information and the above calculation, we find that the matrix of T is A = 4 −1 0 5 14 . 347 (b) Because A has more columns than rows, REF(A) must have an unpivoted column, which implies that nullspace(A) = {0}. Hence, T is not one-to-one. On the other hand, colspace(A) = R2 since the first two columns, for example, are linearly independent vectors in R2 . Thus, T is onto. 13. Show T is a linear transformation: Let x, y ∈ V and c, λ ∈ R where λ = 0. T (x + y) = λ(x + y) = λx + λy = T (x) + T (y), and T (cx) = λ(cx) = c(λx) = cT (x). Thus, T is a linear transformation. Show T is one-to-one: x ∈ Ker(T ) ⇐⇒ T (x) = 0 ⇐⇒ λx = 0 ⇐⇒ x = 0, since λ = 0. Thus, Ker(T ) = {0}. By Theorem 5.4.7, T is one-to-one. Show T is onto: dim[Ker(T )]+ dim[Rng(T )] = dim[V ] =⇒ dim[{0}]+ dim[Rng(T )] = dim[V ] =⇒ 0 + dim[Rng(T )] = dim[V ] =⇒ dim[Rng(T )] = dim[V ] =⇒ Rng(T ) = V . Thus, T is onto by Definition 5.3.3. Find T −1 : 1 1 1 x =λ x = x. T −1 (x) = x since T (T −1 (x)) = T λ λ λ 14. T : P1 → P1 where T (ax + b) = (2b − a)x + (b + a). Show T is one-to-one: Ker(T ) = {p ∈ P1 : T (p) = 0} = {ax + b ∈ P1 : (2b − a)x + (b + a) = 0}. Thus, a and b must satisfy the system 2b − a = 0, b + a = 0. The only solution to this system is a = b = 0. Consequently, Ker(T ) = {0}, so T is one-to-one. Show T is onto: Since Ker(T ) = {0}, dim[Ker(T )] = 0. Thus, dim[Ker(T )]+ dim[Rng(T )] = dim[P1 ] =⇒ dim[Rng(T )] = dim[P1 ], and since Rng(T ) is a subspace of P1 , it follows that Rng(T ) = P1 . Consequently, T is onto by Theorem 5.4.7. Determine T −1 : Since T is one-to-one and onto, T −1 exists. T (ax + b) = (2b − a)x + (b + a) 2B − A A+B and a = =⇒ T −1 [(2b − a)x + (b + a)] = ax + b. If we let A = 2b − a and B = a + b, then b = 3 3 1 1 −1 so that T [Ax + B ] = (2B − A)x + (A + B ). 3 3 15. T is not one-to-one: Ker(T ) = {p ∈ P2 : T (p) = 0} = {ax2 + bx + c : c + (a − b)x = 0}. Thus, a, b, and c must satisfy the system c = 0 and a − b = 0. Consequently, Ker(T ) = {r(x2 + x) : r ∈ R} = {0}. Thus, by Theorem 5.4.7, T is not one-to-one. T −1 does not exist because T is not one-to-one. T is onto: Since Ker(T ) = {r(x2 + x) : r ∈ R}, we see that dim[Ker(T )] = 1. Thus, dim[Ker(T )]+ dim[Rng(T )] = dim[P2 ] =⇒ 1+ dim[Rng(T )] = 3 =⇒ dim[Rng(T )] = 2. Since Rng(T ) is a subspace of P1 , 2 = dim[Rng(T )] ≤ dim[P1 ] = 2, and so equality holds: Rng(T ) = P1 . Thus, T is onto by Theorem 5.4.7. 16. {v1 , v2 } is a basis for V and T : V → V is a linear transformation. Show T is one-to-one: Any v ∈ V can be expressed as v = av1 + bv2 where a, b ∈ R. T (v) = 0 ⇐⇒ T (av1 + bv2 ) = 0 ⇐⇒ aT (v1 ) + bT (v2 ) = 0 ⇐⇒ a(v1 + 2v2 ) + b(2v1 − 3v2 ) = 0 ⇐⇒ (a + 2b)v1 + (2a − 3b)v2 = 0 ⇐⇒ a + 2b = 0 and 2a − 3b = 0 ⇐⇒ a = b = 0 ⇐⇒ v = 0. Hence 348 Ker(T ) = {0}. Therefore T is one-to-one. Show T is onto: dim[Ker(T )] = dim[{0}] = 0 and dim[Ker(T )]+ dim[Rng(T )] = dim[V ] implies that 0+ dim[Rng(T )] = 2 or dim[Rng(T )] = 2. Since Rng(T ) is a subspace of V , it follows that Rng(T ) = V . Thus, T is onto by Theorem 5.4.7. Determine T −1 : Since T is one-to-one and onto, T −1 exists. T (av1 + bv2 ) = (a + 2b)v1 + (2a − 3b)v2 1 =⇒ T −1 [(a + 2b)v1 + (2a − 3b)v2 ] = (av1 + bv2 ). If we let A = a + 2b and B = 2a − 3b, then a = (3A + 2B ) 7 1 1 1 and b = (2A − B ). Hence, T −1 (Av1 + B v2 ) = (3A + 2B )v1 + (2A − B )v2 . 7 7 7 17. Let v ∈ V . Then there exists a, b ∈ R such that v = av1 + bv2 . a b (v1 + v2 ) + (v1 − v2 ) (T1 T2 )v = T1 [aT2 (v1 ) + bT2 (v2 )] = T1 2 2 a+b a−b a+b a−b T1 (v1 ) + T1 (v2 ) = (v1 + v2 ) + (v1 − v2 ) = 2 2 2 2 a+b a−b a+b a−b = + v1 + − v2 = av1 + bv2 = v. 2 2 2 2 (T2 T1 )v = T2 [aT1 (v1 ) + bT1 (v2 )] = T2 [a(v1 + v2 ) + b(v1 − v2 )] = T2 [(a + b)v1 + (a − b)v2 ] = (a + b)T2 (v1 ) + (a − b)T2 (v2 ) a−b a+b (v1 + v2 ) + (v1 − v2 ) = av1 + bv2 = v. = 2 2 − Since (T1 T2 )v = v and (T2 T1 )v = v for all v ∈ V , it follows that T2 is the inverse of T1 , thus T2 = T1 1 . 18. An arbitrary vector in P1 can be written as p(x) = ap0 (x) + bp1 (x), where p0 (x) = 1, and p1 (x) = x denote the standard basis vectors in P1 . Hence, we can define an isomorphism T : R2 → P1 by T (a, b) = a + bx. 19. Let S denote the subspace of M2 (R) consisting of all upper triangular matrices. An arbitrary vector in S can be written as ab 10 01 00 =a +b +c . 0c 00 00 01 Therefore, we define an isomorphism T : R3 → S by T (a, b, c) = ab 0c . 20. Let S denote the subspace of M2 (R) consisting of all skew-symmetric matrices. An arbitrary vector in S can be written as 0a 01 =a . −a 0 −1 0 Therefore, we can define an isomorphism T : R → S by T (a) = 0a −a 0 . 349 21. Let S denote the subspace of M2 (R) consisting of all symmetric matrices. An arbitrary vector in S can be written as ab 10 01 00 =a +b +c . bc 00 10 01 Therefore, we can define an isomorphism T : R3 → S by T (a, b, c) = ab 0 e 22. A typical vector in V takes the form A = 0 0 00 ab bc c f h 0 . d g . Therefore, we can define T : V → R10 via i j T (A) = (a, b, c, d, e, f, g, h, i, j ). It is routine to verify that T is an invertible linear transformation. Therefore, we have n = 10. 23. A typical vector in V takes the form p = a0 + a2 x2 + a4 x4 + a6 x6 + a8 x8 . Therefore, we can define T : V → R5 via T (p) = (a0 , a2 , a4 , a6 , a8 ). It is routine to verify that T is an invertible linear transformation. Therefore, we have n = 5. 24. We have − T1 1 (x) = A−1 x = 25. We have −2 −2 1 4 x. 11/14 −8/7 1/14 − 5/7 1/7 x. T3 1 (x) = A−1 x = −3/7 3/14 1/7 −1/14 27. We have 28. The matrix of T2 T1 is x. −1/3 −1/6 1/3 2/3 − T2 1 (x) = A−1 x = 26. We have 3 −1 −2 1 8 −29 3 − 19 −2 x. T4 1 (x) = A−1 x = −5 2 −8 1 −4 −1 2 2 1 2 1 3 = −6 −7 6 8 . The matrix of T1 T2 is 1 2 1 3 −4 −1 2 2 . 351 29. The matrix of T4 T3 is 1 2 1 2 6 7 11 3 351 10 25 23 0 1 2 1 2 1 = 5 14 15 3 5 −1 267 12 19 1 11 3 6 01 2= 4 3 5 −1 23 . 13 18 8 6 . 43 11 The matrix of T3 T4 is = 350 30. We have (T2 T1 )(cu1 ) = T2 (T1 (cu1 )) = T2 (cT1 (u1 )) = c(T2 (T1 (u1 ))) = c(T2 T1 )(u1 ), as needed. 31. We first prove that if T is one-to-one, then T is onto. The assumption that T is one-to-one implies that dim[Ker(T )] = 0. Hence, by Theorem 5.3.8, dim[W ] = dim[V ] = dim[Rng(T )], which implies that Rng(T ) = W . That is, T is onto. Next we show that if T is onto, then T is one-to-one. We have Rng(T ) = W , so that dim[Rng(T )] = dim[W ] = dim[V ]. Hence, by Theorem 5.3.8, we have dim[Ker(T )] = 0. Hence, Ker(T ) = {0}, which implies that T is one-to-one. 32. If T : V → W is linear, then show that T −1 : W → V is also linear. Let y, z ∈ W, c ∈ R, and T (u) = y and T (v) = z where u, v ∈ V . Then T −1 (y) = u and T −1 (z) = v. Thus, T −1 (y + z) = T −1 (T (u) + T (v)) = T −1 (T (u + v)) = u + v = T −1 (y) + T −1 (z), and T −1 (cy) = T −1 (cT (u)) = T −1 (T (cu)) = cu = cT −1 (y). Hence, T −1 is a linear transformation. 33. Let T : V → V be a one-to-one linear transformation. Since T is one-to-one, it follows from Theorem 5.4.7 that Ker(T ) = {0}, so dim[Ker(T )] = 0. By Theorem 5.3.8 and substitution, dim[Ker(T )]+ dim[Rng(T )] = dim[V ] =⇒ 0 + dim[Rng(T )] = dim[V ] =⇒ dim[Rng(T )] = dim[V ], and since Rng(T ) is a subspace of V , it follows that Rng(T ) = V , thus T is onto. T −1 exists because T is both one-to-one and onto. 34. To show that {T (v1 ), T (v2 ), . . . , T (vk )} is linearly independent, assume that c1 T (v1 ) + c2 T (v2 ) + · · · + ck T (vk ) = 0. We must show that c1 = c2 = · · · = ck = 0. Using the linearity properties of T , we can write T (c1 v1 + c2 v2 + · · · + ck vk ) = 0. Now, since T is one-to-one, we can conclude that c1 v1 + c2 v2 + · · · + ck vk = 0, and since {v1 , v2 , . . . , vk } is linearly independent, we conclude that c1 = c2 = · · · = ck = 0 as desired. 35. To prove that T is onto, let w be an arbitrary vector in W . We must find a vector v in V such that T (v) = w. Since {w1 , w2 , . . . , wm } spans W , we can write w = c1 w1 + c2 w2 + · · · + cm wm for some scalars c1 , c2 , . . . , cm . Therefore T (c1 v1 + c2 v2 + · · · + cm vm ) = c1 T (v1 ) + c2 T (v2 ) + · · · + cm T (vm ) = c1 w1 + c2 w2 + · · · + cm wm = w, 351 which shows that v = c1 v1 + c2 v2 + · · · + cm vm maps under T to w, as desired. 36. Since T is a linear transformation, Rng(T ) is a subspace of W , but dim[W ] = dim[Rng(T )] = n, so W = Rng(T ), which means, by definition, that T is onto. 37. This problem is not correct. 38. Suppose that x belongs to Rng(T ). This means that there exists a vector v in V such that T (v) = x. Applying T to both sides of this equation, we have T (T (v)) = T (x), or T 2 (v) = T (x). However, since T 2 = 0, we conclude that T 2 (v) = 0, and hence, T (x) = 0. By definition, this means that x belongs to Ker(T ). Thus, since every vector in Rng(T ) belongs to Ker(T ), Rng(T ) is a subset of Ker(T ). In addition, Rng(T ) is closed under addition and scalar multiplication, by Theorem 5.3.5, and therefore, Rng(T ) forms a subspace of Ker(T ). 39. (a) To show that T2 T1 : V1 → V3 is one-to-one, we show that Ker(T2 T1 ) = {0}. Suppose that v1 ∈ Ker(T2 T1 ). This means that (T2 T1 )(v1 ) = 0. Hence, T2 (T1 (v1 )) = 0. However, since T2 is one-to-one, we conclude that T1 (v1 ) = 0. Next, since T1 is one-to-one, we conclude that v1 = 0, which shows that the only vector in Ker(T2 T1 ) is 0, as expected. (b) To show that T2 T1 : V1 → V3 is onto, we begin with an arbitrary vector v3 in V3 . Since T2 : V2 → V3 is onto, there exists v2 in V2 such that T2 (v2 ) = v3 . Moreover, since T1 : V1 → V2 is onto, there exists v1 in V1 such that T1 (v1 ) = v2 . Therefore, (T2 T1 )(v1 ) = T2 (T1 (v1 )) = T2 (v2 ) = v3 , and therefore, we have found a vector, namely v1 , in V1 that is mapped under T2 T1 to v3 . Hence, T2 T1 is onto. (c) This follows immediately from parts (a) and (b). Solutions to Section 5.5 True-False Review: 1. FALSE. The matrix representation is an m × n matrix, not an n × m matrix. 2. FALSE. The matrix [T ]B would only make sense if C was a basis for V and B was a basis for W , and C this would only be true if V and W were the same vector space. Of course, in general V and W can be different, and so this statement does not hold. 3. FALSE. The correct equation is given in (5.5.2). 4. FALSE. The correct statement is [T ]C B −1 = [T −1 ]B . C 5. TRUE. Many examples are possible. A fairly simple one is the following. Let T1 : R2 → R2 be given by T1 (x, y ) = (x, y ), and let T2 : R2 → R2 be given by T2 (x, y ) = (y, x). Clearly, T1 and T2 are different 10 linear transformations. Now if B1 = {(1, 0), (0, 1)} = C1 , then [T1 ]C1 = . If B2 = {(1, 0), (0, 1)} and B1 01 10 C2 = {(0, 1), (1, 0)}, then [T2 ]C2 = . Thus, although T1 = T2 , we found suitable bases B1 , C1 , B2 , C2 B2 01 such that [T1 ]C1 = [T2 ]C2 . B1 B2 6. TRUE. This is the content of part (b) of Corollary 5.5.10. Problems: 352 1. (a): We must determine T (1), T (x), and T (x2 ), and find the components of the resulting vectors in R2 relative to the basis C . We have T (1) = (1, 2), T (x) = (0, 1), T (x2 ) = (−3, −2). Therefore, relative to the standard basis C on R2 , we have [T (1)]C = 1 2 , [T (x)]C = 0 1 [T (x2 )]C = , −3 −2 . Therefore, [T ]C = B 1 2 0 −3 1 −2 . (b): We must determine T (1), T (1 + x), and T (1 + x + x2 ), and find the components of the resulting vectors in R2 relative to the basis C . We have T (1) = (1, 2), T (1 + x + x2 ) = (−2, 1). T (1 + x) = (1, 3), Setting T (1) = (1, 2) = c1 (1, −1) + c2 (2, 1) and solving, we find c1 = −1 and c2 = 1. Thus, [T (1)]C = −1 1 . Next, setting T (1 + x) = (1, 3) = c1 (1, −1) + c2 (2, 1) −5/3 4/3 and solving, we find c1 = −5/3 and c2 = 4/3. Thus, [T (1 + x)]C = . Finally, setting T (1 + x + x2 ) = (−2, 1) = c1 (1, −1) + c2 (2, 1) −4/3 −1/3 into the columns of [T ]C , we obtain B and solving, we find c1 = −4/3 and c2 = −1/3. Thus, [T (1 + x + x2 )]C = [T (1)]C , [T (1 + x)]C , and [T (1 + x + x2 )]C [T ]C = B 5 −1 − 3 4 1 3 −4 3 −1 3 . Putting the results for . 2. (a): We must determine T (E11 ), T (E12 ), T (E21 ), and T (E22 ), and find the components of the resulting vectors in P3 relative to the basis C . We have T (E11 ) = 1 − x3 , T (E12 ) = 3x2 , T (E21 ) = x3 , Therefore, relative to the standard basis C on P3 , we have 1 0 0 0 0 0 [T (E11 )]C = 0 , [T (E12 )]C = 3 , [T (E21 )]C = 0 −1 0 1 T (E22 ) = −1. , −1 0 [T (E22 )]C = 0 . 0 353 Putting these results into the columns of [T ]C , we obtain B 10 00 [T ]C = B 03 −1 0 0 −1 0 0 . 0 0 1 0 (b): The values of T (E21 ), T (E11 ), T (E22 ), and T (E12 ) were determined in part (a). We must express those results in terms of the ordered basis C given in (b). We have 0 0 0 0 0 1 −1 0 [T (E21 )]C = , [T (E11 )]C = 1 −1 , [T (E22 )]C = 0 , [T (E12 )]C = 0 . 0 0 0 3 Putting these results into the columns of [T ]C , we obtain B 0 0 00 0 1 −1 0 [T ]C = B 1 −1 00 0 0 03 . 3. (a): We must determine T (1, 0, 0), T (0, 1, 0), and T (0, 0, 1), and find the components of the resulting vectors relative to the basis C . We have T (1, 0, 0) = cos x, T (0, 0, 1) = −2 cos x + sin x. T (0, 1, 0) = 3 sin x, Therefore, relative to the basis C , we have [T (1, 0, 0)]C = 1 0 , [T (0, 1, 0)]C = 0 3 , [T (0, 0, 1)]C = −2 1 . Putting these results into the columns of [T ]C , we obtain B [T ]C = B 1 0 0 −2 3 1 . (b): We must determine T (2, −1, −1), T (1, 3, 5), and T (0, 4, −1), and find the components of the resulting vectors relative to the basis C . We have T (2, −1, −1) = 4 cos x − 4 sin x, T (1, 3, 5) = −9 cos x + 14 sin x, T (0, 4, −1) = 2 cos x + 11 sin x. Setting 4 cos x − 4 sin x = c1 (cos x − sin x) + c2 (cos x + sin x) and solving, we find c1 = 4 and c2 = 0. Therefore [T (2, −1, −1)]C = 4 0 . Next, setting −9 cos x + 14 sin x = c1 (cos x − sin x) + c2 (cos x + sin x) 354 and solving, we find c1 = −23/2 and c2 = 5/2. Therefore, [T (1, 3, 5)]C = −23/2 5/2 . Finally, setting 2 cos x + 11 sin x = c1 (cos x − sin x) + c2 (cos x + sin x) and solving, we find c1 = −9/2 and c2 = 13/2. Therefore [T (0, 4, −1)]C = into the columns of [T ]C , B −9/2 13/2 . Putting these results we obtain [T ]C = B 4 − 23 2 5 0 2 −9 2 13 2 . 4. (a): We must determine T (1), T (x), and T (x2 ), and find the components of the resulting vectors relative to the standard basis C on P3 . We have T (1) = 1 + x, T (x) = x + x2 , T (x2 ) = x2 + x3 . Therefore, 1 1 [T (1)]C = , 0 0 0 1 [T (x)]C = , 1 0 Putting these results into the columns of [T ]C , we obtain B 10 1 1 [T ]C = B 0 1 00 0 0 [T (x2 )]C = . 1 1 0 0 . 1 1 (b): We must determine T (1), T (x − 1), and T ((x − 1)2 ), and find the components of the resulting vectors relative to the given basis C on P3 . We have T (1) = 1 + x, T (x − 1) = −1 + x2 , T ((x − 1)2 ) = 1 − 2x − x2 + x3 . Setting 1 + x = c1 (1) + c2 (x − 1) + c3 (x − 1)2 + c4 (x − 1)3 2 1 and solving, we find c1 = 2, c2 = 1, c3 = 0, and c4 = 0. Therefore [T (1)]C = . Next, setting 0 0 −1 + x2 = c1 (1) + c2 (x − 1) + c3 (x − 1)2 + c4 (x − 1)3 0 2 and solving, we find c1 = 0, c2 = 2, c3 = 1, and c4 = 0. Therefore, [T (x − 1)]C = . Finally, setting 1 0 1 − 2x − x2 + x3 = c1 (1) + c2 (x − 1) + c3 (x − 1)2 + c4 (x − 1)3 355 −1 −1 and solving, we find c1 = −1, c2 = −1, c3 = 2, and c4 = 1. Therefore, [T ((x − 1)2 )]C = 2 . Putting 1 these results into the columns of [T ]C , we obtain B 2 0 −1 1 2 −1 . [T ]C = B 0 1 2 00 1 5. (a): We must determine T (1), T (x), T (x2 ), and T (x3 ), and find the components of the resulting vectors relative to the standard basis C on P2 . We have T (1) = 0, T (x) = 1, T (x2 ) = 2x, T (x3 ) = 3x2 . Therefore, if C is the standard basis on P2 , then we have 0 1 0 [T (1)]C = 0 , [T (x)]C = 0 , [T (x2 )]C = 2 , 0 0 0 0 [T (x3 )]C = 0 . 3 Putting these results into the columns of [T ]C , we obtain B 0100 [T ]C = 0 0 2 0 . B 0003 (b): We must determine T (x3 ), T (x3 + 1), T (x3 + x), and T (x3 + x2 ), and find the components of the resulting vectors relative to the given basis C on P2 . We have T (x3 ) = 3x2 , T (x3 + 1) = 3x2 , T (x3 + x) = 3x2 + 1, T (x3 + x2 ) = 3x2 + 2x. Setting 3x2 = c1 (1) + c2 (1 + x) + c3 (1 + x + x2 ) 0 and solving, we find c1 = 0, c2 = −3, and c3 = 3. Therefore [T (x3 )]C = −3 . Likewise, [T (x3 + 1)]C = 3 0 −3 . Next, setting 3 3x2 + 1 = c1 (1) + c2 (1 + x) + c3 (1 + x + x2 ) 1 and solving, we find c1 = 1, c2 = −3, and c3 = 3. Therefore, [T (x3 + x)]C = −3 . Finally, setting 3 3x2 + 2x = c1 (1) + c2 (1 + x) + c3 (1 + x + x2 ) 356 and solving, we find c1 = −2, c2 = −1, and c3 = 3. Therefore, [T (x3 + 2x)]C −2 = −1 . Putting these 3 results into the columns of [T ]C , we obtain B 0 0 1 −2 [T ]C = −3 −3 −3 −1 . B 3 3 3 3 6. (a): We must determine T (E11 ), T (E12 ), T (E21 ), and T (E22 ), and find the components of the resulting vectors relative to the standard basis C on R2 . We have T (E11 ) = (1, 1), T (E12 ) = (0, 0), T (E21 ) = (0, 0), T (E22 ) = (1, 1). 2 Therefore, since C is the standard basis on R , we have [T (E11 )]C = 1 1 , [T (E12 )]C = 0 0 , 0 0 [T (E21 )]C = , [T (E22 )]C = 1 1 . Putting these results into the columns of [T ]C , we obtain B [T ]C = B 1 1 0 0 0 0 1 1 . (b): Let us denote the four matrices in the ordered basis for B as A1 , A2 , A3 , and A4 , respectively. We must determine T (A1 ), T (A2 ), T (A3 ), and T (A4 ), and find the components of the resulting vectors relative to the standard basis C on R2 . We have T (A1 ) = (−4, −4), T (A2 ) = (3, 3), T (A3 ) = (−2, −2), T (A4 ) = (0, 0). Therefore, since C is the standard basis on R2 , we have [T (A1 )]C = −4 −4 , [T (A2 )]C = 3 3 , −2 −2 [T (A3 )]C = , [T (A4 )]C = 0 0 . Putting these results into the columns of [T ]C , we obtain B [T ]C = B −4 −4 3 −2 3 −2 0 0 . 7. (a): We must determine T (E11 ), T (E12 ), T (E21 ), and T (E22 ), and find the components of the resulting vectors relative to the standard basis C on M2 (R). We have T (E11 ) = E11 , T (E12 ) = 2E12 − E21 , T (E21 ) = 2E21 − E12 , T (E22 ) = E22 . Therefore, we have 1 0 [T (E11 )]C = , 0 0 0 2 [T (E12 )]C = −1 , 0 0 −1 [T (E21 )]C = 2 , 0 0 0 [T (E22 )]C = . 0 1 357 Putting these results into the columns of [T ]C , we obtain B 1 0 00 0 2 −1 0 C [T ]B = 0 −1 20 0 0 01 . (b): Let us denote the four matrices in the ordered basis for B as A1 , A2 , A3 , and A4 , respectively. We must determine T (A1 ), T (A2 ), T (A3 ), and T (A4 ), and find the components of the resulting vectors relative to the standard basis C on M2 (R). We have T (A1 ) = −1 −2 −2 −3 , T (A2 ) = 1 3 0 2 , 0 −8 7 −2 T (A3 ) = , T (A4 ) = 07 −2 0 . Therefore, we have −1 −2 [T (A1 )]C = −2 , −3 1 0 [T (A2 )]C = , 3 2 0 −8 [T (A3 )]C = 7 , −2 Putting these results into the columns of [T ]C , we obtain B −1 1 0 0 −2 0 −8 7 C [T ]B = −2 3 7 −2 −3 2 −2 0 0 7 [T (A4 )]C = −2 . 0 . 8. (a): We must determine T (e2x ) and T (e−3x ), and find the components of the resulting vectors relative to the given basis C . We have T (e2x ) = 2e2x and T (e−3x ) = −3e−3x . Therefore, we have [T ]C = B 2 0 0 −3 . (b): We must determine T (e2x − 3e−3x ) and T (2e−3x ), and find the components of the resulting vectors relative to the given basis C . We have T (e2x − 3e−3x ) = 2e2x + 9e−3x and T (2e−3x ) = −6e−3x . Now, setting 2e2x + 9e−3x = c1 (e2x + e−3x ) + c2 (−e2x ) and solving, we find c1 = 9 and c2 = 7. Therefore, [T (e2x − 3e−3x )]C = −6e−3x = c1 (e2x + e−3x ) + c2 (−e2x ) 9 7 . Finally, setting 358 −6 −6 and solving, we find c1 = c2 = −6. Therefore, [T (2e−3x )]C = . Putting these results into the columns of [T ]C , we obtain B 9 −6 7 −6 [T ]C = B . 9. (a): Let us first compute [T ]C . We must determine T (1) and T (x), and find the components of the resulting B vectors relative to the standard basis C = {E11 , E12 , E21 , E22 } on M2 (R). We have T (1) = 1 0 0 −1 and −1 −2 T (x) = 0 1 . Therefore 1 0 [T (1)]C = 0 −1 −1 0 and [T (x)]C = −2 . 1 Hence, 1 −1 0 0 [T ]C = B 0 −2 . −1 1 Now, −2 3 [p(x)]B = [−2 + 3x]B = . Therefore, we have 1 −1 0 0 [T (p(x))]C = [T ]C [p(x)]B = B 0 −2 −1 1 −5 0 = −6 . 5 −2 3 Thus, T (p(x)) = −5 −6 0 5 . (b): We have T (p(x)) = T (−2 + 3x) = −5 −6 0 5 . 10. (a): Let us first compute [T ]C . We must determine T (1, 0, 0), T (0, 1, 0), and T (0, 0, 1), and find the compoB nents of the resulting vectors relative to the standard basis C = {1, x, x2 , x3 } on P3 . We have T (1, 0, 0) = 2 − x − x3 , T (0, 1, 0) = −x, T (0, 0, 1) = x + 2x3 . 359 Therefore, 2 −1 [T (1, 0, 0)]C = 0 , −1 0 −1 [T (0, 1, 0)]C = 0 , 0 0 1 [T (0, 0, 1)]C = . 0 2 Putting these results into the columns of [T ]C , we obtain B 2 00 −1 −1 1 . [T ]C = B 0 0 0 −1 02 Now, 2 [v]B = [(2, −1, 5)]B = −1 , 5 and therefore, 4 2 00 2 −1 −1 1 C −1 = 4 [T (v)]C = [T ]B [v]B = 0 0 0 0 5 8 −1 02 . Therefore, T (v) = 4 + 4x + 8x3 . (b): We have T (v) = T (2, −1, 5) = 4 + 4x + 8x3 . 11. (a): Let us first compute [T ]C . We must determine T (E11 ), T (E12 ), T (E21 ), and T (E22 ), and find the B components of the resulting vectors relative to the standard basis C = {E11 , E12 , E21 , E22 }. We have T (E11 ) = 2 −1 0 −1 , T (E12 ) = −1 0 0 −1 , T (E21 ) = 00 03 , T (E22 ) = 1 0 3 0 Therefore, 2 −1 [T (E11 )]C = 0 , −1 −1 0 [T (E12 )]C = 0 , −1 0 0 [T (E21 )]C = , 0 3 Putting these results into the columns of [T ]C , we obtain B 2 −1 0 −1 00 C [T ]B = 0 00 −1 −1 3 1 3 . 0 0 1 3 [T (E22 )]C = . 0 0 . 360 Now, −7 2 [A]B = 1 , −3 and therefore, 2 −1 0 1 −7 −1 0 0 3 2 C [T (A)]C = [T ]B [A]B = 0 0 0 0 1 −1 −1 3 0 −3 −19 −2 = . 0 8 Hence, T (A) = −19 −2 0 8 . T (A) = −19 −2 0 8 . (b): We have 12. (a): Let us first compute [T ]C . We must determine T (1), T (x), and T (x2 ), and find the components of the B resulting vectors relative to the standard basis C = {1, x, x2 , x3 , x4 }. We have T (1) = x2 , T (x) = x3 , T (x2 ) = x4 . Therefore, [T (1)]C = 0 0 1 0 0 , [T (x)]C = 0 0 0 1 0 Putting these results into the columns of [T ]C , we obtain B 00 0 0 [T ]C = 1 0 B 0 1 00 [T (x2 )]C = , 0 0 0 0 1 0 0 0 0 1 . . Now, −1 [p(x)]B = [−1 + 5x − 6x2 ]B = 5 , −6 and therefore, [T (p(x))]C = [T ]C [p(x)]B B = 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 −1 5 = −6 0 0 −1 5 −6 . 361 Hence, T (p(x)) = −x2 + 5x3 − 6x4 . (b): We have T (p(x)) = −x2 + 5x3 − 6x4 . 13. (a): Let us first compute [T ]C . We must determine T (Eij ) for 1 ≤ i, j ≤ 3, and find components of the B resulting vectors relative to the standard basis C = {1}. We have T (E11 ) = T (E22 ) = T (E33 ) = 1 and T (E12 ) = T (E13 ) = T (E23 ) = T (E21 ) = T (E31 ) = T (E32 ) = 0. Therefore [T (E11 )]C = [T (E22 ]C = [T (E33 ]C = [1] and all other component vectors are [0]. Putting these results into the columns of [T ]C , we obtain B [T ]C = B 10 00 10 00 1 . Now, [A]B = 2 −6 0 1 4 −4 0 0 −3 . and therefore, [T (A)]C = [T ]C [A]B = B 10 00 10 00 1 2 −6 0 1 4 −4 0 0 −3 = [3]. Hence, T (A) = 3. (b): We have T (A) = 3. 14. (a): Let us first compute [T ]C . We must determine T (1), T (x), T (x2 ), T (x3 ), and T (x4 ), and find the B components of the resulting vectors relative to the standard basis C = {1, x, x2 , x3 }. We have T (1) = 0, T (x) = 1, T (x2 ) = 2x, T (x3 ) = 3x2 , T (x4 ) = 4x3 . 362 Therefore, 0 0 [T (1)]C = , 0 0 1 0 [T (x)]C = , 0 0 0 2 [T (x2 )]C = , 0 0 Putting these results into the columns of [T ]C , we obtain B 010 0 0 2 C [T ]B = 0 0 0 000 0 0 [T (x3 )]C = , 3 0 0 0 [T (x4 )]C = . 0 4 0 0 . 0 4 0 0 3 0 Now, [p(x)]B = [3 − 4x + 6x + 6x − 2x ]B = 2 3 4 3 −4 6 6 −2 , and therefore, 0 0 [T (p(x))]C = [T ]C [p(x)]B = B 0 0 1 0 0 0 0 2 0 0 3 0 −4 −4 0 6 = 12 18 0 6 −8 4 −2 0 0 3 0 . Therefore, T (p(x)) = T (3 − 4x + 6x2 + 6x3 − 2x4 ) = −4 + 12x + 18x2 − 8x3 . (b): We have T (p(x)) = p (x) = −4 + 12x + 18x2 − 8x3 . 15. (a): Let us first compute [T ]C . We must determine T (1), T (x), T (x2 ), and T (x3 ), and find the components B of the resulting vectors relative to the standard basis C = {1}. We have T (1) = 1, T (x) = 2, T (x2 ) = 4, T (x3 ) = 8. Therefore, [T (1)]C = [1], [T (x)]C = [2], [T (x2 )]C = [4], Putting these results into the columns of [T ]C , we obtain B [T ]C = B 1 24 8 . Now, 0 2 [p(x)]B = [2x − 3x2 ]B = −3 , 0 [T (x3 )]C = [8]. 363 and therefore, [T (p(x))]C = [T ]C [p(x)]B = B 1 2 4 0 2 −3 = [−8]. 0 8 Therefore, T (p(x)) = −8. (b): We have p(2) = 2 · 2 − 3 · 22 = −8. 16. The linear transformation T2 T1 : P4 → R is given by (T2 T1 )(p(x)) = p (2). Let A denote the standard basis on P4 , let B denote the standard basis on P3 , and let C denote the standard basis on R. (a): To determine [T2 T1 ]C , we compute A (T2 T1 )(1) = 0, (T2 T1 )(x) = 1, (T2 T1 )(x2 ) = 4, (T2 T1 )(x3 ) = 12, (T2 T1 )(x4 ) = 32. Therefore [(T2 T1 )(1)]C = [0], [(T2 T1 )(x2 )]C = [4], [(T2 T1 )(x)]C = [1], [(T2 T1 )(x3 )]C = [12], [(T2 T1 )(x4 )]C = [32]. Putting these results into the columns of [T2 T1 ]C , we obtain A [T2 T1 ]C = A 0 1 4 12 32 0 2 0 0 0 0 3 0 0 0 = 0 4 0 . (b): We have [T2 ]C [T1 ]B = B A 1 2 4 8 0 0 0 0 1 0 0 0 14 12 32 = [T2 T1 ]C . A (c): Let px) = 2 + 5x − x2 + 3x4 . Then the component vector of p(x) relative to the standard basis A is ( 2 5 [p(x)]A = −1 . Thus, 0 3 [(T2 T1 )(p(x))]C = [T2 T1 ]C [p(x)]A A = 0 1 4 12 Therefore, (T2 T1 )(2 + 5x − x2 + 3x4 ) = 97. 32 2 5 −1 0 3 = [97]. 364 Of course, p (2) = 5 − 2 · 2 + 12 · 23 = 97 by direct calculation as well. 17. The linear transformation T2 T1 : P1 → R2 is given by (T2 T1 )(a + bx) = (0, 0). Let A denote the standard basis on P1 , let B denote the standard basis on M2 (R), and let C denote the standard basis on R2 . (a): To determine [T2 T1 ]C , we compute A (T2 T1 )(1) = (0, 0) and (T2 T1 )(x) = (0, 0). Therefore, we obtain [T2 T1 ]C = 02 . A (b): We have [T2 ]C [T1 ]B = B A 10 10 01 01 1 −1 0 0 C 0 −2 = 02 = [T2 T1 ]A . −1 1 (c): The component vector of p(x) = −3 + 8x relative to the standard basis A is [p(x)]A = [(T2 T1 )(p(x))]C = [T2 T1 ]C [p(x)]A = 02 [p(x)]A = A 0 0 −3 8 . Thus, . Therefore, (T2 T1 )(−3 + 8x) = (0, 0). Of course (T2 T1 )(−3 + 8x) = T1 −11 −16 0 11 = (0, 0) by direct calculation as well. 18. The linear transformation T2 T1 : P2 → P2 is given by (T2 T1 )(p(x)) = [(x + 1)p(x)] . Let A denote the standard basis on P2 , let B denote the standard basis on P3 , and let C denote the standard basis on P2 . (a): To determine [T2 T1 ]C , we compute as follows: A (T2 T1 )(1) = 1, (T2 T1 )(x) = 1 + 2x, (T2 T1 )(x2 ) = 2x + 3x2 . Therefore, 1 [(T2 T1 )(1)]C = 0 , 0 1 [(T2 T1 )(x)]C = 2 , 0 0 [(T2 T1 )(x2 )]C = 2 . 3 365 Putting these results into the columns of [T2 T1 ]C , we obtain A 110 [T2 T1 ]C = 0 2 2 . A 003 (b): We have 0 [T2 ]C [T1 ]B = 0 B A 0 1 0 0 0 2 0 1 0 1 0 0 3 0 0 1 1 0 0 11 0 = 02 1 00 1 0 2 = [T2 T1 ]C . A 3 (c): The component vector of p(x) = 7 − x + 2x2 relative to the standard basis A is [p(x)]A 7 = −1 . 2 Thus, 11 [(T2 T1 )(p(x))]C = [T2 T1 ]C [p(x)]A = 0 2 A 00 0 7 6 2 −1 = 2 . 3 2 6 Therefore, (T2 T1 )(7 − x + 2x2 ) = 6 + 2x + 6x2 . Of course (T2 T1 )(7 − x + 2x2 ) = T2 ((x + 1)(7 − x + 2x2 )) = T2 (7 + 6x + x2 + 2x3 ) = 6 + 2x + 6x2 by direct calculation as well. (d): YES. Since the matrix [T2 T1 ]C computed in part (a) is invertible, T2 T1 is invertible. A 19. NO. The matrices [T ]C obtained in Problem 2 are not invertible (they contain rows of zeros), and B therefore, the corresponding linear transformation T is not invertible. 20. YES. We can explain this answer by using a matrix representation of T . Let B = {1, x, x2 } and let C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Then T (1) = (1, 1, 1), T (x) = (0, 1, 2), and T (x2 ) = (0, 1, 4), and so 100 [T ]C = 1 1 1 . B 124 Since this matrix is invertible, the corresponding linear transformation T is invertible. 21. Note that w belongs to Rng(T ) ⇐⇒ w = T (v) for some v in V ⇐⇒ [w]C = [T (v)]C for some v in V ⇐⇒ [w]C = [T ]C [v]B for some v in V . B The right-hand side of this last expression can be expressed as a linear combination of the columns of [T ]C , B and therefore, w belongs to Rng(T ) if and only if [w]C can be expressed as a linear combination of the columns of [T ]C . That is, if and only if [w]C belongs to colspace([T ]C ). B B 366 Solutions to Section 5.6 True-False Review: 1. FALSE. If v = 0, then Av = λv = 0, but by definition, an eigenvector must be a nonzero vector. 2. TRUE. When we compute det(A − λI ) for an upper or lower triangular matrix A, the determinant is the product of the entries lying along the main diagonal of A − λI : det(A − λI ) = (a11 − λ)(a22 − λ) . . . (ann − λ). The roots of this characteristic equation are precisely the values a11 , a22 , . . . , ann along the main diagonal of the matrix A. 3. TRUE. The eigenvalues of a matrix are precisely the set of roots of its characteristic equation. Therefore, two matrices A and B that have the same characteristic equation will have the same eigenvalues. 00 01 and B = . 00 00 2 2 2 We have det(A − λI ) = (−λ) = λ = det(B − λI ). Note that every nonzero vector in R is an eigenvector a of A corresponding to λ = 0. However, only vectors of the form with a = 0 are eigenvectors of B . 0 Therefore, A and B do not have precisely the same set of eigenvectors. In this case, every eigenvector of B is also an eigenvector of A, but not conversely. 4. FALSE. Many examples of this can be found. As a simple one, consider A = 5. TRUE. Geometrically, all nonzero points v = (x, y ) in R2 are oriented in a different direction from the origin after a 90◦ rotation than they are initially. Therefore, the vectors v and Av are not parallel. 6. TRUE. The characteristic equation of an n × n matrix A, det(A − λI ) is a polynomial of degree n i the indeterminate λ. Since such a polynomial always possesses n roots (with possible repeated or complex roots) by the Fundamental Theorem of Algebra, the statement is true. 7. FALSE. This is not true, in general, when the linear combination formed involves eigenvectors cor10 responding to different eigenvalues. For example, let A = , with eigenvalues λ = 1 and λ = 2. 02 1 It is easy to see that corresponding eigenvectors to these eigenvalues are, respectively, v1 = and 0 0 v2 = . However, note that 1 1 A(v1 + bf v 2 ) = , 2 which is not of the form λ(v1 + v2 ), and therefore v1 + v2 is not an eigenvector of A. As a more trivial illustration, note that if v is an eigenvector of A, then 0v is a linear combination of {v} that is no longer an eigenvector of A. 8. TRUE. This is basically a fact about roots of polynomials. Complex roots of real polynomials always occur in complex conjugate pairs. Therefore, if λ = a + ib (b = 0) is an eigenvalue of A, then so is λ = a − ib. 9. TRUE. If λ is an eigenvalue of A, then we have Av = λv for some eigenvector v of A corresponding to λ. Then A2 v = A(Av) = A(λv) = λ(Av) = λ(λv) = λ2 v, which shows that v is also an eigenvector of A2 , this time corresponding to the eigenvalue λ2 . 367 Problems: 1 2 1. Av = 3 2 1 1 = 4 4 =4 1 1 = λv . 1 −2 −6 2 6 2 2 −5 1 = 3 = 3 1 = λv. 2. Av = −2 2 1 8 −1 3 −1 1 4 c1 + 4c2 3. Since v = c1 0 + c2 −3 = −3c2 , it follows that −3 0 −3c1 14 1 c1 + 4c2 −2c1 − 8c2 c1 + 4c2 = −2 −3c2 = λv. 1 −3c2 = 6c2 Av = 3 2 3 4 −1 −3c1 6c1 −3c1 41 1 1 2 λ1 = λ1 =⇒ = =⇒ λ1 = 2. 23 −2 −2 −4 −2λ1 41 1 1 5 λ2 Av2 = λ2 v2 =⇒ = λ2 =⇒ = =⇒ λ2 = 5. 23 1 1 5 λ2 Thus λ1 = 2 corresponds to v1 and λ2 = 5 corresponds to v2 . 4. Av1 = λ1 v1 =⇒ 5. The only vectors that are mapped into a scalar multiple of themselves under a reflection in the x-axis are those vectors that either point along the x-axis, or that point along the y -axis. Hence, the eigenvectors are of the form (a, 0) or (0, b) where a and b are arbitrary nonzero real numbers. A vector that points along the x-axis will have neither its magnitude nor its direction altered by a reflection in the x-axis. Hence, the eigenvectors of the form (a, 0) correspond to the eigenvalue λ = 1. A vector of the form (0, b) will be mapped into the vector (0, −b) = −1(0, b) under a reflection in the x-axis. Consequently, the eigenvectors of the form (0, b) correspond to the eigenvalue λ = −1. 6. Any vectors lying along the line y = x are unmoved by the action of T , and hence, any vector (t, t) with t = 0 is an eigenvector with corresponding eigenvalue λ = 1. On the other hand, any vector lying along the line y = −x will be reflected across the line y = x, thereby experiencing a 180◦ change of direction. Therefore, any vector (t, −t) with t = 0 is an eigenvector with corresponding eigenvalue λ = −1. All vectors that do not lie on the line y = x or the line y = −x are not eigenvectors of this linear transformation. 7. If θ = 0, π , there are no vectors that are mapped into scalar multiples of themselves under the rotation, and consequently, there are no real eigenvalues and eigenvectors in this case. If θ = 0, then every vector is mapped onto itself under the rotation, therefore λ = 1, and every nonzero vector in R2 is an eigenvector. If θ = π , then every vector is mapped onto its negative under the rotation, therefore λ = −1, and once again, every nonzero vector in R2 is an eigenvector. 8. Any vectors lying on the y -axis are unmoved by the action of T , and hence, any vector (0, y, 0) with y not zero is an eigenvector of T with corresponding eigenvalue λ = 1. On the other hand, any vector lying in the xz -plane, say (x, 0, z ) is transformed under T to (0, 0, 0). Thus, any vector (x, 0, z ) with x and z not both zero is an eigenvector with corresponding eigenvalue λ = 0. 3−λ −1 = 0 ⇐⇒ λ2 − 2λ − 8 = 0 −5 −1 − λ ⇐⇒ (λ + 2)(λ − 4) = 0 ⇐⇒ λ = −2 or λ = 4. 5 −1 v1 0 If λ = −2 then (A − λI )v = 0 assumes the form = −5 1 v2 0 =⇒ 5v1 − v2 = 0 =⇒ v2 = 5v1 . If we let v1 = t ∈ R, then the solution set of this system is {(t, 5t) : t ∈ R} 9. det(A − λI ) = 0 ⇐⇒ 368 so the eigenvectors corresponding to λ = −2 are v = t(1, 5) where t ∈ R. −1 −1 v1 0 If λ = 4 then (A − λI )v = 0 assumes the form = −5 −5 v2 0 =⇒ −v1 − v2 = 0 =⇒ v2 = −v1 . If we let v2 = r ∈ R, then the solution set of this system is {(r, −r) : r ∈ R} so the eigenvectors corresponding to λ = 4 are v = r(1, −1) where r ∈ R. 1−λ 6 = 0 ⇐⇒ λ2 + 2λ − 15 = 0 2 −3 − λ ⇐⇒ (λ − 3)(λ + 5) = 0 ⇐⇒ λ = 3 or λ = −5. −2 6 v1 0 = If λ = 3 then (A − λI )v = 0 assumes the form 2 −6 v2 0 =⇒ v1 − 3v2 = 0. If we let v2 = r ∈ R, then the solution set of this system is {(3r, r) : r ∈ R} so the eigenvectors corresponding to λ = 3 are v = r(3, 1) where r ∈ R. 66 v1 0 If λ = −5 then (A − λI )v = 0 assumes the form = 22 v2 0 =⇒ v1 + v2 = 0. If we let v2 = s ∈ R, then the solution set of this system is {(−s, s) : s ∈ R} so the eigenvectors corresponding to λ = −5 are v = s(−1, 1) where s ∈ R. 10. det(A − λI ) = 0 ⇐⇒ 7−λ 4 = 0 ⇐⇒ λ2 − 10λ + 25 = 0 −1 3−λ ⇐⇒ (λ − 5)2 = 0 ⇐⇒ λ = 5 of multiplicity two. 2 4 v1 0 = If λ = 5 then (A − λI )v = 0 assumes the form −1 −2 v2 0 =⇒ v1 + 2v2 = 0 =⇒ v1 = −2v2 . If we let v2 = t ∈ R, then the solution set of this system is {(−2t, t) : t ∈ R} so the eigenvectors corresponding to λ = 5 are v = t(−2, 1) where t ∈ R. 11. det(A − λI ) = 0 ⇐⇒ 12. det(A − λI ) = 0 ⇐⇒ 2−λ 0 0 2−λ = 0 ⇐⇒ (2 − λ)2 = 0 ⇐⇒ λ = 2 of multiplicity two. 00 v1 0 = 00 v2 0 Thus, if we let v1 = s and v2 = t where s, t ∈ R, then the solution set of this system is {(s, t) : s, t ∈ R} so the eigenvectors corresponding to λ = 2 are v = s(1, 0) + t(0, 1) where s, t ∈ R. If λ = 2 then (A − λI )v = 0 assumes the form 13. det(A − λI ) = 0 ⇐⇒ 3−λ 4 −2 −1 − λ = 0 ⇐⇒ λ2 − 2λ + 5 = 0 ⇐⇒ λ = 1 ± 2i. 2 + 2i −2 v1 0 = 4 −2 + 2i v2 0 =⇒ (1 + i)v1 − v2 = 0. If we let v1 = s ∈ C, then the solution set of this system is {(s, (1 + i)s) : s ∈ C} so the eigenvectors corresponding to λ = 1 − 2i are v = s(1, 1 + i) where s ∈ C. By Theorem 5.6.8, since the entries of A are real, λ = 1 + 2i has corresponding eigenvectors of the form v = t(1, 1 − i) where t ∈ C. If λ = 1 − 2i then (A − λI )v = 0 assumes the form 14. det(A − λI ) = 0 ⇐⇒ 2−λ −3 3 2−λ = 0 ⇐⇒ (2 − λ)2 = −9 ⇐⇒ λ = 2 ± 3i. 3i 3 v1 0 = −3 3i v2 0 =⇒ v2 = −iv1 . If we let v1 = t ∈ C, then the solution set of this system is {(t, −it) : t ∈ C} so the eigenvectors corresponding to λ = 2 − 3i are v = t(1, −i) where t ∈ C. By Theorem 5.6.8, since the entries of A are real, λ = 2 + 3i has corresponding eigenvectors of the form v = r(1, i) where r ∈ C. If λ = 2 − 3i then (A − λI )v = 0 assumes the form 369 15. det(A − λI ) = 0 ⇐⇒ 10 − λ 0 −8 −12 2−λ 12 8 0 −6 − λ = 0 ⇐⇒ (λ − 2)3 = 0 ⇐⇒ λ = 2 of multiplicity three. 8 −12 8 v1 0 0 0 v2 = 0 If λ = 2 then (A − λI )v = 0 assumes the form 0 −8 12 −8 v3 0 =⇒ 2v1 − 3v2 + 2v3 = 0. Thus, if we let v2 = 2s and v3 = t where s, t ∈ R, then the solution set of this system is {(3s − t, 2s, t) : s, t ∈ R} so the eigenvectors corresponding to λ = 2 are v = s(3, 2, 0) + t(−1, 0, 1) where s, t ∈ R. 3−λ 0 1 0 −1 = 0 ⇐⇒ (λ − 1)(λ − 3)2 = 0 ⇐⇒ λ = 1 or λ = 3 of 2−λ multiplicity two. 2 0 0 v1 0 1 −1 v2 = 0 If λ = 1 then (A − λI )v = 0 assumes the form 0 1 −1 1 v3 0 =⇒ v1 = 0 and v2 − v3 = 0. Thus, if we let v3 = s where s ∈ R, then the solution set of this system is {(0, s, s) : s ∈ R} so the eigenvectors corresponding to λ = 1 are = s(0, 1, where s ∈ R. v 1) 0 0 0 v1 0 If λ = 3 then (A − λI )v = 0 assumes the form 0 −1 −1 v2 = 0 1 −1 −1 v3 0 =⇒ v1 = 0 and v2 + v3 = 0. Thus, if we let v3 = t where t ∈ R, then the solution set of this system is {(0, −t, t) : t ∈ R} so the eigenvectors corresponding to λ = 3 are v = t(0, −1, 1) where t ∈ R. 16. det(A − λI ) = 0 ⇐⇒ 0 2−λ −1 17. det(A − λI ) = 0 ⇐⇒ 1−λ 0 2 0 3−λ −2 18. det(A − λI ) = 0 ⇐⇒ 6−λ −5 0 3 −2 − λ 0 0 2 = 0 ⇐⇒ (λ − 1)3 = 0 ⇐⇒ λ = 1 of multiplicity three. −1 − λ 0 0 0 v1 0 2 2 v2 = 0 If λ = 1 then (A − λI )v = 0 assumes the form 0 2 −2 −2 v3 0 =⇒ v1 = 0 and v2 + v3 = 0. Thus, if we let v3 = s where s ∈ R, then the solution set of this system is {(0, −s, s) : s ∈ R} so the eigenvectors corresponding to λ = 1 are v = s(0, −1, 1) where s ∈ R. or λ = 3. −4 2 −1 − λ = 0 ⇐⇒ (λ − 1)(λ + 1)(λ − 3) = 0 ⇐⇒ λ = −1, λ = 1, 7 3 −4 v1 0 2 v2 = 0 If λ = −1 then (A − λI )v = 0 assumes the form −5 −1 0 0 0 v3 0 =⇒ v1 = v3 − v2 and 4v2 − 3v3 = 0. Thus, if we let v3 = 4r where r ∈ R, then the solution set of this system is {(r, 3r, 4r) : r ∈ R} so the eigenvectors corresponding to λ = −1 v r(1, 3, 4) where r ∈ R. are = 5 3 −4 v1 0 2 v2 = 0 If λ = 1 then (A − λI )v = 0 assumes the form −5 −3 0 0 −2 v3 0 =⇒ 5v1 + 3v2 = 0 and v3 = 0. Thus, if we let v2 = −5s where s ∈ R, then the solution set of this system is {(3s, −5s, 0) : s ∈ R} so the eigenvectors corresponding to λ = 1 are v = s(3, −5, 0) where s ∈ R. 370 3 3 −4 v1 0 2 v2 = 0 If λ = 3 then (A − λI )v = 0 assumes the form −5 −5 0 0 −4 v3 0 =⇒ v1 + v2 = 0 and v3 = 0. Thus, if we let v2 = t where t ∈ R, then the solution set of this system is {(−t, t, 0) : t ∈ R} so the eigenvectors corresponding to λ = 3 are v = t(−1, 1, 0) where t ∈ R. 7−λ 8 0 −8 −9 − λ 0 6 6 = 0 ⇐⇒ (λ + 1)3 = 0 ⇐⇒ λ = −1 of multiplicity −1 − λ three. 8 −8 6 v1 0 If λ = −1 then (A − λI )v = 0 assumes the form 8 −8 6 v2 = 0 0 00 v3 0 =⇒ 4v1 − 4v2 + 3v3 = 0. Thus, if we let v2 = r and v3 = 4s where r, s ∈ R, then the solution set of this system is {(r − 3s, r, 4s) : r, s ∈ R} so the eigenvectors corresponding to λ = −1 are v = r(1, 1, 0)+ s(−3, 0, 4) where r, s ∈ R. 19. det(A − λI ) = 0 ⇐⇒ 20. det(A − λI ) = 0 ⇐⇒ −λ 1 0 2−λ 2 −1 multiplicity two. −1 0 3−λ = 0 ⇐⇒ (λ − 1)(λ − 2)2 = 0 ⇐⇒ λ = 1 or λ = 2 of −1 1 −1 v1 0 1 0 v2 = 0 If λ = 1 then (A − λI )v = 0 assumes the form 0 2 −1 2 v3 0 =⇒ v2 = 0 and v1 + v3 = 0. Thus, if we let v3 = r where r ∈ R, then the solution set of this system is {(−r, 0, r) : r ∈ R} so the eigenvectors corresponding to λ = 1 are = r(−1, 0, 1) where r ∈ R. v −2 1 −1 v1 0 0 0 v2 = 0 If λ = 2 then (A − λI )v = 0 assumes the form 0 2 −1 1 v3 0 =⇒ 2v1 − v2 + v3 = 0. Thus, if we let v1 = s and v3 = t where s, t ∈ R, then the solution set of this system is {(s, 2s + t, t) : s, t ∈ R} so the eigenvectors corresponding to λ = 2 are v = s(1, 2, 0) + t(0, 1, 1) where s, t ∈ R. 21. det(A − λI ) = 0 ⇐⇒ 1−λ 0 0 0 −λ −1 0 1 −λ = 0 ⇐⇒ (1 − λ)(1 + λ2 ) = 0 ⇐⇒ λ = 1 or λ = ±i. 0 0 0 v1 0 1 v2 = 0 If λ = 1 then (A − λI )v = 0 assumes the form 0 −1 0 −1 −1 v3 0 =⇒ −v2 + v3 = 0 and −v2 − v3 = 0. The solution set of this system is {(r, 0, 0) : r ∈ C} so the eigenvectors corresponding to λ = 1 are v = r(1, 0, 0) where r C. ∈ 1+i 00 v1 0 i 1 v2 = 0 If λ = −i then (A − λI )v = 0 assumes the form 0 0 −1 i v3 0 =⇒ v1 = 0 and −v2 + iv3 = 0. The solution set of this system is {(0, si, s) : s ∈ C} so the eigenvectors corresponding to λ = −i are v = s(0, i, 1) where s ∈ C. By Theorem 5.6.8, since the entries of A are real, λ = i has corresponding eigenvectors of the form v = t(0, −i, 1) where t ∈ C. 22. det(A − λI ) = 0 ⇐⇒ −2 − λ 1 1 1 −1 − λ 3 0 −1 −3 − λ = 0 ⇐⇒ (λ + 2)(λ2 + 4λ + 5) = 0 ⇐⇒ λ = −2 or 371 λ = −2 ± i. 01 0 v1 0 If λ = −2 then (A − λI )v = 0 assumes the form 1 1 −1 v2 = 0 1 3 −1 v3 0 =⇒ v2 = 0 and v1 − v3 = 0. Thus, if we let v3 = r where r ∈ C, then the solution set of this system is {(r, 0, r) : r ∈ C} so the eigenvectors corresponding to λ = −2 are v = r(1, , where r∈ C. 0 1) −i 1 0 v1 0 −1 v2 = 0 If λ = −2 + i then (A − λI )v = 0 assumes the form 1 1 − i 1 3 −1 − i v3 0 −2 + i −1 − 2i =⇒ v1 + v3 = 0 and v2 + v3 = 0. Thus, if we let v3 = 5s where s ∈ C, then the solution 5 5 set of this system is {((2 − i)s, (1 + 2i)s, 5s) : s ∈ C} so the eigenvectors corresponding to λ = −2 + i are v = s(2 − i, 1 + 2i, 5) where s ∈ C. By Theorem 5.6.8, since the entries of A are real, λ = −2 − i has corresponding eigenvectors of the form v = t(2 + i, 1 − 2i, 5) where t ∈ C. 2−λ 3 2 −1 1−λ −1 24. det(A − λI ) = 0 ⇐⇒ 5−λ 0 0 0 5−λ 0 25. det(A − λI ) = 0 ⇐⇒ −λ 2 2 2 −λ 2 2 2 −λ 3 0 = 0 ⇐⇒ λ(λ − 2)(λ − 4) = 0 ⇐⇒ λ = 0, λ = 2, or λ = 4. 3−λ 2 −1 3 v1 0 1 0 v2 = 0 If λ = 0 then (A − λI )v = 0 assumes the form 3 2 −1 3 v3 0 =⇒ v1 + 2v2 − 3v3 = 0 and −5v2 + 9v3 = 0. Thus, if we let v3 = 5r where r ∈ R, then the solution set of this system is {(−3r, 9r, 5r) : r ∈ R} so the eigenvectors corresponding to = are = r(−3, 9, 5) where r ∈ R. λ0 v 0 −1 3 v1 0 If λ = 2 then (A − λI )v = 0 assumes the form 3 −1 0 v2 = 0 2 −1 1 v3 0 =⇒ v1 − v3 = 0 and v2 − 3v3 = 0. Thus, if we let v3 = s where s ∈ R, then the solution set of this system is {(s, 3s, s) : s ∈ R} so the eigenvectors corresponding to λ = 2 are v = s(1, 3, 1) where s ∈ R. −2 −1 3 v1 0 0 v2 = 0 If λ = 4 then (A − λI )v = 0 assumes the form 3 −3 2 −1 −1 v3 0 =⇒ v1 − v3 = 0 and v2 − v3 = 0. Thus, if we let v3 = t where t ∈ R, then the solution set of this system is {(t, t, t) : t ∈ R} so the eigenvectors corresponding to λ = 4 are v = t(1, 1, 1) where t ∈ R. 23. det(A − λI ) = 0 ⇐⇒ 0 0 = 0 ⇐⇒ (λ − 5)3 = 0 ⇐⇒ λ = 5 of multiplicity three. 5−λ 000 v1 0 If λ = 5 then (A − λI )v = 0 assumes the form 0 0 0 v2 = 0 000 v3 0 Thus, if we let v1 = r, v2 = s, and v3 = t where r, s, t ∈ R, then the solution set of this system is {(r, s, t) : r, s, t ∈ R} so the eigenvectors corresponding to λ = 5 are v = r(1, 0, 0) + s(0, 1, 0) + t(0, 0, 1) where r, s, t ∈ R. That is, every nonzero vector in R3 is an eigenvector of A corresponding to λ = 5. two. = 0 ⇐⇒ (λ − 4)(λ + 2)2 = 0 ⇐⇒ λ = 4 or λ = −2 of multiplicity −4 2 2 v1 0 2 v2 = 0 If λ = 4 then (A − λI )v = 0 assumes the form 2 −4 2 2 −4 v3 0 372 =⇒ v1 − v3 = 0 and v2 − v3 = 0. Thus, if we let v3 = r where r ∈ R, then the solution set of this system is {(r, r, r) : r ∈ R} so the eigenvectors corresponding to λ = 4 are v = (1, 1, 1) where r ∈ R. r 222 v1 0 If λ = −2 then (A − λI )v = 0 assumes the form 2 2 2 v2 = 0 222 v3 0 =⇒ v1 + v2 + v3 = 0. Thus, if we let v2 = s and v3 = t where s, t ∈ R, then the solution set of this system is {(−s − t, s, t) : s, t ∈ R} so the eigenvectors corresponding to λ = −2 are v = s(−1, 1, 0) + t(−1, 0, 1) where s, t ∈ R. 1−λ 2 3 4 4 3−λ 2 1 26. det(A − λI ) = 0 ⇐⇒ = 0 ⇐⇒ λ4 − 14λ3 − 32λ2 = 0 4 5 6−λ 7 7 6 5 4−λ ⇐⇒ λ2 (λ − 16)(λ + 2) = 0 ⇐⇒ λ = 16, λ = −2, λ = 0 of multiplicity two. or −15 2 3 4 v1 0 4 −13 2 1 v2 0 = If λ = 16 then (A − λI )v = 0 assumes the form 4 5 −10 7 v3 0 7 6 5 −12 v4 0 =⇒ v1 − 1841v3 + 2078v4 = 0, v2 + 82v3 − 93v4 = 0, and 31v3 − 35v4 = 0. Thus, if we let v4 = 31r where r ∈ R, then the solution set of this system is {(17r, 13r, 35r, 31r) : r ∈ R} so the eigenvectors corresponding to λ = 16 are v = r(17, 13, 35, 31) where r ∈ R. 3234 v1 0 4 5 2 1 v2 0 If λ = −2 then (A − λI )v = 0 assumes the form 4 5 8 7 v3 = 0 7656 v4 0 =⇒ v1 + v4 = 0, v2 − v4 = 0, and v3 + v4 = 0. Thus, if we let v4 = s where s ∈ R, then the solution set of this system is {(−s, s, −s, s) : s ∈ R} so the eigenvectors corresponding to λ = −2 are v = s(−1, 1, −1, 1) where s ∈ R. 1234 v1 0 4 3 2 1 v2 0 If λ = 0 then (A − λI )v = 0 assumes the form 4 5 6 7 v3 = 0 7654 v4 0 =⇒ v1 − v3 − 2v4 = 0 and v2 + 2v3 + 3v4 = 0. Thus, if we let v3 = a and v4 = b where a, b ∈ R, then the solution set of this system is {(a + 2b, −2a − 3b, a, b) : a, b ∈ R} so the eigenvectors corresponding to λ = 0 are v = a(1, −2, 1, 0) + b(2, −3, 0, 1) where a, b ∈ R. 27. det(A − λI ) = 0 ⇐⇒ −λ 1 0 0 −1 −λ 0 0 0 0 −λ −1 0 0 1 −λ = 0 ⇐⇒ (λ2 + 1)2 = 0 ⇐⇒ λ = ±i, where each root is of multiplicity two. i10 0 v1 0 −1 i 0 0 v2 0 If λ = −i then (A − λI )v = 0 assumes the form 0 0 i −1 v3 = 0 001 i v4 0 =⇒ v1 − iv2 = 0 and v3 + iv4 = 0. Thus, if we let v2 = r and v4 = s where r, s ∈ C, then the solution set of this system is {(ir, r, −is, s) : r, s ∈ C} so the eigenvectors corresponding to λ = −i are v = r(i, 1, 0, 0) + s(0, 0, −i, 1) where r, s ∈ C. By Theorem 5.6.8, since the entries of A are real, λ = i has corresponding eigenvectors v = a(−i, 1, 0, 0) + b(0, 0, i, 1) where a, b ∈ C. 373 28. This matrix is lower triangular, and therefore, the eigenvalues appear along the main diagonal of the matrix: λ = 1 + i, 1 − 3i, 1. Note that the eigenvalues do not occur in complex conjugate pairs, but this does not contradict Theorem 5.6.8 because the matrix does not consist entirely of real elements. 1−λ 2 29. (a) p(λ) = det(A − λI2 ) = −1 4−λ = λ2 − 5λ + 6. (b) A2 − 5A + 6I2 = = 1 −1 2 4 1 −1 2 4 −1 −5 10 14 + 1 −1 2 4 −5 −5 5 −10 −20 + 1 0 6 0 0 6 0 1 = +6 0 0 0 0 1 6 1 6 . = 02 . (c) Using part (b) of this problem: A2 − 5A + 6I2 = 02 ⇐⇒ A−1 (A2 − 5A + 6I2 ) = A−1 · 02 ⇐⇒ A − 5I2 + 6A−1 = 02 ⇐⇒ 6A−1 = 5I2 − A 1 6 1 = 6 ⇐⇒ A−1 = 5 ⇐⇒ A−1 1 0 4 −2 30. (a) det(A − λI2 ) = 1−λ 2 2 −2 − λ 0 1 1 1 − 1 −1 2 4 , or A−1 = 2 3 1 −3 = 0 ⇐⇒ λ2 + λ − 6 = 0 ⇐⇒ (λ − 2)(λ + 3) = 0 ⇐⇒ λ = 2 or λ = −3. 1 2 10 ∼ = B. 0 −6 01 1−λ 0 det(B − λI2 ) = 0 ⇐⇒ = 0 ⇐⇒ (λ − 1)2 = 0 ⇐⇒ λ = 1 of multiplicity two. Matrices A 0 1−λ and B do not have the same eigenvalues. (b) 1 2 2 −2 ∼ 31. A(3v1 − v2 ) = 3Av1 − Av2 = 3(2v1 ) − (−3v2 ) = 6v1 + 3v2 =6 1 −1 +3 2 1 = 6 −6 + 6 3 = 12 −3 . 32. (a) Let a, b, c ∈ R. If v = av1 + bv2 + cv3 , then 5, 0, 3) = a(1, −1, 1) + b(2, 1, 3) + c(−1, −1, 2), or a + 2b − c = 5 −a + b − c = 0 (5, 0, 3) = (a + 2b − c, −a + b − c, a + 3b + 2c). The last equality results in the system: a + 3b + 2c = 3. This system has the solution a = 2, b = 1, and c = −1. Consequently, v = 2v1 + v2 − v3 . (b) Using part (a): Av = A(2v1 + v2 − v3 ) = 2Av1 + Av2 − Av3 = 2(2v1 ) + (−2v2 ) − (3v3 ) = 4v1 − 2v2 − 3v3 = 4(1, −1, 1) − 2(2, 1, 3) − 3(−1, −1, 2) = (3, −3, −8). 33. 374 A(c1 v1 + c2 v2 + c3 v3 ) = A(c1 v1 ) + A(c2 v2 ) + A(c3 v3 ) = c1 (Av1 ) + c2 (Av2 ) + c3 (Av3 ) = c1 (λv1 ) + c2 (λv2 ) + c3 (λv3 ) = λ(c1 v1 + c2 v2 + c3 v3 ). Thus, c1 v1 + c2 v2 + c3 v3 is an eigenvector of A corresponding to the eigenvalue λ. 34. Recall that the determinant of an upper (lower) triangular matrix is just the product of its main diagonal elements. Let A be an n × n upper (lower) triangular matrix. It follows that A − λIn is an upper (lower) triangular matrix with main diagonal element aii − λ, i = 1, 2, . . . , n. Consequently, n det(A − λIn ) = 0 ⇐⇒ (aii − λ) = 0. i=1 This implies that λ = a11 , a22 , . . . , ann . 35. Any scalar λ such that det(A − λI ) = 0 is an eigenvalue of A. Therefore, if 0 is an eigenvalue of A, then det(A − 0 · I ) = 0, or det(A) = 0, which implies that A is not invertible. On the other hand, if 0 is not an eigenvalue of A, then det(A − 0 · I ) = 0, or det(A) = 0, which implies that A is invertible. 36. A is invertible, so A−1 exists. Also, λ is an eigenvalue of A so that Av = λv. Thus, 1 A−1 (Av) = A−1 (λv) =⇒ (A−1 A)v = λA−1 v =⇒ In v = λA−1 v =⇒ v = λA−1 v =⇒ v = A−1 v. λ 1 Therefore is an eigenvalue of A−1 provided that λ is an eigenvalue of A. λ 37. By assumption, we have Av = λv and B v = µv. (a) Therefore, (AB )v = A(B v) = A(µv) = µ(Av) = µ(λv) = (λµ)v, which shows that v is an eigenvector of AB with corresponding eigenvalue λµ. (b) Also, (A + B )v = Av + B v = λv + µv = (λ + µ)v, which shows that v is an eigenvector of A + B with corresponding eigenvalue λ + µ. 38. Recall that a matrix and its transpose have the same determinant. Thus, det(A − λIn ) = det([A − λIn ]T ) = det(AT − λIn ). Since A and AT have the same characteristic polynomial, it follows that both matrices also have the same eigenvalues. 39. (a) v = r + is is an eigenvector with eigenvalue λ = a + bi, b = 0 =⇒ Av = λv =⇒ A(r + is) = (a + bi)(r + is) = (ar − bs) + i(as + br) =⇒ Ar = ar − bs and As = as + br. Now if r = 0, then A0 = a0 − bs =⇒ 0 = 0 − bs =⇒ 0 = bs =⇒ s = 0 since b = 0. This would mean that v = 0 so v could not be an eigenvector. Thus, it must be that r = 0. Similarly, if s = 0, then r = 0, and again, this would contradict the fact that v is an eigenvector. Hence, it must be the case that r = 0 and s = 0. (b) As in part (a), Ar = ar − bs and As = as + br. Let c1 , c2 ∈ R. Then if c1 r + c2 s = 0, (39.1) 375 we have A(c1 r + c2 s) = 0 =⇒ c1 Ar + c2 As = 0 =⇒ c1 (ar − bs) + c2 (as + br) = 0. Hence, (c1 a + c2 b)r + (c2 a − c1 b)s = 0 =⇒ a(c1 r + c2 s) + b(c2 r − c1 s) = 0 =⇒ b(c2 r − c1 s) = 0 where we have used (39.1). Since b = 0, we must have c2 r − c1 s = 0. Combining this with (39.1) yields c1 = c2 = 0. Therefore, it follows that r and s are linearly independent vectors. 40. λ1 = 2, v = r(−1, 1). λ2 = 5, v = s(1, 2). 41. λ1 = −2 (multiplicity two), v = r(1, 1, 1). λ2 = −5, v = s(20, 11, 14). 42. λ1 = 3 (multiplicity two), v = r(1, 0, −1) + s(0, 1, −1). λ2 = 6, v = t(1, 1, 1). √ √ √ √ √ √ √ √ 43. λ1 = 3 − 6, v = r( 6, −1 + 6, −5 + 6). λ2 = 3 + 6, v = s( 6, 1 + 6, 5 + 6), λ3 = −2, v = t(−1, 3, 0). 44. λ1 = 0, v = r(2, 2, 1). λ2 = 3i, v = s(−4 − 3i, 5, −2 + 6i), λ3 = −3i, v = t(−4 + 3i, 5, −2 − 6i). 45. λ1 = −1 (multiplicity four), v = a(−1, 0, 0, 1, 0) + b(−1, 0, 1, 0, 0) + c(−1, 0, 0, 0, 1) + d(−1, 1, 0, 0, 0). Solutions to Section 5.7 True-False Review: 1. TRUE. This is the definition of a nondefective matrix. 2. TRUE. The eigenspace Eλ is equal to the null space of the n × n matrix A − λI , and this null space is a subspace of Rn . 3. TRUE. The dimension of an eigenspace never exceeds the algebraic multiplicity of the corresponding eigenvalue. 4. TRUE. Eigenvectors corresponding to distinct eigenspaces are linearly independent. Therefore if we choose one (nonzero) vector from each distinct eigenspace, the chosen vectors will form a linearly independent set. 5. TRUE. Since each eigenvalue of the matrix A occurs with algebraic multiplicity 1, we can simply choose one eigenvector from each eigenspace to obtain a basis of eigenvectors for A. Thus, A is nondefective. 6. FALSE. Many examples will show that this statement is false, including the n × n identity matrix In for n ≥ 2. The matrix In is not defective, and yet, has λ = 1 occurring with algebraic multiplicity n. 7. TRUE. Eigenvectors corresponding to distinct eigenvalues are always linearly independent, as proved in the text in this section. Problems: 1. det(A − λI ) = 0 ⇐⇒ 1−λ 2 4 3−λ = 0 ⇐⇒ λ2 − 4λ − 5 = 0 ⇐⇒ (λ − 5)(λ + 1) = 0 ⇐⇒ λ = 5 or λ = −1. −4 4 v1 0 = =⇒ v1 − v2 = 0. The solution 2 −2 v2 0 set of this system is {(r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = 5 is E1 = {v ∈ R2 : v = r(1, 1), r ∈ R}. A basis for E1 is {(1, 1)}, and dim[E1 ] = 1. 24 v1 0 If λ2 = −1 then (A − λI )v = 0 assumes the form = =⇒ v1 + 2v2 = 0. The solution 24 v2 0 set of this system is {(−2s, s) : s ∈ R}, so the eigenspace corresponding to λ2 = −1 is E2 = {v ∈ R2 : v = s(−2, 1), s ∈ R}. A basis for E2 is {(−2, 1)}, and dim[E2 ] = 1. A complete set of eigenvectors for A is given by {(1, 1), (−2, 1)}, so A is nondefective. If λ1 = 5 then (A − λI )v = 0 assumes the form 376 2. det(A − λI ) = 0 ⇐⇒ 3−λ 0 0 3−λ = 0 ⇐⇒ (3 − λ)2 = 0 ⇐⇒ λ = 3 of multiplicity two. 00 v1 0 = . 00 v2 0 The solution set of this system is {(r, s) : r, s ∈ R}, so the eigenspace corresponding to λ1 = 3 is E1 = {v ∈ R2 : v = r(1, 0) + s(0, 1), r, s ∈ R}. A basis for E1 is {(1, 0), (0, 1)}, and dim[E1 ] = 2. A is nondefective. If λ1 = 3 then (A − λI )v = 0 assumes the form 3. det(A − λI ) = 0 ⇐⇒ 1−λ −2 2 5−λ = 0 ⇐⇒ (λ − 3)2 = 0 ⇐⇒ λ = 3 of multiplicity two. −2 2 v1 0 = =⇒ v1 − v2 = 0. The solution −2 2 v2 0 set of this system is {(r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = 3 is E1 = {v ∈ R2 : v = r(1, 1), r ∈ R}. A basis for E1 is {(1, 1)}, and dim[E1 ] = 1. A is defective since it does not have a complete set of eigenvectors. If λ1 = 3 then (A − λI )v = 0 assumes the form 4. det(A − λI ) = 0 ⇐⇒ 5−λ −2 5 −1 − λ = 0 ⇐⇒ λ2 − 4λ + 5 = 0 ⇐⇒ λ = 2 ± i. 3+i 5 v1 0 = =⇒ −2v1 +(−3+i)v2 = −2 −3 + i v2 0 0. The solution set of this system is {((−3 + i)r, 2r) : r ∈ C}, so the eigenspace corresponding to λ1 = 2 − i is E1 = {v ∈ C2 : v = r(−3 + i, 2), r ∈ C}. A basis for E1 is {(−3 + i, 2)}, and dim[E1 ] = 1. If λ2 = 2 + i then from Theorem 5.6.8, the eigenvectors corresponding to λ2 = 2 + i are v = s(−3 − i, 2) where s ∈ C, so the eigenspace corresponding to λ2 = 2 + i is E2 = {v ∈ C2 : v = s(−3 − i, 2), s ∈ C}. A basis for E2 is {(−3 − i, 2)}, and dim[E2 ] = 1. A is nondefective since it has a complete set of eigenvectors, namely {(−3 + i, 2), (−3 − i, 2)}. If λ1 = 2 − i then (A−λI )v = 0 assumes the form 5. det(A − λI ) = 0 ⇐⇒ 3−λ 0 0 −4 −1 − λ −4 −1 −1 2−λ = 0 ⇐⇒ (λ + 2)(λ − 3)2 = 0 ⇐⇒ λ = −2 or λ = 3 of multiplicity two. 5 −4 −1 v1 0 1 −1 v2 = 0 . If λ1 = −2 then (A − λI )v = 0 assumes the form 0 0 −4 4 v3 0 The solution set of this system is {(r, r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = −2 is E1 = {v ∈ R3 : v = r(1, 1, 1), r ∈ R}. A basis for E1 is (1, 1, 1)}, and E1= 1. { dim[ ] 0 −4 −1 v1 0 If λ2 = 3 then (A − λI )v = 0 assumes the form 0 −4 −1 v2 = 0 =⇒ 4v2 + v3 = 0. The 0 −4 −1 v3 0 solution set of this system is {(s, t, −4t) : s, t ∈ R}, so the eigenspace corresponding to λ2 = 3 is E2 = {v ∈ R3 : v = s(1, 0, 0) + t(0, 1, −4), s, t ∈ R}. A basis for E2 is {(1, 0, 0), (0, 1, −4)}, and dim[E2 ] = 2. A complete set of eigenvectors for A is given by {(1, 1, 1), (1, 0, 0), (0, 1, −4)}, so A is nondefective. 6. det(A − λI ) = 0 ⇐⇒ 4−λ 0 0 0 2−λ −2 0 −3 1−λ = 0 ⇐⇒ (λ + 1)(λ − 4)2 = 0 ⇐⇒ λ = −1 or λ = 4 of multiplicity two. 5 0 0 v1 0 3 −3 v2 = 0 If λ1 = −1 then (A − λI )v = 0 assumes the form 0 0 −2 2 v3 0 377 =⇒ v1 = 0 and v2 − v3 = 0. The solution set of this system is {(0, r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = −1 is E1 = {v ∈ R3 : v = r(0, 1, 1), r ∈ R}. A basis for E1 is {(0, 1, 1)}, and dim[E1 ] = 1. 0 0 0 v1 0 If λ2 = 4 then (A − λI )v = 0 assumes the form 0 −2 −3 v2 = 0 =⇒ 2v2 + 3v3 = 0. 0 −2 −3 v3 0 The solution set of this system is {(s, 3t, −2t) : s, t ∈ R}, so the eigenspace corresponding to λ2 = 4 is E2 = {v ∈ R3 : v = s(1, 0, 0) + t(0, 3, −2), s, t ∈ R}. A basis for E2 is {(1, 0, 0), (0, −1, 1)}, and dim[E2 ] = 2. A complete set of eigenvectors for A is given by {(1, 0, 0), (0, 3, −2), (0, 1, 1)}, so A is nondefective. 7. det(A − λI ) = 0 ⇐⇒ 3−λ −1 0 1 5−λ 0 0 0 4−λ = 0 ⇐⇒ (λ − 4)3 = 0 ⇐⇒ λ = 4 of multiplicity three. −1 1 0 v1 0 If λ1 = 4 then (A − λI )v = 0 assumes the form −1 1 0 v2 = 0 000 v3 0 =⇒ v1 − v2 = 0 and v3 ∈ R. The solution set of this system is {(r, r, s) : r, s ∈ R}, so the eigenspace corresponding to λ1 = 4 is E1 = {v ∈ R3 : v = r(1, 1, 0) + s(0, 0, 1), r, s ∈ R}. A basis for E1 is {(1, 1, 0), (0, 0, 1)}, and dim[E1 ] = 2. A is defective since it does not have a complete set of eigenvectors. 8. det(A − λI ) = 0 ⇐⇒ 3−λ 2 1 0 0 −λ −4 4 −λ = 0 ⇐⇒ (λ − 3)(λ2 + 16) = 0 ⇐⇒ λ = 3 or λ = ±4i. 0 0 0 v1 0 If λ1 = 3 then (A − λI )v = 0 assumes the form 2 −3 −4 v2 = 0 1 4 −3 v3 0 =⇒ 11v1 − 25v3 = 0 and 11v2 − 2v3 . The solution set of this system is {(25r, 2r, 11r) : r ∈ C}, so the eigenspace corresponding to λ1 = 3 is E1 = {v ∈ C3 : v = r(25, 2, 11), r ∈ C}. A basis for E1 is {(25, 2, 11)}, and dim[E1 ] = 1. 3 + 4i 0 0 v1 0 2 4i −4 v2 = 0 If λ2 = −4i then (A − λI )v = 0 assumes the form 1 4 4i v3 0 =⇒ v1 = 0 and iv2 − v3 = 0. The solution set of this system is {(0, s, is) : s ∈ C}, so the eigenspace corresponding to λ2 = −4i is E2 = {v ∈ C3 : v = s(0, 1, i), s ∈ C}. A basis for E2 is {(0, 1, i)}, and dim[E2 ] = 1. If λ3 = 4i then from Theorem 5.6.8, the eigenvectors corresponding to λ3 = 4i are v = t(0, 1, −i) where t ∈ C, so the eigenspace corresponding to λ3 = 4i is E3 = {v ∈ C3 : v = t(0, 1, −i), t ∈ C}. A basis for E3 is {(0, 1, −i)}, and dim[E3 ] = 1. A complete set of eigenvectors for A is given by {(25, 2, 11), (0, 1, i), (0, 1, −i)}, so A is nondefective. 9. det(A − λI ) = 0 ⇐⇒ 4−λ −4 0 1 6 −λ −7 0 −3 − λ = 0 ⇐⇒ (λ + 3)(λ − 2)2 = 0 ⇐⇒ λ = −3 or λ = 2 of multiplicity two. 71 6 v1 0 −4 3 −7 v2 = 0 If λ1 = −3 then (A − λI )v = 0 assumes the form 00 0 v3 0 =⇒ v1 + v3 = 0 and v2 − v3 = 0. The solution set of this system is {(−r, r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = −3 is E1 = {v ∈ R3 : v = r(−1, 1, 1), r ∈ R}. A basis for E1 is {(−1, 1, 1)}, and 378 dim[E1 ] = 1. 2 1 6 v1 0 If λ2 = 2 then (A − λI )v = 0 assumes the form −4 −2 −7 v2 = 0 0 0 5 v3 0 =⇒ 2v1 + v2 = 0 and v3 = 0. The solution set of this system is {(−s, 2s, 0) : s ∈ R}, so the eigenspace corresponding to λ2 = 2 is E2 = {v ∈ R3 : v = s(−1, 2, 0), s ∈ R}. A basis for E2 is {(−1, 2, 0)}, and dim[E2 ] = 1. A is defective because it does not have a complete set of eigenvectors. 2−λ 0 0 0 0 = 0 ⇐⇒ (λ − 2)3 = 0 ⇐⇒ λ = 2 of multiplicity three. 2−λ 000 v1 0 If λ1 = 2 then (A − λI )v = 0 assumes the form 0 0 0 v2 = 0 . The solution set of this 000 v3 0 system is {(r, s, t) : r, s, t ∈ R}, so the eigenspace corresponding to λ1 = 2 is E1 = {v ∈ R3 : v = r(1, 0, 0) + s(0, 1, 0) + t(0, 0, 1), r, s, t ∈ R}. A basis for E1 is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and dim[E1 ] = 3. A is nondefective since it has a complete set of eigenvectors. 10. det(A − λI ) = 0 ⇐⇒ 0 2−λ 0 7−λ 8 0 −8 −9 − λ 0 6 6 = 0 ⇐⇒ (λ + 1)3 = 0 ⇐⇒ λ = −1 of multiplicity −1 − λ three. 8 −8 6 v1 0 If λ1 = −1 then (A − λI )v = 0 assumes the form 8 −8 6 v2 = 0 =⇒ 4v1 − 4v2 + 3v3 = 0. 0 00 v3 0 The solution set of this system is {(r − 3s, r, 4s) : r, s ∈ R}, so the eigenspace corresponding to λ1 = −1 is E1 = {v ∈ R3 : v = r(1, 1, 0) + s(−3, 0, 4), r, s ∈ R}. A basis for E1 is {(1, 1, 0), (−3, 0, 4)}, and dim[E1 ] = 2. A is defective since it does not have a complete set of eigenvectors. 11. det(A − λI ) = 0 ⇐⇒ −1 −1 = 0 ⇐⇒ (λ − 2)λ2 = 0 ⇐⇒ λ = 2 or λ = 0 of −1 − λ multiplicity two. 0 2 −1 v1 0 If λ1 = 2 then (A − λI )v = 0 assumes the form 2 −1 −1 v2 = 0 2 3 −3 v3 0 =⇒ 4v1 − 3v3 = 0 and 2v2 − v3 = 0. The solution set of this system is {(3r, 2r, 4r) : r ∈ R}, so the eigenspace corresponding to λ1 = 2 is E1 = {v ∈ R3 : v = r(3, 2, 4), r ∈ R}. A basis for E1 is {(3, 2, 4)}, and dim[E1 ] = 1. 2 2 −1 v1 0 If λ2 = 0 then (A − λI )v = 0 assumes the form 2 1 −1 v2 = 0 2 3 −1 v3 0 =⇒ 2v1 − v3 = 0 and v2 = 0. The solution set of this system is {(s, 0, 2s) : s ∈ R}, so the eigenspace corresponding to λ2 = 0 is E2 = {v ∈ R3 : v = s(1, 0, 2), s ∈ R}. A basis for E2 is {(1, 0, 2)}, and dim[E2 ] = 1. A is defective because it does not have a complete set of eigenvectors. 12. det(A − λI ) = 0 ⇐⇒ 2−λ 2 2 2 1−λ 3 379 1−λ 1 1 −1 −1 − λ −1 2 2 = 0 ⇐⇒ (λ − 2)λ2 = 0 ⇐⇒ λ = 2 or λ = 0 of 2−λ multiplicity two. −1 −1 2 v1 0 If λ1 = 2 then (A − λI )v = 0 assumes the form 1 −3 2 v2 = 0 1 −1 0 v3 0 =⇒ v1 − v3 = 0 and v2 − v3 = 0. The solution set of this system is {(r, r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = 2 is E1 = {v ∈ R3 : v = r(1, 1, 1), r ∈ R}. A basis for E1 is {(1, 1, 1)}, and dim[E1 ] = 1. 1 −1 2 v1 0 If λ2 = 0 then (A − λI )v = 0 assumes the form 1 −1 2 v2 = 0 =⇒ v1 − v2 + 2v3 = 0. 1 −1 2 v3 0 The solution set of this system is {(s − 2t, s, t) : s, t ∈ R}, so the eigenspace corresponding to λ2 = 0 is E2 = {v ∈ R3 : v = s(1, 1, 0) + t(−2, 0, 1), s, t ∈ R}. A basis for E2 is {(1, 1, 0), (−2, 0, 1)}, and dim[E2 ] = 2. A is nondefective because it has a complete set of eigenvectors. 13. det(A − λI ) = 0 ⇐⇒ 14. det(A − λI ) = 0 ⇐⇒ 2−λ −1 −2 3 0 −λ 1 −1 4 − λ = 0 ⇐⇒ (λ − 2)3 = 0 ⇐⇒ λ = 2 of multiplicity three. 0 30 v1 0 If λ1 = 2 then (A − λI )v = 0 assumes the form −1 −2 1 v2 = 0 −2 −1 2 v3 0 =⇒ v1 − v3 = 0 and v2 = 0. The solution set of this system is {(r, 0, r) : r ∈ R}, so the eigenspace corresponding to λ1 = 2 is E1 = {v ∈ R3 : v = r(1, 0, 1), r ∈ R}. A basis for E1 is {(1, 0, 1)}, and dim[E1 ] = 1. A is defective since it does not have a complete set of eigenvectors. 15. det(A − λI ) = 0 ⇐⇒ −λ −1 −1 −1 −λ −1 −1 −1 −λ = 0 ⇐⇒ (λ + 2)(λ − 1)2 = 0 ⇐⇒ λ = −2 or λ = 1 of multiplicity two. 2 −1 −1 v1 0 2 −1 v2 = 0 If λ1 = −2 then (A − λI )v = 0 assumes the form −1 −1 −1 2 v3 0 =⇒ v1 − v3 = 0 and v2 − v3 = 0. The solution set of this system is {(r, r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = −2 is E1 = {v ∈ R3 : v = r(1, 1, 1), r ∈ R}. A basis for E1 is {(1, 1, 1)}, and dim[E1 ] = 1. −1 −1 −1 v1 0 If λ2 = 1 then (A − λI )v = 0 assumes the form −1 −1 −1 v2 = 0 =⇒ v1 + v2 + v3 = 0. −1 −1 −1 v3 0 The solution set of this system is {(−s − t, s, t) : s, t ∈ R}, so the eigenspace corresponding to λ2 = 1 is E2 = {v ∈ R3 : v = s(−1, 1, 0) + t(−1, 0, 1), s, t ∈ R}. A basis for E2 is {(−1, 1, 0), (−1, 0, 1)}, and dim[E2 ] = 2. A is nondefective because it has a complete set of eigenvectors. 16. (λ − 4)(λ + 1) = 0 ⇐⇒ λ = 4 or λ = −1. Since A has two distinct eigenvalues, it has two linearly independent eigenvectors and is, therefore, nondefective. 17. (λ − 1)2 = 0 ⇐⇒ λ = 1 of multiplicity two. 380 5 5 v1 0 = =⇒ v1 + v2 = 0. The solution −5 −5 v2 0 set of this system is {(−r, r) : r ∈ R}, so the eigenspace corresponding to λ1 = 1 is E1 = {v ∈ R2 : v = r(−1, 1), r R}. A basis for E1 is {(1, 1)}, and dim[E1 ] = 1. A is defective since it does not have a complete set of eigenvectors. If λ1 = 1 then (A − λI )v = 0 assumes the form 18. λ2 − 4λ + 13 = 0 ⇐⇒ λ = 2 ± 3i. Since A has two distinct eigenvalues, it has two linearly independent eigenvectors and is, therefore, nondefective. 19. (λ − 2)2 (λ + 1) = 0 ⇐⇒ λ = −1 or λ = 2 of multiplicity two. To determine whether A is nondefective, all we require is the dimension of the eigenspacecorresponding λ = 2. to −1 −3 1 v1 0 If λ = 2 then (A − λI )v = 0 assumes the form −1 −3 1 v2 = 0 =⇒ −v1 − 3v2 + v3 = 0. −1 −3 1 v3 0 The solution set of this system is {(−3s + t, s, t) : s, t ∈ R}, so the eigenspace corresponding to λ = 2 is E = {v ∈ R3 : v = s(−3, 1, 0) + t(1, 0, 1), s, t ∈ R}. Since dim[E ] = 2, A is nondefective. 20. (λ − 3)3 = 0 ⇐⇒ λ = 3 of multiplicity three. −4 2 2 v1 0 If λ = 3 then (A − λI )v = 0 assumes the form −4 2 2 v2 = 0 =⇒ −2v1 + v2 + v3 = 0 and −4 2 2 v3 0 v2 = 0. The solution set of this system is {(r, 2r − s, s) : r, s ∈ R}, so the eigenspace corresponding to λ = 3 is E = {v ∈ R3 : v = r(1, 2, 0) + s(0, −1, 1), r, s ∈ R}. A is defective since it does not have a complete set of eigenvectors. 21. det(A − λI ) = 0 ⇐⇒ 2−λ 3 1 4−λ = 0 ⇐⇒ (λ − 1)(λ − 5) = 0 ⇐⇒ λ = 1 or λ = 5. 11 v1 0 = =⇒ v1 + v2 = 0. The eigenspace 33 v2 0 corresponding to λ1 = 1 is E1 = {v ∈ R2 : v = r(−1, 1), r ∈ R}. A basis for E1 is {(−1, 1)}. −3 1 v1 0 If λ2 = 5 then (A − λI )v = 0 assumes the form = =⇒ 3v1 − v2 = 0. The 3 −1 v2 0 eigenspace corresponding to λ2 = 5 is E2 = {v ∈ R2 : v = s(1, 3), s ∈ R}. A basis for E2 is {(1, 3)}. If λ1 = 1 then (A − λI )v = 0 assumes the form v2 3 2 1 E1 E2 v1 -2 -1 -1 1 2 Figure 0.0.73: Figure for Exercise 21 22. det(A − λI ) = 0 ⇐⇒ 2−λ 0 3 2−λ = 0 ⇐⇒ (2 − λ)2 = 0 ⇐⇒ λ = 2 of multiplicity two. 381 03 v1 0 = =⇒ v1 ∈ R and v2 = 0. The 00 v2 0 eigenspace corresponding to λ1 = 2 is E1 = {v ∈ R2 : v = r(1, 0), r ∈ R}. A basis for E1 is {(1, 0)}. If λ1 = 2 then (A − λI )v = 0 assumes the form v2 E1 v1 Figure 0.0.74: Figure for Exercise 22 23. det(A − λI ) = 0 ⇐⇒ 5−λ 0 0 5−λ = 0 ⇐⇒ (5 − λ)2 = 0 ⇐⇒ λ = 5 of multiplicity two. 00 v1 0 = =⇒ v1 , v2 ∈ R. The eigenspace 00 v2 0 corresponding to λ1 = 5 is E1 = {v ∈ R2 : v = r(1, 0) + s(0, 1), r, s ∈ R}. A basis for E1 is {(1, 0), (0, 1)}. If λ1 = 5 then (A − λI )v = 0 assumes the form v2 E1 is the whole of R2 v1 Figure 0.0.75: Figure for Exercise 23 24. det(A − λI ) = 0 ⇐⇒ multiplicity two. 3−λ 1 −1 1 3−λ −1 −1 −1 3−λ = 0 ⇐⇒ (λ − 5)(λ − 2)2 = 0 ⇐⇒ λ = 5 or λ = 2 of −2 1 −1 v1 0 If λ1 = 5 then (A − λI )v = 0 assumes the form 1 −2 −1 v2 = 0 −1 −1 −2 v3 0 =⇒ v1 + v3 = 0 and v2 + v3 = 0. The eigenspace corresponding to λ1 = 5 is E1 = {v ∈ R3 : v = r(1, 1, −1), r ∈ R}. A basis for E1 is {(1, 1, −1)}. 1 1 −1 v1 0 1 −1 v2 = 0 =⇒ v1 + v2 − v3 = 0. If λ2 = 2 then (A − λI )v = 0 assumes the form 1 −1 −1 1 v3 0 The eigenspace corresponding to λ2 = 2 is E2 = {v ∈ R3 : v = s(−1, 1, 0) + t(1, 0, 1), s, t ∈ R}. A basis for E2 is {(−1, 1, 0), (1, 0, 1)}. 382 v3 E2 (1, 0, 1) (-1, 1, 0) v2 E1 V1 (1, 1, -1) Figure 0.0.76: Figure for Exercise 24 25. det(A − λI ) = 0 ⇐⇒ −3 − λ −1 0 1 −1 − λ 0 three. 0 2 −2 − λ = 0 ⇐⇒ (λ + 2)3 = 0 ⇐⇒ λ = −2 of multiplicity −1 1 0 v1 0 If λ1 = −2 then (A − λI )v = 0 assumes the form −1 1 2 v2 = 0 =⇒ v1 − v2 = 0 and 000 v3 0 v3 = 0. The eigenspace corresponding to λ1 = −2 is E1 = {v ∈ R3 : v = r(1, 1, 0), r ∈ R}. A basis for E1 is {(1, 1, 0)}. v3 E1 v2 V1 (1, 1, 0) Figure 0.0.77: Figure for Exercise 25 1 −2 3 v1 0 26. (a) If λ1 = 1 then (A − λI )v = 0 assumes the form 1 −2 3 v2 = 0 1 −2 3 v3 0 =⇒ v1 − 2v2 + 3v3 = 0. The eigenspace corresponding to λ1 = 1 is E1 = {v ∈ R3 : v = r(2, 1, 0) + s(−3, 0, 1), r, s ∈ R}. A basis for E1 is {(2, 1, 0), (−3, 0, 1)}. Now apply the Gram-Schmidt process where v1 = (−3, 0, 1), and v2 = (2, 1, 0). Let u1 = v1 so that v2 , u1 = (2, 1, 0), (−3, 0, 1) = 2(−3) + 1 · 0 + 0 · 1 = −6 and ||u1 ||2 = (−3)2 + 02 + 12 = 10. v2 , u1 6 1 u2 = v2 − u1 = (2, 1, 0) + (−3, 0, 1) = (1, 5, 3). 2 ||u1 || 10 5 Thus, {(−3, 0, 1), (1, 5, 3)} is an orthogonal basis for E1 . −1 −2 3 v1 0 (b) If λ2 = 3 then (A − λI )v = 0 assumes the form 1 −4 3 v2 = 0 =⇒ v1 − v2 = 0 and 1 −2 1 v3 0 383 v2 − v3 = 0. The eigenspace corresponding to λ2 = 3 is E2 = {v ∈ R3 : v = r(1, 1, 1), r ∈ R}. A basis for E2 is {(1, 1, 1)}. To determine the orthogonality of the vectors, consider the following inner products: (−3, 0, 1), (1, 1, 1) = −3 + 0 + 1 = −2 = 0 and (1, 5, 3), (1, 1, 1) = 1 + 5 + 3 = 9 = 0. Thus, the vectors in E1 are not orthogonal to the vectors in E2 . −1 −1 1 v1 0 1 v2 = 0 27. (a) If λ1 = 2 then (A − λI )v = 0 assumes the form −1 −1 1 1 −1 v3 0 =⇒ v1 + v2 − v3 = 0. The eigenspace corresponding to λ1 = 2 is E1 = {v ∈ R3 : v = r(−1, 1, 0) + s(1, 0, 1), r, s ∈ R}. A basis for E1 is {(−1, 1, 0), (1, 0, 1)}. Now apply the Gram-Schmidt process where v1 = (1, 0, 1), and v2 = (−1, 1, 0). Let u1 = v1 so that v2 , u1 = (−1, 1, 0), (1, 0, 1) = −1 · 1 + 1 · 0 + 0 · 1 = −1 and ||u1 ||2 = 12 + 02 + 12 = 2. 1 1 v2 , u1 u1 = (−1, 1, 0) + (1, 0, 1) = (−1, 2, 1). u2 = v2 − ||u1 ||2 2 2 Thus, {(1, 0, 1), (−1, 2, 1)} is an orthogonal basis for E1 . 2 −1 1 v1 0 2 1 v2 = 0 . The eigenspace (b) If λ2 = −1 then (A − λI )v = 0 assumes the form −1 1 12 v3 0 corresponding to λ2 = −1 is E2 = {v ∈ R3 : v = r(−1, −1, 1), r ∈ R}. A basis for E2 is {(−1, −1, 1)}. To determine the orthogonality of the vectors, consider the following inner products: (1, 0, 1), (−1, −1, 1) = −1 + 0 + 1 = 0 and (−1, 2, 1), (−1, −1, 1) = 1 − 2 + 1 = 0. Thus, the vectors in E1 are orthogonal to the vectors in E2 . 28. We are given that the eigenvalues of A are λ1 = 0 (multiplicity two), and λ2 = a + b + c. cases to consider: λ1 = λ2 or λ1 = λ2 . ab If λ1 = λ2 then λ1 = 0 is of multiplicity three, and (A − λI )v = 0 assumes the form a b ab 0 0 , or equivalently, 0 av1 + bv2 + cv3 = 0. There are two c v1 c v2 = c v3 (28.1) The only way to have three linearly independent eigenvectors for A is if a = b = c = 0. If λ1 = λ2 then λ1 = 0 is of multiplicity two, and λ2 = a + b + c = 0 are distinct eigenvalues. By Theorem 5.7.11, E1 must have dimension two for A to possess a complete set of eigenvectors. The system for determining the eigenvectors corresponding to λ1 = 0 is once more given by (28.1). Since we can choose two variables freely in (28.1), it follows that there are indeed two corresponding linearly independent eigenvectors. Consequently, A is nondefective in this case. 29. (a) Setting λ = 0 in (5.7.4), we have p(λ) = det(A − λI ) = det(A), and in (5.7.5), we have p(λ) = p(0) = bn . Thus, bn = det(A). The value of det(A − λI ) is the sum of products of its elements, one taken from each row and each column. Expanding det(A − λI ) yields equation (5.7.5). The expression involving λn in p(λ) comes from the product, n (aii − λ), of the diagonal elements. All the remaining products of the determinant have degree not higher i=1 than n − 2, since, if one of the factors of the product is aij , where i = j , then this product cannot contain 384 the factors λ − aii and λ ajj . Hence, n (aii − λ) + (terms of degree not higher than n − 2) p(λ) = i=1 so, p(λ) = (−1)n λn + (−1)n−1 (a11 + a22 + · · · + ann )λn−1 + · · · + an . Equating like coefficients from (5.7.5), it follows that b1 = (−1)n−1 (a11 + a22 + · · · + ann ). n i=1 n therefore bn = n (λi − 0) or p(0) = (b) Letting λ = 0, we have from (5.7.6) that p(0) = λi , but from (5.7.5), p(0) = bn , i=1 λi . Letting λ = 1, we have from (5.7.6) that i=1 n n (λ − λi ) = (−1)n [λn − (λ1 + λ2 + · · · + λn )λn−1 + · · · + bn ]. (λi − λ) = (−1)n p(λ) = i=1 i=1 Equating like coefficients with (5.7.5), it follows that b1 = (−1)n−1 (λ1 + λ2 + · · · + λn ). (c) From (a), bn = det(A), and from (b), det(A) = λ1 λ2 · · · λn , so det(A) is the product of the eigenvalues of A. From (a), b1 = (−1)n−1 (a11 + a22 + · · · + ann ) and from (b), b1 = (−1)n−1 (λ1 + λ2 + · · · + λn ), thus a11 + a22 + · · · + ann = λ1 + λ2 + · · · + λn . That is, tr(A) is the sum of the eigenvalues of A. 30. (a) We have det(A) = 19 and tr(A) = 3, so the product of the eigenvalues of A is 19, and the sum of the eigenvalues of A is 3. (b) We have det(A) = −69 and tr(A) = 1, so the product of the eigenvalues of A is -69, and the sum of the eigenvalues of A is 1. (c) We have det(A) = −607 and tr(A) = 24, so the product of the eigenvalues of A is -607, and the sum of the eigenvalues of A is 24. 31. Note that Ei = ∅ since 0 belongs to Ei . Closure under Addition: Let v1 , v2 ∈ Ei . Then A(v1 + v2 ) = Av1 + Av2 = λi v1 + λi v2 = λi (v1 + v2 ) =⇒ v1 + v2 ∈ Ei . Closure under Scalar Multiplication: Let c ∈ C and v1 ∈ Ei . Then A(cv1 ) = c(Av1 ) = c(λi v1 ) = λi (cv1 ) =⇒ cv1 ∈ Ei . Thus, by Theorem 4.3.2, Ei is a subspace of C n . 32. The condition c1 v1 + c2 v2 = 0 (32.1) =⇒ A(c1 v1 ) + A(c2 v2 ) = 0 =⇒ c1 Av1 + c2 Av2 = 0 =⇒ c1 (λ1 v1 ) + c2 (λ2 v2 ) = 0. (32.2) Substituting c2 v2 from (32.1) into (32.2) yields (λ1 − λ2 )c1 v1 = 0. Since λ2 = λ1 , we must have c1 v1 = 0, but v1 = 0, so c1 = 0. Substituting into (32.1) yields c2 = 0 also. Consequently, v1 and v2 are linearly independent. 385 33. Consider c1 v1 + c2 v2 + c3 v3 = 0. (33.1) If c1 = 0, then the preceding equation can be written as w1 + w2 = 0, where w1 = c1 v1 and w2 = c2 v2 + c3 v3 . But this would imply that {w1 , w2 } is linearly dependent, which would contradict Theorem 5.7.5 since w1 and w2 are eigenvectors corresponding to different eigenvalues. Consequently, we must have c1 = 0. But then (33.1) implies that c2 = c3 = 0 since {v1 , v2 } is a linearly independent set by assumption. Hence {v1 , v2 , v3 } is linearly independent. 34. λ1 = 1 (multiplicity 3), basis: {(0, 1, 1)}. 35. λ1 = 0 (multiplicity 2), basis: {(−1, 1, 0), (−1, 0, 1)}. λ2 = 3, basis: {(1, 1, 1)}. √ √ √ 36. λ1 = 2, basis: {(1, −2 2, 1)}. λ2 = 0, basis: {(1, 0, −1)}. λ3 = 7, basis: {( 2, 1, 2)}. 37. λ1 = −2, basis: {(2, 1, −4)}. λ2 = 3 (multiplicity 2), basis: {(0, 2, 1), (3, 11, 0)}. 38. λ1 = 0 (multiplicity 2), basis: {(0, 1, 0, −1), (1, 0, −1, 0)}. λ2 = 6, basis: {(1, 1, 1, 1)}. λ3 = −2, basis: {1, −1, 1, −1)}. 39. A has eigenvalues: λ1 = 1 3 a+ 2 2 a2 + 8b2 , λ2 = 3 1 a− 2 2 a2 + 8b2 , λ3 = 0. Provided a = ±b, these eigenvalues are distinct, and therefore the matrix is nondefective. If a = b = 0, then the eigenvalue λ = 0 has multiplicity two. A basis for the corresponding eigenspace is {(−1, 0, 1), (−1, 1, 0)}. Since this is two-dimensional, the matrix is nondefective in this case. If a = −b = 0, then the eigenvalue λ = 0 once more has multiplicity two. A basis for the corresponding eigenspace is {(0, 1, 1), (1, 1, 0)}, therefore the matrix is nondefective in this case also. If a = b = 0, then A = 02 , so that λ = 0 (multiplicity three), and the corresponding eigenspace is all of R3 . Hence A is nondefective. 40. If a = b = 0, then A = 03 , which is nondefective. We now assume that at least one of either a or b is nonzero. A has eigenvalues: λ1 = b, λ2 = a − b, and λ3 = 3a + b. Provided a = 0, 2b, −b, the eigenvalues are distinct, and therefore A is nondefective. If a = 0, then the eigenvalue λ = b has multiplicity two. In this case, a basis for the corresponding eigenspace is {(1, 0, 1), (0, 1, 0)}, so that A is nondefective. If a = 2b, then the eigenvalue λ = b has multiplicity two. In this case, a basis for the corresponding eigenspace is {(−2, 1, 0), (−1, 0, 1)}, so that A is nondefective. If a = −b, then the eigenvalue λ = −2b has multiplicity two. In this case, a basis for the corresponding eigenspace is {(0, 1, 1), (1, 0, −1)}, so that A is nondefective. Solutions to Section 5.8 True-False Review: 386 1. TRUE. The terms “diagonalizable” and “nondefective” are synonymous. The diagonalizability of a matrix A hinges on the ability to form an invertible matrix S with a full set of linearly independent eigenvectors of the matrix as its columns. This, in turn, requires the original matrix to be nondefective. 2. TRUE. If we assume that A is diagonalizable, then there exists an invertible matrix S and a diagonal matrix D such that S −1 AS = D. Since A is invertible, we can take the inverse of each side of this equation to obtain D−1 = (S −1 AS )−1 = S −1 A−1 S, and since D−1 is still a diagonal matrix, this equation shows that A−1 is diagonalizable. 1 0 multiplicity 2). However, A and B are not similar. [Reason: If for some invertible matrix S , but since A = I2 , this would imply above.] 3. FALSE. For instance, the matrices A = I2 and B = 1 both have eigenvalue λ = 1 (with 1 A and B were similar, then S −1 AS = B that B = I2 , contrary to our choice of B 4. FALSE. An n × n matrix is diagonalizable if and only if it has n linearly independent eigenvectors. Besides, every matrix actually has infinitely many eigenvectors, obtained by taking scalar multiples of a single eigenvector v. 5. TRUE. Assume A is an n × n matrix such that p(λ) = det(A − λI ) has no repeated roots. This implies that A has n distinct eigenvalues. Corresponding to each eigenvalue, we can select an eigenvector. Since eigenvectors corresponding to distinct eigenvalues are linearly independent, this yields n linearly independent eigenvectors for A. Therefore, A is nondefective, and hence, diagonalizable. 6. TRUE. Assuming that A is diagonalizable, then there exists an invertible matrix S and a diagonal matrix D such that S −1 AS = D. Therefore, D2 = (S −1 AS )2 = (S −1 AS )(S −1 AS ) = S −1 ASS −1 AS = S −1 A2 S. Since D2 is still a diagonalizable matrix, this equation shows that A2 is diagonalizable. − 7. TRUE. Since In 1 AIn = A, A is similar to itself. 8. TRUE. The sum of the dimensions of the eigenspaces of such a matrix is even, and therefore not equal to n. This means we cannot obtain n linearly independent eigenvectors for A, and therefore, A is defective (and not diagonalizable). Problems: −1 − λ −2 = 0 ⇐⇒ λ2 − λ − 6 = 0 ⇐⇒ (λ − 3)(λ + 2) = 0 ⇐⇒ λ = 3 or −2 2−λ λ = −2. A is diagonalizable because it has two distinct eigenvalues. −4 −2 v1 0 If λ1 = 3 then (A − λI )v = 0 assumes the form = =⇒ 2v1 + v2 = 0. If we let −2 −1 v2 0 v1 = r ∈ R, then the solution set of this system is {(−r, 2r) : r ∈ R}, so the eigenvectors corresponding to λ1 = 3 are v1 = r(−1, 2) where r ∈ R. 1 −2 v1 0 If λ2 = −2 then (A − λI )v = 0 assumes the form = =⇒ v1 − 2v2 = 0. If we let −2 4 v2 0 v2 = s ∈ R, then the solution set of this system is {(2s, s) : s ∈ R}, so the eigenvectors corresponding to λ2 = −2 are v2 = s(2, 1) where s ∈ R. −1 2 Thus, the matrix S = satisfies S −1 AS = diag(3, −2). 21 1. det(A − λI ) = 0 ⇐⇒ 387 2. det(A − λI ) = 0 ⇐⇒ −7 − λ −4 4 1−λ = 0 ⇐⇒ λ2 + 6λ + 9 = 0 ⇐⇒ (λ + 3)2 = 0 ⇐⇒ λ = −3 of multiplicity two. −4 4 v1 0 = =⇒ v1 − v2 = 0. If we let −4 4 v2 0 v1 = r ∈ R, then the solution set of this system is {(r, r) : r ∈ R}, so the eigenvectors corresponding to λ = −3 are v = r(1, 1) where r ∈ R. A has only one linearly independent eigenvector, so by Theorem 5.8.4, A is not diagonalizable. If λ = −3 then (A − λI )v = 0 assumes the form 3. det(A − λI ) = 0 ⇐⇒ 1−λ 2 −8 −7 − λ = 0 ⇐⇒ λ2 + 6λ + 9 = 0 ⇐⇒ (λ + 3)2 = 0 ⇐⇒ λ = −3 of multiplicity two. 4 −8 v1 0 = =⇒ v1 − 2v2 = 0. If we let 2 −4 v2 0 v2 = r ∈ R, then the solution set of this system is {(2r, r) : r ∈ R}, so the eigenvectors corresponding to λ = −3 are v = r(2, 1) where r ∈ R. A has only one linearly independent eigenvector, so by Theorem 5.8.4, A is not diagonalizable. If λ = −3 then (A − λI )v = 0 assumes the form 4. det(A − λI ) = 0 ⇐⇒ −λ 4 −4 −λ = 0 ⇐⇒ λ2 + 16 = 0 ⇐⇒ λ = ±4i. A is diagonalizable because it has two distinct eigenvalues. 4i 4 v1 0 = =⇒ v1 − iv2 = 0. If we let −4 4i v2 0 v2 = r ∈ C, then the solution set of this system is {(ir, r) : r ∈ C}, so the eigenvectors corresponding to λ = −4i are v = r(i, 1) where r ∈ C. Since the entries of A are real, it follows from Theorem 5.6.8 that v2 = (−i, 1) is an eigenvector corresponding to λ = 4i. i −i Thus, the matrix S = satisfies S −1 AS = diag(−4i, 4i). 1 1 If λ = −4i then (A − λI )v = 0 assumes the form 5. det(A − λI ) = 0 ⇐⇒ λ = 4. 1−λ 0 1 0 3−λ 1 0 7 −3 − λ = 0 ⇐⇒ (1 − λ)(λ + 4)(λ − 4) = 0 ⇐⇒ λ = 1, λ = −4 or 00 0 v1 0 7 v2 = 0 If λ = 1 then (A − λI )v = 0 assumes the form 0 2 1 1 −4 v3 0 =⇒ 2v1 − 15v3 = 0 and 2v2 + 7v3 = 0. If we let v3 = 2r where r ∈ R, then the solution set of this system is {(15r, −7r, 2r) : r ∈ R} so the eigenvectors corresponding to λ = 1 are v1 r(15, −7, 2) where r ∈ R. = 500 v1 0 If λ = −4 then (A − λI )v = 0 assumes the form 0 7 7 v2 = 0 111 v3 0 =⇒ v1 = 0 and v2 + v3 = 0. If we let v2 = s ∈ R, then the solution set of this system is {(0, s, −s) : s ∈ R} so the eigenvectors corresponding to λ = −4 are v2 = s(0, 1, −1) where s ∈ R. −3 0 0 v1 0 0 −1 7 v2 = 0 If λ = 4 then (A − λI )v = 0 assumes the form 1 1 −7 v3 0 =⇒ v1 = 0 and v2 − 7v3 = 0. If we let v3 = t ∈ R, then the solution set of this system is {(0, 7t, t) : t ∈ R} so the eigenvectors corresponding to λ = 4 are v3 = t(0, 7, 1) where t ∈ R. 388 15 0 0 1 satisfies S −1 AS = diag(1, −4, 4). Thus, the matrix S = −7 7 2 1 −1 6. det(A − λI ) = 0 ⇐⇒ 1−λ 2 2 −2 −3 − λ −2 7. det(A − λI ) = 0 ⇐⇒ −λ −2 −2 −2 −λ −2 −2 −2 −λ 0 0 = 0 ⇐⇒ (λ + 1)3 = 0 ⇐⇒ λ = −1 of multiplicity three. −2 − λ 2 −2 0 v1 0 If λ = −1 then (A − λI )v = 0 assumes the form 2 −2 0 v2 = 0 =⇒ v1 − v2 = 0 and 2 −2 0 v3 0 v3 ∈ R. If we let v2 = r ∈ R and v3 = s ∈ R, then the solution set of this system is {(r, r, s) : r, s ∈ R}, so the eigenvectors corresponding to λ = −1 are v1 = r(1, 1, 0) and v2 = s(0, 0, 1) where r, s ∈ R. A has only two linearly independent eigenvectors, so by Theorem 5.8.4, A is not diagonalizable. two. = 0 ⇐⇒ (λ − 2)2 (λ + 4) = 0 ⇐⇒ λ = −4 or λ = 2 of multiplicity 4 −2 −2 v1 0 4 −2 v2 = 0 If λ = −4 then (A − λI )v = 0 assumes the form −2 −2 −2 4 v3 0 =⇒ v1 − v3 = 0 and v2 − v3 = 0. If we let v3 = r ∈ R, then the solution set of this system is {(r, r, r) : r ∈ R} so the eigenvectors corresponding to λ = −4 are v1 = r(1, 1, 1) where r ∈ R. −2 −2 −2 v1 0 If λ = 2 then (A − λI )v = 0 assumes the form −2 −2 −2 v2 = 0 =⇒ v1 + v2 + v3 = 0. If −2 −2 −2 v3 0 we let v2 = s ∈ R and v3 = t ∈ R, then the solution set of this system is {(−s − t, s, t) : s, t ∈ R}, so two linearly independent eigenvectors corresponding to λ = 2 are v2 = s(−1, 1, 0) and v3 = t(−1, 0, 1). 1 −1 −1 0 1 satisfies S −1 AS = diag(−4, 2, 2). Thus, the matrix S = 1 1 1 0 −2 − λ −2 −2 4 4 = 0 ⇐⇒ λ2 (λ − 3) = 0 ⇐⇒ λ = 3, or λ = 0 of multiplicity 4−λ two. −5 14 v1 0 If λ = 3 then (A − λI )v = 0 assumes the form −2 −2 4 v2 = 0 −2 11 v3 0 =⇒ v1 − v3 = 0 and v2 − v3 = 0. If we let v3 = r ∈ R, then the solution set of this system is {(r, r, r) : r ∈ R}, so the eigenvectors corresponding to λ = 3 are 1 = r(1, 1, 1)where r ∈ R. v −2 1 4 v1 0 If λ = 0 then (A − λI )v = 0 assumes the form −2 1 4 v2 = 0 =⇒ −2v1 + v2 + 4v3 = 0. If −2 1 4 v3 0 we let v1 = s ∈ R and v3 = t ∈ R, then the solution set of this system is {(s, 2s − 4t, t) : s, t ∈ R}, so two linearly independent eigenvectors corresponding to λ = 0 are v2 = s(1, 2, 0) and v3 = t(0, −4, 1). 11 0 Thus, the matrix S = 1 2 −4 satisfies S −1 AS = diag(3, 0, 0). 10 1 8. det(A − λI ) = 0 ⇐⇒ 1 1−λ 1 389 9. det(A − λI ) = 0 ⇐⇒ 2−λ 0 2 0 1−λ −1 0 0 1−λ = 0 ⇐⇒ (λ − 2)(λ − 1)2 = 0 ⇐⇒ λ = 2 or λ = 1 of multiplicity two. 1 00 v1 0 0 0 v2 = 0 =⇒ v1 = v2 = 0 and If λ = 1 then (A − λI )v = 0 assumes the form 0 2 −1 0 v3 0 v3 ∈ R. If we let v3 = r ∈ R then the solution set of this system is {(0, 0, r) : r ∈ R}, so there is only one corresponding linearly independent eigenvector. Hence, by Theorem 5.8.4, A is not diagonalizable. 10. det(A − λI ) = 0 ⇐⇒ 4−λ 3 0 11. det(A − λI ) = 0 ⇐⇒ −λ 2 −1 −2 −λ −2 1 2 −λ 0 −1 − λ 2 0 −1 = 0 ⇐⇒ (λ2 + 1)(λ − 4) = 0 ⇐⇒ λ = 4, or λ = ±i. 1−λ 0 0 0 v1 0 If λ = 4 then (A − λI )v = 0 assumes the form 3 −5 −1 v2 = 0 0 2 −3 v3 0 =⇒ 6v1 − 17v3 = 0 and 2v2 − 3v3 = 0. If we let v3 = 6r ∈ C, then the solution set of this system is {(17r, 9r, 6r) : r ∈ C}, so the eigenvectors corresponding to λ = 4 are v1 r(17, , 6) r ∈ C. = 9 where 4−i 0 0 v1 0 −1 − i −1 v2 = 0 =⇒ v1 = 0 and If λ = i then (A − λI )v = 0 assumes the form 3 0 2 1−i v3 0 2v2 + (1 − i)v3 = 0. If we let v3 = −2s ∈ C, then the solution set of this system is {(0, (1 − i)s, 2s) : s ∈ C}, so the eigenvectors corresponding to λ = i are v2 = s(0, 1 − i, 2) where s ∈ C. Since the entries of A are real, v3 = t(0, 1 + i, 2) where t ∈ C are the eigenvectors corresponding to λ = −i by Theorem 5.6.8. 17 0 0 Thus, the matrix S = 9 1 − i 1 + i satisfies S −1 AS = diag(4, i, −i). 6 2 2 = 0 ⇐⇒ λ(λ2 + 9) = 0 ⇐⇒ λ = 0, or λ = ±3i. 0 2 −1 v1 0 −2 0 −2 v2 = 0 If λ = 0 then (A − λI )v = 0 assumes the form 12 0 v3 0 =⇒ v1 + v3 = 0 and 2v2 − v3 = 0. If we let v3 = 2r ∈ C, then the solution set of this system is {(−2r, r, 2r) : r ∈ C}, so the eigenvectors corresponding to λ = 0 are v1 = (−2, 1, 2) where r ∈ C. r 3i 2 −1 v1 0 If λ = −3i then (A − λI )v = 0 assumes the form −2 3i −2 v2 = 0 12 3i v3 0 =⇒ 5v1 +(−4+3i)v3 = 0 and 5v2 +(2+6i)v3 = 0. If we let v3 = 5s ∈ C, then the solution set of this system is {((4 − 3i)s, (−2 − 6i)s, 5s) : s ∈ C}, so the eigenvectors corresponding to λ = −3i are v2 = s(4 − 3i, −2 − 6i, 5) where s ∈ C. Since the entries of A are real, v3 = t(4 + 3i, −2 + 6i, 5) where t ∈ C are the eigenvectors corresponding to λ = 3i by Theorem 5.6.8. −2 4 + 3i 4 − 3i Thus, the matrix S = 1 −2 + 6i −2 − 6i satisfies S −1 AS = diag(0, 3i, −3i). 2 5 5 12. det(A − λI ) = 0 ⇐⇒ 1−λ −2 0 −2 1−λ 0 0 0 3−λ = 0 ⇐⇒ (λ − 3)2 (λ + 1) = 0 ⇐⇒ λ = 1 or λ = 3 of 390 multiplicity two. 2 −2 0 v1 0 2 0 v2 = 0 If λ = −1 then (A − λI )v = 0 assumes the form −2 0 04 v3 0 =⇒ 2v1 − v2 = 0 and v3 = 0. If we let v1 = r ∈ R, then the solution set of this system is {(r, r, 0) : r ∈ R} so the eigenvectors corresponding to λ = −1 are v1 = r(1, 1, 0) where r ∈ . R −2 −2 0 v1 0 If λ = 3 then (A − λI )v = 0 assumes the form −2 −2 0 v2 = 0 =⇒ v1 + v2 = 0 and 0 00 v3 0 v3 ∈ R. If we let v2 = s ∈ R and v3 = t ∈ R, then the solution set of this system is {(−s, s, t) : s, t ∈ R}, so the eigenvectors corresponding to λ 3 are v2 = s(−1, 1, 0) and v3 = t(0, 0, 1). = 1 −1 0 1 0 satisfies S −1 AS = diag(−1, 3, 3). Thus, the matrix S = 1 0 01 13. λ1 = 2 (multiplicity 2), basis for eigenspace: {(−3, 1, 0), (3, 0, 1)}. λ2 = 1, basis for eigenspace: {(1, 2, 2)}. −3 3 3 Set S = 1 0 2 . Then S −1 AS = diag(2, 2, 1). 012 14. λ1 = 0 (multiplicity 2), basis for eigenspace: {(0, 1, 0, −1), (1, 0, −1, 0)}. λ2 = 2, basis for eigenspace: (1, 1, 1, 1)}. λ3 = 10, basis for eigenspace: {(−1, 1, −1, 1)}. { 0 1 1 −1 1 01 1 −1 Set S = 0 −1 1 −1 . Then S AS = diag(0, 0, 2, 10). −1 01 1 14 . 23 A has eigenvalues λ1 = −1, λ2 = 5 with corresponding linearly independent eigenvectors v1 = (−2, 1) and −2 1 v2 = (1, 1). If we set S = , then S −1 AS = diag(−1, 5), therefore, under the transformation 11 x = S y, the given system of differential equations simplifies to 15. The given system can be written as x = Ax, where A = y1 y2 = −1 0 05 y1 y2 . Hence, y1 = −y1 and y2 = 5y2 . Integrating these equations, we obtain y1 (t) = c1 e−t , y2 (t) = c2 e5t . Returning to the original variables, we have x = Sy = −2 1 1 1 c1 e−t c2 e5t = −2c1 e−t + c2 e5t c1 e−t + c2 e5t . Consequently, x1 (t) = −2c1 e−t + c2 e5t and x2 (t) = c1 e−t + c2 e5t . 6 −2 . −2 6 A has eigenvalues λ1 = 4, λ2 = 8 with corresponding linearly independent eigenvectors v1 = (1, 1) and 16. The given system can be written as x = Ax, where A = 391 1 −1 , then S −1 AS = diag(4, 8), therefore, under the transformation 1 1 x = S y, the given system of differential equations simplifies to v2 = (−1, 1). If we set S = y1 y2 = 4 0 0 8 y1 y2 . Hence, y1 = 4y1 and y2 = 8y2 . Integrating these equations, we obtain y1 (t) = c1 e4t , y2 (t) = c2 e8t . Returning to the original variables, we have x = Sy = c1 e4t c2 e8t 1 −1 1 1 = c1 e4t − c2 e8t c1 e4t + c2 e8t . Consequently, x1 (t) = c1 e4t − c2 e8t and x2 (t) = c1 e4t + c2 e8t . 9 6 . −10 −7 A has eigenvalues λ1 = −1, λ2 = 3 with corresponding linearly independent eigenvectors v1 = (3, −5) and 3 −1 v2 = (−1, 1). If we set S = , then S −1 AS = diag(−1, 3), therefore, under the transformation −5 1 x = S y, the given system of differential equations simplifies to 17. The given system can be written as x = Ax, where A = y1 y2 = −1 0 03 y1 y2 . Hence, y1 = −y1 and y2 = 3y2 . Integrating these equations, we obtain y1 (t) = c1 e−t , y2 (t) = c2 e3t . Returning to the original variables, we have x = Sy = c1 e−t c2 e3t 3 −1 −5 1 = 3c1 e−t − c2 e3t −5c1 e−t + c2 e3t . Consequently, x1 (t) = 3c1 e−t − c2 e3t and x2 (t) = −5c1 e−t + c2 e3t . −12 −7 . 16 10 A has eigenvalues λ1 = 2, λ2 = −4 with corresponding linearly independent eigenvectors v1 = (1, −2) and 1 7 v2 = (7, −8). If we set S = , then S −1 AS = diag(2, −4), therefore, under the transformation −2 −8 x = S y, the given system of differential equations simplifies to 18. The given system can be written as x = Ax, where A = y1 y2 = 2 0 0 −4 y1 y2 . Hence, y1 = 2y1 and y2 = −4y2 . Integrating these equations, we obtain y1 (t) = c1 e2t , y2 (t) = c2 e−4t . 392 Returning to the original variables, we have c1 e2t c2 e−4t 1 7 −2 −8 x = Sy = = c1 e2t + 7c2 e−4t −2c1 e2t − 8c2 e−4t . Consequently, x1 (t) = c1 e2t + 7c2 e−4t and x2 (t) = −2c1 e2t − 8c2 e−4t . 01 . −1 0 A has eigenvalues λ1 = i, λ2 = −i with corresponding linearly independent eigenvectors v1 = (1, i) and 1 1 v2 = (1, −i). If we set S = , then S −1 AS = diag(i, −i), therefore, under the transformation i −i x = S y, the given system of differential equations simplifies to 19. The given system can be written as x = Ax, where A = y1 y2 = i 0 0 −i y1 y2 . Hence, y1 = iy1 and y2 = −iy2 . Integrating these equations, we obtain y1 (t) = c1 eit , y2 (t) = c2 e−it . Returning to the original variables, we have x = Sy = 1 1 i −i c1 eit c2 e−it = c1 eit + c2 e−it i(c1 eit − c2 e−it ) . Consequently, x1 (t) = c1 eit + c2 e−it and x2 (t) = i(c1 eit − c2 e−it ). Using Eulers formula, these expressions can be written as x1 (t) = (c1 + c2 ) cos t + i(c1 − c2 ) sin t, x2 (t) = i(c1 − c2 ) cos t − (c1 + c2 ) sin t, or equivalently, x1 (t) = a cos t + b sin t, x2 (t) = b cos t − a sin t, where a = c1 + c2 , and b = i(c1 − c2 ). 3 −4 −1 20. The given system can be written as x = Ax, where A = 0 −1 −1 . 0 −4 2 A has eigenvalue λ1 = −2 with corresponding eigenvector v1 = (1, 1, 1), and eigenvalue λ2 3 with corre= 11 0 1 , sponding linearly independent eigenvectors v2 = (1, 0, 0) and v3 = (0, 1, −4). If we set S = 1 0 1 0 −4 then S −1 AS = diag(−2, 3, 3), therefore, under the transformation x = S y, the given system of differential equations simplifies to y1 −2 0 0 y1 y2 = 0 3 0 y2 . 003 y3 y3 Hence, y1 = −2y1 , y2 = 3y2 , and y3 = 3y3 . Integrating these equations, we obtain y1 (t) = c1 e−2t , y2 (t) = c2 e3t , y3 (t) = c3 e3t . 393 Returning to the original variables, we 1 x = Sy = 1 1 have 1 0 c1 e−2t c1 e−2t + c2 e3t 0 1 c2 e3t = c1 e−2t + c3 e3t . 0 −4 c3 e3t c1 e−2t − 4c3 e3t Consequently, x1 (t) = c1 e−2t + c2 e3t , x2 (t) = c1 e−2t + c3 e3t , and c1 e−2t − 4c3 e3t . 1 1 −1 1 . 21. The given system can be written as x = Ax, where A = 1 1 −1 1 1 A has eigenvalue λ1 = −1 with corresponding eigenvector v1 = (−1, 1, −1), and eigenvalue λ2 = 2 with corre −1 0 1 0 , sponding linearly independent eigenvectors v2 = (0, 1, 1) and v3 = (1, 0, −1). If we set S = 1 1 −1 1 −1 then S −1 AS = diag(−1, 2, 2), therefore, under the transformation x = S y, the given system of differential equations simplifies to y1 −1 0 0 y1 y2 = 0 2 0 y2 . 002 y3 y3 Hence, y1 = −y1 , y2 = 2y2 , and y3 = 2y3 . Integrating these equations, we obtain y1 (t) = c1 e−t , y2 (t) = c2 e2t , y3 (t) = c3 e2t . Returning to the original variables, −1 x = Sy = 1 −1 we have 0 1 c1 e−t −c1 e−t + c3 e2t . 1 0 c2 e2t = c1 e−t + c2 e2t 2t 1 −1 c3 e −c1 e−t + c2 e2t − c3 e2t Consequently, x1 (t) = −c1 e−t + c3 e2t , x2 (t) = c1 e−t + c2 e2t , and −c1 e−t + (c2 − c3 )e2t . 22. A2 = (SDS −1 )(SDS −1 ) = SD(S −1 S )DS −1 = SDIn DS −1 = SD2 S −1 . We now use mathematical induction to establish the general result. Suppose that for k = m > 2 that Am = SDm S −1 . Then Am+1 = AAm = (SDS −1 )(SDm S −1 = SD(S −1 S )Dm S −1 ) = SDm+1 S −1 . It follows by mathematical induction that Ak = SDk S −1 for k = 1, 2, . . . 23. Let A = diag(a1 , a2 , . . . , an ) and let B = diag(b1 , b2 , . . . , bn ). Then from the index form of the matrix product, n 0, if i = j, (AB )ij = aik bkj = aii bij = ai bi , if i = j. k=1 Consequently, AB = diag(a1 b1 , a2 b2 , . . . , an bn ). Applying this result to the matrix D = diag(λ1 , λ2 , . . . , λk ), it follows directly that Dk = diag(λk , λk , . . . , λk ). n 1 2 24. The matrix A has eigenvalues λ1 = 5, λ2 = −1, with corresponding eigenvectors v1 = (1, −3) and 1 2 v2 = (2, −3). Thus, if we set S = , then S −1 AS = D, where D = diag(5, −1). −3 −3 394 Equivalently, A = SDS −1 . It follows from the results of the previous two examples that 1 2 −3 −3 A3 = SD3 S −1 = 125 0 0 −1 whereas A5 = SD5 S −1 = 1 2 −3 −3 3125 0 0 −1 −1 − 2 3 1 1 3 = −1 − 2 3 1 1 3 −127 −84 378 251 , −3127 −2084 9378 6251 = . 25. (a) This is self-evident from matrix multiplication. Another perspective on this is that when we multiply a matrix B on the left by a diagonal matrix√ , the ith row of B gets multiplied by the ith√ D√ diagonal element √ of D. Thus, if we multiply the ith row of D, λi , by the ith diagonal element of D, λi , the result in √√ √ √√ the ith row of the product is λi λi = λi . Therefore, D D = D, which means that D is a square root of D. (b) We have √ √ √ √√ (S DS −1 )2 = (S DS −1 )(S DS −1 ) = S ( DI D)S −1 = SDS −1 = A, as required. (c) We begin by diagonalizing A. We have det(A − λI ) = det 6−λ −3 −2 7−λ = (6 − λ)(7 − λ) − 6 = λ2 − 13λ + 36 = (λ − 4)(λ − 9), so the eigenvalues of A are λ = 4 and λ = 9. An eigenvector of A corresponding to λ = 4 is eigenvector of A corresponding to λ = 9 is S= √ 20 03 root of A is given by We take D= 2 −3 1 2 1 −3 , and an . Thus, we can form and D= . A fast computation shows that S −1 = √ 1 1 √ A = S DS −1 = 4 0 0 9 . 3/5 2/5 1/5 −1/5 12/5 −2/5 −3/5 13/5 . By part (b), one square . Directly squaring this result confirms this matrix as a square root of A. 26. (a) Show: A ∼ A. The identity matrix, I , is invertible, and I = I −1 . Since IA = AI =⇒ A = I −1 AI , it follows that A ∼ A. (b) Show: A ∼ B ⇐⇒ B ∼ A. A ∼ B =⇒ there exists an invertible matrix S such that B = S −1 AS =⇒ A = SBS −1 = (S −1 )−1 BS −1 . But S −1 is invertible since S is invertible. Consequently, B ∼ A. (c) Show A ∼ B and B ∼ C ⇐⇒ A ∼ C . A ∼ B =⇒ there exists an invertible matrix S such that B = S −1 AS ; moreover, B ∼ C =⇒ there exists an invertible matrix P such that C = P −1 BP . Thus, C = P −1 BP = P −1 (S −1 AS )P = (P −1 S −1 )A(SP ) = (SP )−1 A(SP ) where SP is invertible. Therefore A ∼ C. 395 27. Let A ∼ B mean that A is similar to B . Show: A ∼ B ⇐⇒ AT ∼ B T . A ∼ B =⇒ there exists an invertible matrix S such that B = S −1 AS =⇒ B T = (S −1 AS )T = S T AT (S −1 )T = S T AT (S T )−1 . S T is invertible because S is invertible, [since det(S ) = det(S T )]. Thus, AT ∼ B T . 28. We are given that Av = λv and B = S −1 AS . B (S −1 v) = (S −1 AS )(S −1 v) = S −1 A(SS −1 )v = S −1 AI v = S −1 Av = S −1 (λv) = λ(S −1 v). Hence, S −1 v is an eigenvector of B corresponding to the eigenvalue λ. 29. (a) S −1 AS = diag(λ1 , λ2 , . . . , λn ) =⇒ det(S −1 AS ) = λ1 λ2 · · · λn =⇒ det(A) det(S −1 ) det(S ) = λ1 λ2 · · · λn =⇒ det(A) = λ1 λ2 · · · λn . Since all eigenvalues are nonzero, it follows that det(A) = 0. Consequently, A is invertible. (b) S −1 AS = diag(λ1 , λ2 , . . . , λn ) =⇒ [S −1 AS ]−1 = [diag(λ1 , λ2 , . . . , λn )]−1 11 1 =⇒ S −1 A−1 (S −1 )−1 = diag , ,..., λ 1 λ2 λn 11 1 −1 −1 =⇒ S A S = diag , ,..., . λ 1 λ2 λn 30. (a) S −1 AS = diag(λ1 , λ2 , . . . , λn ) =⇒ (S −1 AS )T = [diag(λ1 , λ2 , . . . , λn )]T =⇒ S T AT (S −1 )T = diag(λ1 , λ2 , . . . , λn ) =⇒ S T AT (S T )−1 = diag(λ1 , λ2 , . . . , λn ). Since we have that Q = (ST )−1 , this implies that Q−1 AT Q = diag(λ1 , λ2 , . . . , λn ). (b) Let MC = [v1 , v2 , v3 , . . . , vn ] where MC denotes the matrix of cofactors of S . We see from part (a) that AT is nondefective, which means it possesses a complete set of eigenvectors. Also from part (a), Q−1 AT Q = diag(λ1 , λ2 , . . . , λn ) where Q = (S T )−1 , so AT Q = Q diag(λ1 , λ2 , . . . , λn ) (30.1) If we let MC denote the matrix of cofactors of S , then S −1 = T adj(S ) MC = =⇒ (S −1 )T = det(S ) det(S ) T MC det(S ) T =⇒ (S T )−1 = MC det(S ) =⇒ Q = MC . det(S ) Substituting this result into Equation (30.1), we obtain MC MC AT = diag(λ1 , λ2 , . . . , λn ) det(S ) det(S ) =⇒ AT MC = MC diag(λ1 , λ2 , . . . , λn ) =⇒ AT [v1 , v2 , v3 , . . . , vn ] = [v1 , v2 , v3 , . . . , vn ] diag(λ1 , λ2 , . . . , λn ) =⇒ [AT v1 , AT v2 , AT v3 , . . . , AT vn ] = [λ1 v1 , λ2 v2 , λ3 v3 , . . . , λn vn ] =⇒ AT vi = λvi for i ∈ {1, 2, 3, . . . , n}. Hence, the column vectors of MC are linearly independent eigenvectors of AT . −2 − λ 4 = 0 ⇐⇒ (λ + 3)(λ − 2) = 0 ⇐⇒ λ1 = −3 or λ2 = 2. 1 1−λ If λ1 = −3 the corresponding eigenvectors are of the v1 = r(−4, 1) where r ∈ R. If λ2 = 2 the corresponding eigenvectors are of the v2 = s(1, 1) where s ∈ R. 41 Thus, a complete set of eigenvectors is {(4, 1), (1, 1)} so that S = . From problem 17, if MC denotes 11 1 −1 the matrix of cofactors of S , then MC = . Consequently, (1, −1) is an eigenvector corresponding −1 4 to λ = −3 and (−1, 4) is an eigenvector corresponding to λ = 2 for the matrix AT . 31. det(A − λI ) = 0 ⇐⇒ 396 λ1 0λ =⇒ [Av1 , Av2 ] = [λv1 , v1 + λv2 ] =⇒ Av1 = λv1 and Av2 = v1 + λv2 =⇒ (A − λI )v1 = 0 and (A − λI )v2 = v1 . 32. S −1 AS = Jλ ⇐⇒ AS = SJλ ⇐⇒ A[v1 , v2 ] = [v1 , v2 ] 2−λ 1 = 0 ⇐⇒ (λ − 3)2 = 0 ⇐⇒ λ1 = 3 of multiplicity two. −1 4−λ If λ1 = 3 the corresponding eigenvectors are of the v1 = r(1, 1) where r ∈ R. Consequently, A does not have a complete set of eigenvectors, so it is a defective matrix. 31 By the preceding problem, J3 = is similar to A. Hence, there exists S = [v1 , v2 ] such that 03 S −1 AS = J3 . From the first part of the problem, we can let v1 = (1, 1). Now consider (A − λI )v2 = v1 where v1 = (a, b) for a, b ∈ R. Upon substituting, we obtain 33. det(A − λI ) = 0 ⇐⇒ −1 −1 Thus, S takes the form S = 34. λ S −1 AS = 0 0 1 λ 0 1 1 1 b−1 1 b a b = 1 1 =⇒ −a + b = 1. where b ∈ R, and if b = 0, then S = 0 λ 1 ⇐⇒ A[v1 , v2 , v3 ] = [v1 , v2 , v3 ] 0 λ 0 1 −1 1 0 . 0 1 λ 1 λ 0 ⇐⇒ [Av1 , Av2 , Av3 ] = [λv1 , v1 + λv2 , v2 + λv3 ] ⇐⇒ Av1 = λv1 , Av2 = v1 + λv2 , and Av3 = v2 + λv3 ⇐⇒ (A − λI )v1 = 0, (A − λI )v2 = v1 , and (A − λI )v3 = v2 . n n 35. (a) From (5.8.15), c1 f1 + c2 f2 + · · · + cn fn = 0 ⇐⇒ i=1 n n j =1 i=1 ⇐⇒ sji ej = 0 ci j =1 n sji ci ej = 0 ⇐⇒ sji ci = 0, j = 1, 2, . . . , n, since {ei } is a linearly independent set. The i=1 latter equation is just the component form of the linear system S c = 0. Since {fi } is a linearly independent set, the only solution to this system is the trivial solution c = 0, so det(S ) = 0. Consequently, S is invertible. (b) From (5.8.14) and (5.8.15), we have n T (fk ) = n bik i=1 n n j =1 i=1 sji ej = j =1 sji bik ej , k = 1, 2, . . . , n. Replacing i with j and j with i yields n T (fk ) = n sij bjk ei , k = 1, 2, . . . , n. i=1 j =1 (c) From (5.8.15) and (5.8.13), we have n T (fk ) = n sjk T (ej ) = j =1 n sjk j =1 aij ei , i=1 (∗) 397 that is, n T (fk ) = n aij sjk ei , k = 1, 2, . . . , n. i=1 (∗∗) j =1 (d) Subtracting (∗) from (∗∗) yields n (sij bjk − aij sjk ) ei = 0. i=1 n j =1 Thus, since {e1 } is a linearly independent set, n n sij bjk = j =1 aij sjk , i = 1, 2, . . . , n. j =1 But this is just the index form of the matrix equation SB = AS . Multiplying both sides of the preceding equation on the left by S −1 yields B = S −1 AS . Solutions to Section 5.9 True-False Review: 1. TRUE. In the definition of the matrix exponential function eAt , we see that powers of the matrix A must be computed: (At)3 (At)2 eAt = In + (At) + + + .... 2! 3! In order to do this, A must be a square matrix. 2. TRUE. We see this by plugging in t = 1 into the definition of the matrix exponential function. All terms containing A3 , A4 , A5 , . . . must be zero, leaving us with the result given in this statement. 3. FALSE. The inverse of the matrix exponential function eAt is the matrix exponential function e−At , and this will exist for all square matrices A, not just invertible ones. 4. TRUE. The matrix exponential function eAt converges to a matrix the same size as A, for all t ∈ R. This is asserted, but not proven, directly beneath Definition 5.9.1. 5. FALSE. The correct statement is (SDS −1 )k = SDk S −1 . The matrices S and S −1 on the right-hand side of this equation do not get raised to the power k . 6. FALSE. According to Property 1 of the Matrix Exponential Function, we have (eAt )2 = (eAt )(eAt ) = e2At . Problems: 1. det(A − λI ) = 0 ⇐⇒ 2−λ 2 2 −1 − λ = 0 ⇐⇒ λ2 − λ − 6 = 0 ⇐⇒ (λ + 2)(λ − 3) = 0 ⇐⇒ λ = −2 or λ = 3. If λ = −2 then (A − λI )v = 0 assumes the form 4 2 2 1 v1 v2 = 0 0 =⇒ 2v1 + v2 = 0. v1 = (1, −2), is 398 an eigenvector corresponding to λ = −2 and w1 = v1 = ||v1 || 2 1 √ , −√ 5 5 is a unit eigenvector corresponding to λ = −2. −1 2 2 −4 v2 is an eigenvector corresponding to λ = 3 and w2 = = ||v2 || to λ = 3. 1 2 √ √ 5 5 T Thus, S = 2 1 and S AS = diag(−2, 3). √ −√ 5 5 If λ = 3 then (A − λI )v = 0 assumes the form 2. det(A − λI ) = 0 ⇐⇒ 4−λ 6 6 9−λ v1 = v2 1 2 √ ,√ 5 5 0 0 =⇒ v1 + 2v2 = 0. v2 = (2, 1), is a unit eigenvector corresponding = 0 ⇐⇒ λ2 − 13λ = 0 ⇐⇒ λ(λ − 13) = 0 ⇐⇒ λ = 0 or λ = 13. 46 v1 0 = =⇒ 2v1 + 3v2 = 0. v1 = (−3, 2), is 69 v2 0 3 v1 2 = −√ , √ an eigenvector corresponding to λ = 0 and w1 = is a unit eigenvector corresponding ||v1 || 13 13 to λ = 0. −9 6 v1 0 If λ = 13 then (A − λI )v = 0 assumes the form = =⇒ −3v1 + 2v2 = 0. 6 −4 v2 0 2 3 v2 = √ ,√ is a unit eigenvector v2 = (2, 3), is an eigenvector corresponding to λ = 13 and w2 = ||v2 || 13 13 corresponding to λ = 13. 2 3 √ −√ 13 13 T Thus, S = 2 3 and S AS = diag(0, 13). √ √ 13 13 If λ = 0 then (A − λI )v = 0 assumes the form 3. det(A − λI ) = 0 ⇐⇒ 1−λ 2 2 1−λ = 0 ⇐⇒ (λ − 1)2 − 4 = 0 ⇐⇒ (λ + 1)(λ − 3) = 0 ⇐⇒ λ = −1 or λ = 3. 22 v1 0 = =⇒ v1 + v2 = 0. v1 = (−1, 1), is 22 v2 0 1 v1 1 is a unit eigenvector corresponding an eigenvector corresponding to λ = −1 and w1 = = −√ , √ ||v1 || 2 2 to λ = −1. −2 2 v1 0 If λ = 3 then (A − λI )v = 0 assumes the form = =⇒ v1 − v2 = 0. v2 = (1, 1), is 2 −2 v2 0 1 v2 1 is a unit eigenvector corresponding to an eigenvector corresponding to λ = 3 and w2 = = √ ,√ ||v2 || 2 2 λ = 3. 1 1 √ −√ 2 2 T Thus, S = 1 and S AS = diag(−1, 3). 1 √ √ 2 2 If λ = −1 then (A − λI )v = 0 assumes the form 4. det(A − λI ) = 0 ⇐⇒ −λ 0 0 −2 − λ 3 0 3 0 −λ = 0 ⇐⇒ (λ + 2)(λ − 3)(λ + 3) = 0 ⇐⇒ λ = −3, λ = −2, or 399 λ = 3. 3 v1 0 0 v2 = 0 =⇒ v1 + v3 = 0 and v2 = 0. 3 v3 0 1 1 v1 is a unit v1 = (−1, 0, 1), is an eigenvector corresponding to λ = −3 and w1 = = − √ , 0, √ ||v1 || 2 2 eigenvector corresponding to λ = −3. 203 v1 0 If λ = −2 then (A − λI )v = 0 assumes the form 0 0 0 v2 = 0 =⇒ v1 = v3 = 0 and v2 ∈ R. 302 v3 0 v2 = (0, 1, 0) is a unit eigenvector v2 = (0, 1, 0), is an eigenvector corresponding to λ = −2 and w2 = ||v2 || corresponding to λ = −2. −3 0 3 v1 0 0 v2 = 0 =⇒ v1 − v3 = 0 and If λ = 3 then (A − λI )v = 0 assumes the form 0 −5 3 0 −3 v3 0 1 1 v3 v2 = 0. v3 = (1, 0, 1), is an eigenvector corresponding to λ = 3 and w3 = = √ , 0, √ is a unit ||v3 || 2 2 eigenvector corresponding to λ = 3. 1 1 −√ 0√ 2 2 01 0 and S T AS = diag(−3, −2, 3). Thus, S = 1 1 √ 0√ 2 2 3 If λ = −3 then (A − λI )v = 0 assumes the form 0 3 5. det(A − λI ) = 0 ⇐⇒ 1−λ 2 1 2 4−λ 2 1 2 1−λ 0 1 0 = 0 ⇐⇒ λ3 − 6λ2 = 0 ⇐⇒ λ2 (λ − 6) = 0 ⇐⇒ λ = 6 or λ = 0 of multiplicity two. 121 v1 0 If λ = 0 then (A − λI )v = 0 assumes the form 2 4 2 v2 = 0 =⇒ v1 + 2v2 + v3 = 0. 121 v3 0 v1 = (−1, 0, 1) and v2 = (−2, 1, 0) are linearly independnet eigenvectors corresponding to λ = 0. v1 and v2 are not orthogonal since v1 , v2 = 2 = 0, so we will use the Gram-Schmidt procedure. Let u1 = v1 = (−1, 0, 1), so u2 = v2 − v2 , u1 2 u1 = (−2, 1, 0) − (−1, 0, 1) = (−1, 1, −1). 2 ||u1 || 2 u1 1 1 = − √ , 0, √ ||u1 || 2 2 corresponding to λ = 0. u2 1 1 1 = −√ , √ , √ are orthonormal eigenvectors ||u2 || 3 3 3 −5 2 1 v1 0 2 v2 = 0 =⇒ v1 − v3 = 0 and If λ = 6 then (A − λI )v = 0 assumes the form 2 −2 1 2 −5 v3 0 1 2 v3 1 is v2 − 2v3 = 0. v3 = (1, 2, 1), is an eigenvector corresponding to λ = 6 and w3 = = √ ,√ ,√ ||v3 || 6 6 6 a unit eigenvector corresponding to λ = 6. Now w1 = and w2 = 400 Thus, S = 1 1 −√ −√ 3 2 1 √ 0 3 1 1 √ −√ 3 2 6. det(A − λI ) = 0 ⇐⇒ 1 √ 6 2 √ 6 1 √ 6 and S T AS = diag(0, 0, 6). 2−λ 0 0 0 3−λ 1 0 1 3−λ = 0 ⇐⇒ (λ − 2)2 (λ − 4) = 0 ⇐⇒ λ = 4 or λ = 2 of multiplicity two. 000 v1 0 If λ = 2 then (A − λI )v = 0 assumes the form 0 1 1 v2 = 0 =⇒ v2 + v3 = 0 and 011 v3 0 v1 ∈ R. v1 = (1, 0, 0) and v2 = (0, −1, 1) are linearly independnet eigenvectors corresponding to λ = 2, and v1 v2 1 1 w1 = = (1, 0, 0) and w2 = = 0, − √ , √ are unit eigenvectors corresponding to λ = 2. w1 ||v1 || ||v2 || 2 2 and w2 are also orthogonal because w1 , w2 = 0. −2 0 0 v1 0 1 v2 = 0 =⇒ v1 = 0 and If λ = 4 then (A − λI )v = 0 assumes the form 0 −1 0 1 −1 v3 0 v3 1 1 v2 − v3 = 0. v3 = (0, 1, 1), is an eigenvector corresponding to λ = 4 and w3 = = 0, √ , √ is a ||v3 || 2 2 unit eigenvector corresponding to λ = 4. 1 0 0 1 1 √ 0 −√ T Thus, S = 2 2 and S AS = diag(2, 2, 4). 1 1 √ √ 0 2 2 7. det(A − λI ) = 0 ⇐⇒ −λ 1 0 1 −λ 0 0 0 1−λ = 0 ⇐⇒ (λ − 1)2 (λ + 1) = 0 ⇐⇒ λ = −1 or λ = 1 of multiplicity two. −1 10 v1 0 If λ = 1 then (A − λI )v = 0 assumes the form 1 −1 0 v2 = 0 =⇒ v1 − v2 = 0 and 0 00 v3 0 v3 ∈ R. v1 = (1, 1, 0) and v2 = (0, 0, 1) are linearly independnet eigenvectors corresponding to λ = 1, and v1 1 1 v2 w1 = = √ , √ , 0 and w2 = = (0, 0, 1) are unit eigenvectors corresponding to λ = 1. w1 ||v1 || ||v2 || 2 2 and w2 are also orthogonal because w1 , w2 = 0. 110 v1 0 If λ = −1 then (A − λI )v = 0 assumes the form 1 1 0 v2 = 0 =⇒ v1 + v2 = 0 and v3 = 0. 002 v3 0 1 v3 1 v3 = (−1, 1, 0), is an eigenvector corresponding to λ = −1 and w3 = = − √ , √ , 0 is a unit ||v3 || 2 2 eigenvector corresponding to λ = −1. 401 Thus, S = 1 √ 2 1 √ 2 0 1 0 −√ 2 1 √ 0 2 01 8. det(A − λI ) = 0 ⇐⇒ and S T AS = diag(−1, 1, 1). 1−λ 1 −1 1 −1 1−λ 1 1 1−λ λ = −1. = 0 ⇐⇒ (λ − 2)2 (λ + 1) = 0 ⇐⇒ λ = 2 of multiplicity two −1 1 −1 v1 0 1 v2 = 0 =⇒ v1 − v2 + v3 = 0. If λ = 2 then (A − λI )v = 0 assumes the form 1 −1 −1 1 −1 v3 0 v1 = (1, 1, 0) and v2 = (−1, 0, 1) are linearly independnet eigenvectors corresponding to λ = 2. v1 and v2 are not orthogonal since v1 , v2 = −1 = 0, so we will use the Gram-Schmidt procedure. Let u1 = v1 = (1, 1, 0), so u2 = v2 − 1 u1 1 = √ , √ ,0 ||u1 || 2 2 corresponding to λ = 2. 1 v2 , u1 u1 = (−1, 0, 1) + (1, 1, 0) = ||u1 ||2 2 11 − , ,1 . 22 u2 1 2 1 = −√ , √ , √ are ||u2 || 6 6 6 2 1 −1 v1 1 v2 = If λ = −1 then (A − λI )v = 0 assumes the form 1 2 −1 1 2 v3 Now w1 = and w2 = orthonormal eigenvectors 0 0 =⇒ v1 − v3 = 0 and 0 v3 1 1 1 v2 + v3 = 0. v3 = (1, −1, 1), is an eigenvector corresponding to λ = −1 and w3 = = √ , −√ , √ ||v3 || 3 3 3 is a unit eigenvector corresponding to λ = −1. 1 1 1 √ √ −√ 2 6 3 1 1 1 √ √ − √ and S T AS = diag(2, 2, −1). Thus, S = 2 6 3 1 2 √ √ 0 6 3 9. det(A − λI ) = 0 ⇐⇒ λ = 2. 1−λ 0 −1 0 1−λ 1 −1 1 −λ = 0 ⇐⇒ (1 − λ)(λ + 1)(λ − 2) = 0 ⇐⇒ λ = −1, λ = 1, or 0 −1 v1 0 2 1 v2 = 0 =⇒ 2v1 − v3 = 0 and 1 1 v3 0 v1 1 1 2 2v2 + v3 = 0. v1 = (1, −1, 2), is an eigenvector corresponding to λ = −1 and w1 = = √ , −√ , √ ||v1 || 6 6 6 is a unit eigenvector corresponding to λ = −1. 0 0 −1 v1 0 1 v2 = 0 =⇒ v1 − v2 = 0 and If λ = 1 then (A − λI )v = 0 assumes the form 0 0 −1 1 −1 v3 0 1 v2 1 v3 = 0. v2 = (1, 1, 0), is an eigenvector corresponding to λ = 1 and w2 = = √ , √ , 0 is a unit ||v2 || 2 2 2 If λ = −1 then (A − λI )v = 0 assumes the form 0 −1 402 eigenvector corresponding to λ = 1. −1 0 −1 v1 0 1 v2 = 0 =⇒ v1 + v3 = 0 and If λ = 2 then (A − λI )v = 0 assumes the form 0 −1 −1 1 −2 v3 0 1 1 1 v3 = −√ , √ , √ v2 − v3 = 0. v3 = (−1, 1, 1), is an eigenvector corresponding to λ = 2 and w3 = ||v3 || 3 3 3 is a unit eigenvector corresponding to λ = 2. 1 1 1 √ √ −√ 3 2 6 1 1 1 −√ √ and S T AS = diag(−1, 1, 2). √ Thus, S = 3 2 6 1 2 √ √ 0 6 3 3−λ 3 4 4 0 = 3−λ or λ = 8. 5 If λ = −2 then (A − λI )v = 0 assumes the form 3 4 10. det(A − λI ) = 0 ⇐⇒ 3 3−λ 0 0 ⇐⇒ (λ − 3)(λ + 2)(λ − 8) = 0 ⇐⇒ λ = −2, λ = 3, 34 v1 0 5 0 v2 = 0 =⇒ 4v1 + 5v3 = 0 and 4v2 − 05 v3 0 1 3 v1 4 3v3 = 0. v1 = (−5, 3, 4), is an eigenvector corresponding to λ = −2 and w1 = = −√ , √ , √ ||v1 || 25252 is a unit eigenvector corresponding to λ = −2. 034 v1 0 If λ = 3 then (A − λI )v = 0 assumes the form 3 0 0 v2 = 0 =⇒ v1 = 0 and 3v2 + 4v3 = 0. 400 v3 0 43 v2 v2 = (0, −4, 3), is an eigenvector corresponding to λ = 3 and w2 = = 0, − , is a unit eigenvector ||v2 || 55 corresponding to λ = 3. −5 3 4 v1 0 0 v2 = 0 =⇒ 4v1 − 5v3 = 0 and If λ = 8 then (A − λI )v = 0 assumes the form 3 −5 4 0 −5 v3 0 3 4 v3 1 4v2 − 3v3 = 0. v3 = (5, 3, 4), is an eigenvector corresponding to λ = 8 and w3 = = √,√,√ ||v3 || 25252 is a unit eigenvector corresponding to λ = 8. 1 1 0√ −√ 2 2 3 4 3 √ √ and S T AS = diag(−2, 3, 8). − Thus, S = 5 5 2 52 4 3 4 √ √ 5 52 52 −3 − λ 2 2 2 2 = 0 ⇐⇒ (λ + 5)2 (λ − 1) = 0 ⇐⇒ λ = −5 of −3 − λ multiplicity two, or λ = 1. −4 2 2 v1 0 2 v2 = 0 =⇒ v1 − v3 = 0 and If λ = 1 then (A − λI )v = 0 assumes the form 2 −4 2 2 −4 v3 0 11. det(A − λI ) = 0 ⇐⇒ 2 −3 − λ 2 403 1 v1 1 1 is a = √ ,√ ,√ ||v1 || 3 3 3 unit eigenvector corresponding to λ = 1. 222 v1 0 If λ = −5 then (A − λI )v = 0 assumes the form 2 2 2 v2 = 0 =⇒ v1 + v2 + v3 = 0. 222 v3 0 v2 = (−1, 1, 0) and v3 = (−1, 0, 1) are linearly independent eigenvectors corresponding to λ = −5. v2 and v3 are not orthogonal since v2 , v3 = −1 = 0, so we will use the Gram-Schmidt procedure. Let u2 = v2 = (−1, 1, 0), so v2 − v3 = 0. v1 = (1, 1, 1) is an eigenvector corresponding to λ = 1 and w1 = u3 = v3 − v3 , u2 1 u2 = (−1, 0, 1) − (−1, 1, 0) = ||u2 ||2 2 11 − ,− ,1 . 22 u2 1 1 u3 1 1 2 = − √ , √ , 0 and w3 = = −√ , −√ , √ ||u2 || ||u3 || 2 2 6 6 6 corresponding to λ = −5. 1 1 1 √ −√ −√ 3 2 6 1 1 1 √ √ − √ and S T AS = diag(1, −5, −5). Thus, S = 3 2 6 1 2 √ √ 0 3 6 Now w2 = 12. det(A − λI ) = 0 ⇐⇒ −λ 1 1 1 −λ 1 1 1 −λ are orthonormal eigenvectors = 0 ⇐⇒ (λ + 1)2 (λ − 2) = 0 ⇐⇒ λ = −1 of multiplicity two, or λ = 2. 111 v1 0 If λ = −1 then (A − λI )v = 0 assumes the form 1 1 1 v2 = 0 =⇒ v1 + v2 + v3 = 0. 111 v3 0 v1 = (−1, 0, 1) and v2 = (−1, 1, 0) are linearly independnet eigenvectors corresponding to λ = −1. v1 and v2 are not orthogonal since v1 , v2 = 1 = 0, so we will use the Gram-Schmidt procedure. Let u1 = v1 = (−1, 0, 1), so u2 = v2 − v2 , u1 1 u1 = (−1, 1, 0) − (−1, 0, 1) = 2 ||u1 || 2 1 1 − , 1, − 2 2 . 1 2 u2 1 are orthonormal eigenvectors = −√ , √ , −√ ||u2 || 6 6 6 −2 1 1 v1 0 1 v2 = 0 =⇒ v1 − v3 = 0 and If λ = 2 then (A − λI )v = 0 assumes the form 1 −2 1 1 −2 v3 0 1 1 v3 1 is a v2 − v3 = 0. v3 = (1, 1, 1), is an eigenvector corresponding to λ = 2 and w3 = = √ ,√ ,√ ||v3 || 3 3 3 unit eigenvector corresponding to = 2. λ 1 1 1 √ −√ −√ 3 6 2 2 1 √ √ and S T AS = diag(−1, −1, 2). 0 Thus, S = 6 3 1 1 1 √ √ −√ 2 6 3 u1 1 1 = − √ , 0, √ ||u1 || 2 2 corresponding to λ = −1. Now w1 = and w2 = 404 13. A has eigenvalues λ1 = 4 and λ2 = −2 with corresponding eigenvectors v1 = (1, 1) and v2 = (−1, 1). 1 1 Therefore, a set of principal axes is √ (1, 1), √ (−1, 1) . Relative to these principal axes, the quadratic 2 2 2 2 form reduces to 4y1 − 2y2 . 14. A has eigenvalues λ1 = 7 and λ2 = 3 with corresponding eigenvectors v1 = (1, 1) and v2 = (1, −1). 1 1 Therefore, a set of principal axes is √ (1, 1), √ (1, −1) . Relative to these principal axes, the quadratic 2 2 2 2 form reduces to 7y1 + 3y2 . 15. A has eigenvalue λ = 2 of multiplicity two with corresponding linearly independent eigenvectors v1 = (1, 0, −1) and v2 = (0, 1, 1). Using the Gram-Schmidt procedure, an orthogonal basis in this eigenspace is {u1 , u2 } where 1 1 u1 = (1, 0, −1), u2 = (0, 1, 1) + (1, 0, −1) = (1, 2, 1). 2 2 1 1 An orthonormal basis for the eigenspace is √ (1, 0, −1), √ (1, 2, 1) . The remaining eigenvalue of A is 2 6 λ = −1, with eigenvector v3 = (−1, 1, −1). Consequently, a set of principal axes for the given quadratic 1 1 1 form is √ (1, 0, −1), √ (1, 2, 1), √ (−1, 1, −1) . Relative to these principal axes, the quadratic form 2 6 3 2 2 2 reduces to 2y1 + 2y2 − y3 . 16. A has eigenvalue λ = 0 of multiplicity two with corresponding linearly independent eigenvectors v1 = (0, 1, 0, −1) and v2 = (1, 0, −1, 0). Notice that these are orthogonal, hence we do not need to apply the Gram-Schmidt procedure. The remaining eigenvalues of A are λ = 4 and λ = 8, with corresponding eigenvectors v3 = (−1, 1, −1, 1) and v4 = (1, 1, 1, 1), respectively. Consequently, a set of principal axes for the given quadratic form is 1 1 1 1 √ (0, 1, 0, −1), √ (1, 0, −1, 0), (−1, 1, −1, 1), (1, 1, 1, 1) . 2 2 2 2 2 2 Relative to these principal axes, the quadratic form reduces to 4y3 + 8y4 . 17. A = ab bc where a, b, c ∈ R. a−λ c = 0 ⇐⇒ λ2 − (a + c)λ + (ac − b2 ) = 0 c b−λ (a + c) ± (a − c)2 + 4b2 ⇐⇒ λ = . Now A has repeated eigenvalues 2 a0 ⇐⇒ (a − c)2 + 4b2 = 0 ⇐⇒ a = c and b = 0 (since a, b, c ∈ R) ⇐⇒ A = 0a ⇐⇒ A = aI2 ⇐⇒ A is a scalar matrix. Consider det(A − λI ) = 0 ⇐⇒ 18.(a) A is a real n × n symmetric matrix =⇒ A possesses a complete set of eigenvectors =⇒ A is similar to a diagonal matrix =⇒ there exists an invertible matrix S such that S −1 AS = diag(λ, λ, λ, . . . , λ) where λ occurs n times. But diag(λ, λ, λ, . . . , λ) = λIn , so S −1 AS = λIn =⇒ AS = S (λIn ) =⇒ AS = λS =⇒ A = λSS −1 =⇒ A = λIn =⇒ A is a scalar matrix (b) Theorem: Let A be a nondefective n × n matrix. If λ is an eigenvalue of multiplicity n, then A is a scalar matrix. Proof: The proof is the same as that in part (a) since A has a complete set of eigenvectors. 405 19. Since real eigenvectors of A that correspond to distinct eigenvalues are orthogonal, it must be the case that if y = (y1 , y2 ) corresponds to λ2 where Ay = λ2 y, then x, y = 0 =⇒ (1, 2), (y1 , y2 ) = 0 =⇒ y1 + 2y2 = 0 =⇒ y1 = −2y2 =⇒ y = (−2y2 , y2 ) = y2 (−2, 1). Consequently, (−2, 1) is an eigenvector corresponding to λ2 . 20. (a) Let A be a real symmetric 2 × 2 matrix with two distinct eigenvalues, λ1 and λ2 , where v1 = (a, b) is an eigenvector corresponding to λ1 . Since real eigenvectors of A that correspond to distinct eigenvalues are orthogonal, it follows that if v2 = (c, d) corresponds to λ2 where av2 = λ2 v2 , then v1 , v2 = 0 =⇒ (a, b), (c, d) = 0, that is, ac + bd = 0. By inspection, we see that v2 = (−b, a). An orthonormal set of eigenvectors for A is 1 1 (a, b), √ (−b, a) . 2 2 + b2 +b a 1 1 √ a −√ b 2+ 2 a2 + b2 , then S T AS = diag(λ , λ ). Thus, if S = a 1 b 1 2 1 √ √ b a 2 + b2 2 + b2 a a √ a2 (b) S T AS = diag(λ1 , λ2 ) =⇒ AS = S diag(λ1 , λ2 ), since S T = S −1 . Thus, 1 a −b λ1 0 ab A = S diag(λ1 , λ2 )S T = 2 a 0 λ2 −b a (a + b2 ) b 1 a −b λ1 a λ 1 b =2 a −λ2 b λ2 a (a + b2 ) b = (a2 1 + b2 ) λ1 a2 + λ2 b2 ab(λ1 − λ2 ) ab(λ1 − λ2 ) λ1 b2 + λ2 a2 . 21. A is a real symmetric 3 × 3 matrix with eigenvalues λ1 and λ2 of multiplicity two. (a) Let v1 = (1, −1, 1) be an eigenvector of A that corresponds to λ1 . Since real eigenvectors of A that correspond to distinct eigenvalues are orthogonal, it must be the case that if v = (a, b, c) corresponds to λ2 where Av = λ2 v, then v1 , v = 0 =⇒ (1, −1, 1), (a, b, c) = 0 =⇒ a − b + c = 0 =⇒ v = r(1, 1, 0)+ s(−1, 0, 1) where r and s are free variables. Consequently, v2 = (1, 1, 0) and v3 = (−1, 0, 1) are linearly independent eigenvectors corresponding to λ2 . Thus, {(1, 1, 0), (−1, 0, 1)} is a basis for E2 . v2 and v3 are not orthogonal since v2 , v3 = −1 = 0, so we will apply the Gram-Schmidt procedure to v2 and v3 . Let u2 = v2 = (1, 1, 0) and 1 v3 , u2 11 u3 = v3 − u2 = (−1, 0, 1) + (1, 1, 0) = − , , 1 . 2 ||u2 || 2 22 1 v1 1 1 is a unit eigenvector corresponding to λ1 , and = √ , −√ , √ ||v1 || 3 3 3 2 1 u3 1 1 u2 1 are orthonormal eigenvectors corresponding = −√ , √ , √ w2 = = √ , √ , 0 , w3 = ||u3 || ||u2 || 6 6 6 2 2 to λ2 . 1 1 1 √ √ −√ 6 2 3 1 1 1 −√ √ √ and S T AS = diag(λ1 , λ2 , λ2 ). Consequently, S = 3 2 6 2 1 √ √ 0 3 6 Now, w1 = 406 (b) Since S is an orthogonal matrix, S T AS = diag(λ1 , λ2 , λ2 ) =⇒ AS = S diag(λ1 , λ2 , λ3 ) T =⇒ A = S diag(λ1 , λ2 , λ3 )S =⇒ 1 1 1 1 1 1 √ √ √ √ −√ −√ 3 3 3 6 2 3 0 1 1 1 λ1 0 1 1 √ −√ √ √ √ 0 0 λ2 0 A= 2 2 6 2 3 0 0 λ2 2 1 1 2 1 √ √ √ √ −√ 0 6 6 6 6 3 λ1 λ1 λ1 1 1 1 √ √ √ √ −√ −√ 3 3 3 6 2 3 λ1 + 2λ2 −λ1 + λ2 λ1 − λ 2 1 1 1 1 λ2 λ2 −√ √ √ √ √ 0 = −λ1 + λ2 λ1 + 2λ2 −λ1 + λ2 . = 3 3 2 6 2 2 λ1 − λ2 −λ1 + λ2 λ1 + 2λ2 1 2 λ 2 2λ 2 λ2 √ √ 0 √ √ −√ 3 6 6 6 6 T 22. (a) Let v1 , v2 ∈ Cn and recall that v1 v2 = [ v1 , v2 ]. T TT T T T T [ Av1 , v2 ] = (Av1 ) v2 = (v1 A )v2 = v1 (−A)v2 = −v1 (A)v2 = −v1 (Av2 ) = −v1 Av2 = [− v1 , Av2 ]. Thus, Av1 , v2 = − v1 , Av2 . (22.1) (b) Let v1 be an eigenvector corresponding to the eigenvalue λ1 , so Av1 = λ1 v1 . (22.2) Taking the inner product of (22.2) with v1 yields Av1 , v1 = λ1 v1 , v1 , that is Av1 , v1 = λ1 ||v1 ||2 . (22.3) Taking the complex conjugate of (22.3) gives Av1 , v1 = λ1 ||v1 ||2 , that is v1 , Av1 = λ1 ||v1 ||2 . (22.4) Adding (22.3) and (22.4), and using (22.1) with v2 = v1 yields (λ1 + λ1 )||v1 ||2 = 0. But ||v1 ||2 = 0, so λ1 + λ1 = 0, or equivalently, λ1 = −λ1 , which means that all nonzero eigenvalues of A are pure imaginary. 23. Let A be an n × n real skew-symmetric matrix where n is odd. Since A is real, the characteristic equation, det(A − λI ) = 0, has real coefficients, so its roots come in conjugate pairs. By problem 20, all nonzero solutions of det(A − λI ) = 0 are pure imaginary, hence when n is odd, zero will be one of the eigenvalues of A. 24. det(A − λI ) = 0 ⇐⇒ −λ 4 −4 −4 −λ −2 4 2 −λ = 0 ⇐⇒ λ3 + 36λ = 0 ⇐⇒ λ = 0 or λ = ±6i. 0 4 −4 v1 0 If λ = 0 then (A − λI )v = 0 assumes the form −4 0 −2 v2 = 0 =⇒ 2v1 + v3 = 0 and 42 0 v3 0 v2 − v3 = 0. If we let v3 = 2r ∈ C, then the solution set of this system is {(−r, 2r, 2r) : r ∈ C} so the eigenvectors corresponding to λ = 0 are v1 = r( 1, 2, 2) where r ∈ C. − 6i 4 −4 v1 0 If λ = −6i then (A−λI )v = 0 assumes the form −4 6i −2 v2 = 0 =⇒ 5v1 +(−2+6i)v3 = 0 42 6i v3 0 407 and 5v2 + (4 + 3i)v3 = 0. If we let v3 = 5s ∈ C, then the solution set of this system is {(2 − 6i)s, (−4 − 3i)s, 5s : s ∈ C}, so the eigenvectors corresponding to λ = −6i are v2 = s(2 − 6i, −4 − 3i, 5) where s ∈ C. By Theorem 5.6.8, since A is a matrix with real entries, the eigenvectors corresponding to λ = 6i are of the form v3 = t(2 + 6i, −4 + 3i, 5) where t ∈ C. 25. det(A − λI ) = 0 ⇐⇒ −λ −1 −6 1 −λ 5 6 −5 −λ √ = 0 ⇐⇒ −λ3 − 62λ = 0 ⇐⇒ λ = 0 or λ = ± 62i. 0 −1 −6 v1 0 0 5 v2 = 0 =⇒ v3 = t, v2 = −6t, v1 = If λ = 0 then (A − λI )v = 0 assumes the form 1 6 −5 0 v3 0 −5t, where t ∈ C. Thus, the solution set of the system is {(−5t, −6t, t) : t ∈ C}, so the eigenvectors corresponding to λ = 0 are v = t(−5, −6, 1), where t ∈ C. For the other eigenvalues, it is best to use technology to generate the corresponding eigenvectors. 26. A is a real n×n orthogonal matrix ⇐⇒ A−1 = AT ⇐⇒ AT A = In . Suppose that A = [v1 , v2 , v3 , . . . , vn ]. T Since the ith row of AT is equal to vi , the matrix multiplication assures us that the ij th entry of AT A is T equal to vi , vi . Thus from the equality AT A = In = [δij ], an n × n matrix A = [v1 , v2 , v3 , . . . , vn ] is T orthogonal if an only if vi , vi = δij , that is, if and only if the columns (rows) of A, {v1 , v2 , v3 , . . . , vn }, form an orthonormal set of vectors. Solutions to Section 5.11 True-False Review: 1. TRUE. See Remark 1 following Definition 5.11.1. 2. TRUE. Each Jordan block corresponds to a cycle of generalized eigenvectors, and each such cycle contains exactly one eigenvector. By construction, the eigenvectors (and generalized eigenvectors) are chosen to form a linearly independent set of vectors. 3. FALSE. For example, in a diagonalizable n × n matrix, the n linearly independent eigenvectors can be arbitrarily placed in the columns of the matrix S . Thus, an ample supply of invertible matrices S can be constructed. 01 4. FALSE. For instance, if J1 = J2 = , then J1 and J2 are in Jordan canonical form, but 00 02 J1 + J2 = is not in Jordan canonical form. 00 5. TRUE. This is simply a restatement of the definition of a generalized eigenvector. 6. FALSE. The number of Jordan blocks corresponding to λ in the Jordan canonical form of A is the number of linearly independent eigenvectors of A corresponding to λ, which is dim[Eλ ], the dimension of the eigenspace corresponding to λ, not dim[Kλ ]. 7. TRUE. This is the content of Theorem 5.11.8. 11 8. FALSE. For instance, if J1 = J2 = 01 12 J1 J2 = is not in Jordan canonical form. 01 , then J1 and J2 are in Jordan canonical form, but 9. TRUE. If we place the vectors in a cycle of generalized eigenvectors of A (see Equation (5.11.3)) in the columns of the matrix S formulated in this section in the order they appear in the cycle, then the 408 corresponding columns of the matrix S −1 AS will form a Jordan block. 10. TRUE. The assumption here is that all Jordan blocks have size 1 × 1, which precisely says that the Jordan canonical form of A is a diagonal matrix. This means that A is diagonalizable. 11. TRUE. Suppose that S −1 AS = B and that J is a Jordan canonical form of A. So there exists an invertible matrix T such that T −1 AT = J . Then B = S −1 AS = S −1 (T JT −1 )S = (T −1 S )−1 J (T −1 S ), and hence, (T −1 S )B (T −1 S )−1 = J, which shows that J is also a Jordan canonical form of B . 01 12. FALSE. For instance, if J = and r = 2, then J is in Jordan canonical form, but rJ = 00 is not in Jordan canonical form. 0 0 2 0 Problems: 1. There are 3 possible Jordan canonical forms: 100 1 0 1 0 , 0 001 0 2. There are 4 possible Jordan canonical forms: 1100 1000 0 1 0 0 0 1 0 0 0 0 3 0 , 0 0 3 0 0003 0003 1 1 0 0 0 , 1 1 0 0 0 1 . 1 1 1 0 , 1 0 0 0 0 1 0 0 0 0 3 0 0 0 , 1 3 , 1 2 0 0 0 0 0 2 0 0 0 0 1 2 0 1 1 0 0 1 0 0 0 0 0 3 0 0 0 . 1 3 2 0 0 0 0 1 2 0 0 0 0 1 2 0 0 0 0 0 2 0 0 0 0 0 2 1 3 0 0 0 0 0 1 3 0 0 0 0 0 0 3 0 0 0 0 0 0 9 0 3. There are 7 possible Jordan canonical forms: 2 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 2 2 0 0 0 0 , 1 2 0 0 0 0 1 2 0 0 2 0 0 0 0 0 0 0 2 0 0 0 0 1 2 1 2 0 0 0 0 0 2 0 0 , 0 0 0 2 0 0 0 0 0 2 2 0 0 0 0 1 2 0 0 0 0 1 2 0 0 2 0 0 0 0 0 0 1 2 0 0 0 0 0 2 , 0 0 0 0 2 2 0 0 0 0 , 1 2 0 0 0 0 1 2 0 0 0 0 1 2 0 0 0 0 1 2 4. There are 10 possible Jordan canonical forms: 3 0 0 0 0 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0 9 0 0 0 0 0 0 9 , 3 0 0 0 0 0 1 3 0 0 0 0 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0 9 0 0 0 0 0 0 9 , 3 0 0 0 0 0 1 3 0 0 0 0 0 0 3 0 0 0 0 0 1 3 0 0 0 0 0 0 9 0 0 0 0 0 0 9 , 3 0 0 0 0 0 0 0 0 0 0 9 , 409 3 0 0 0 0 0 1 3 0 0 0 0 0 1 3 0 0 0 0 0 1 3 0 0 0 0 0 1 9 0 0 0 0 0 0 9 , 3 0 0 0 0 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0 9 0 0 0 0 0 1 9 , 3 0 0 0 0 0 1 3 0 0 0 0 0 1 3 0 0 0 0 0 0 3 0 0 0 0 0 0 9 0 0 0 0 0 1 9 , 3 0 0 0 0 0 1 3 0 0 0 0 0 0 3 0 0 0 0 0 0 3 0 0 0 0 0 0 9 0 0 0 0 0 1 9 , 3 0 0 0 0 0 1 3 0 0 0 0 0 1 3 0 0 0 0 0 1 3 0 0 0 0 0 1 9 0 0 0 0 0 1 9 3 0 0 0 0 0 1 3 0 0 0 0 0 0 3 0 0 0 0 0 1 3 0 0 0 0 0 0 9 0 0 0 0 0 1 9 , . 5. Since λ = 2 occurs with multiplicity 4, it can give rise to the following possible Jordan block sizes: (a) 4 (b) 3,1 (c) 2,2 (d) 2,1,1 (e) 1,1,1,1 Likewise, λ = 6 occurs with multiplicity 4, so it can give rise to the same five possible Jordan block sizes. Finally, λ = 8 occurs with multiplicity 3, so it can give rise to three possible Jordan block sizes: (a) 3 (b) 2,1 (c) 1,1,1 Since the block sizes for each eigenvalue can be independently determined, we have 5 · 5 · 3 = 75 possible Jordan canonical forms. 6. Since λ = 2 occurs with multiplicity 4, it can give rise to the following possible Jordan block sizes: (a) 4 (b) 3,1 (c) 2,2 (d) 2,1,1 (e) 1,1,1,1 Next, λ = 5 occurs with multiplicity 6, so it can give rise to the following possible Jordan block sizes: (a) 6 (b) 5,1 (c) 4,2 (d) 4,1,1 (e) 3,3 (f) 3,2,1 (g) 3,1,1,1 (h) 2,2,2 (i) 2,2,1,1 (j) 2,1,1,1,1 (k) 1,1,1,1,1,1 There are 5 possible Jordan block sizes corresponding to λ = 2 and 11 possible Jordan block sizes corresponding to λ = 5. Multiplying these results, we have 5 · 11 = 55 possible Jordan canonical forms. 410 7. Since (A − 5I )2 = 0, no cycles of generalized eigenvectors corresponding to λ = 5 can have length greater than 2, and hence, only Jordan block sizes 2 or less are possible. Thus, the possible block sizes under this restriction (corresponding to λ = 5) are: 2,2,2 2,2,1,1 2,1,1,1,1 1,1,1,1,1,1 There are four such. There are still five possible block size lists corresponding to λ = 2. Multiplying these results, we have 5 · 4 = 20 possible Jordan canonical forms under this restriction. 8. (a): λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ2 0 0 0 0 0 λ2 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ2 0 0 0 0 1 λ2 , , λ1 0 0 0 0 1 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ2 0 0 0 0 0 λ2 λ1 0 0 0 0 1 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ2 0 0 0 0 1 λ2 , , λ1 0 0 0 0 1 λ1 0 0 0 0 1 λ1 0 0 0 0 0 λ2 0 0 0 0 0 λ2 λ1 0 0 0 0 1 λ1 0 0 0 0 1 λ1 0 0 0 0 0 λ2 0 0 0 0 1 λ2 , (b): The assumption that (A − λ1 I )2 = 0 implies that there can be no Jordan blocks corresponding to λ1 of size 3 × 3 (or greater). Thus, the only possible Jordan canonical forms for this matrix now are λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ2 0 0 0 0 0 λ2 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ2 0 0 0 0 1 λ2 , , λ1 0 0 0 0 1 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ2 0 0 0 0 0 λ2 λ1 0 0 0 0 1 λ1 0 0 0 0 0 λ1 0 0 0 0 0 λ2 0 0 0 0 1 λ2 , 9. The assumption that (A − λI )3 = 0 implies no Jordan blocks of size greater than 3 × 3 are possible. The fact that (A − λI )2 = 0 implies that there is at least one Jordan block of size 3 × 3. Thus, the possible block size combinations for a 6 × 6 matrix with eigenvalue λ of multiplicity 6 and no blocks of size greater than 3 × 3 with at least one 3 × 3 block are: 3,3 3,2,1 3,1,1,1 Thus, there are 3 possible Jordan canonical forms. (We omit the list itself; it can be produced simply from the list of block sizes above.) 411 10. The eigenvalues of the matrix with this characteristic polynomial are λ = 4, 4, −6. The possible Jordan canonical forms in this case are therefore: 40 0 41 0 0 4 0 , 0 4 0 . 0 0 −6 0 0 −6 11. The eigenvalues of the matrix with this characteristic polynomial are λ = 4, 4, 4, −1, −1. The possible Jordan canonical forms in this case are therefore: 4 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 4 0 0 0 −1 0 0 0 −1 4 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 4 0 0 0 −1 1 0 0 −1 , , 4 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 4 0 0 0 −1 0 0 0 −1 4 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 4 0 0 0 −1 1 0 0 −1 , , 4 0 0 0 0 1 4 0 0 0 0 0 0 1 0 0 4 0 0 0 −1 0 0 0 −1 4 0 0 0 0 1 4 0 0 0 0 0 0 1 0 0 4 0 0 0 −1 1 0 0 −1 , . 12. The eigenvalues of the matrix with this characteristic polynomial are λ = −2, −2, −2, 0, 0, 3, 3. The possible Jordan canonical forms in this case are therefore: −2 0 0 0 −2 0 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 3 −2 0 0 0 −2 0 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 3 −2 1 0 0 −2 0 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 3 , , , −2 0 0 0 −2 0 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 00 00 10 0 03 00 0 0 0 0 0 0 3 −2 1 0 0 −2 0 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 3 −2 1 0 0 −2 0 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 3 , , , −2 0 0 0 −2 0 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 3 −2 1 0 0 −2 0 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 00 00 10 0 03 00 0 0 0 0 0 0 3 −2 1 0 0 −2 1 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 3 0 , , , 412 −2 1 0 0 −2 1 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 00 00 10 0 03 00 0 0 0 0 0 0 3 , −2 1 0 0 −2 1 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 3 , −2 1 0 0 −2 1 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 3 . 13. The eigenvalues of the matrix with this characteristic polynomial are λ = −2, −2, 6, 6, 6, 6, 6. The possible Jordan canonical forms in this case are therefore: −2 00 0 −2 0 0 06 0 00 0 00 0 00 0 00 0 0 0 6 0 0 0 −2 100 0 −2 0 0 0 060 0 006 0 000 0 000 0 000 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 −2 0 0 −2 0 0 0 0 0 0 0 0 0 0 −2 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 −2 1 0 −2 0 0 0 0 0 0 0 0 0 0 −2 000 0 −2 0 0 0 060 , 0 006 0 000 0 000 0 000 00000 0 0 0 0 0 6 1 0 0 0 0 6 1 0 0 , 0 0 6 0 0 0 0 0 6 0 00006 00000 0 0 0 0 0 6 1 0 0 0 0 6 1 0 0 , 0 0 6 1 0 0 0 0 6 0 00006 −2 100 0 −2 0 0 0 061 , 0 006 0 000 0 000 0 000 00000 0 0 0 0 0 6 1 0 0 0 0 6 1 0 0 , 0 0 6 0 0 0 0 0 6 0 00006 0 0 0 1 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 −2 0 0 −2 0 0 0 0 0 0 0 0 0 0 −2 0 0 −2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 −2 1 0 −2 0 0 0 0 0 0 0 0 0 0 −2 00 0 −2 0 0 06 , 0 00 0 00 0 00 0 00 00000 0 0 0 0 0 6 1 0 0 0 0 6 1 0 0 0 0 6 0 0 0 0 0 6 1 00006 00000 0 0 0 0 0 6 1 0 0 0 0 6 1 0 0 0 0 6 1 0 0 0 0 6 1 00006 −2 10 0 −2 0 0 06 , 0 00 0 00 0 00 0 00 00000 0 0 0 0 0 6 1 0 0 0 0 6 1 0 0 0 0 6 0 0 0 0 0 6 1 00006 000 000 100 600 061 006 000 0 0 0 0 0 0 6 000 000 100 600 061 006 000 0 0 0 0 0 0 6 , , 413 −2 100 −2 100000 0 −2 0 0 0 0 0 0 −2 0 0 061 0 0 6 1 0 0 0 0 0 0 0 6 1 0 0 , 0 006 000 0 0 0 0 6 1 0 0 0 0 0 0 0 6 0 0 000 0 000 0 000006 000 0 0 0 0 0 0 1 0 0 6 1 0 0 6 1 006 14. Of the Jordan canonical forms in Problem 13, we are asked here to find the ones that contain exactly five Jordan blocks, since there is a correspondence between the Jordan blocks and the linearly independent eigenvectors. There are three: −2 100000 −2 000000 −2 000000 0 −2 0 0 0 0 0 0 −2 0 0 0 0 0 0 −2 0 0 0 0 0 0 0 6 1 0 0 0 0 0 6 1 0 0 0 0 0 6 1 0 0 0 0 0 0 6 0 0 0 , 0 0 0 6 0 0 0 , 0 0 0 6 1 0 0 0 0 0 0 6 0 0 0 0 0 0 6 1 0 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 6 0 0 0 0 0 0 6 0 0 000006 0 000006 0 000006 15. Many examples are possible here. Let A = 0 is not an eigenvector since Av = 1 eigenvector of A corresponding to λ = 0. v= 1 0 0 0 1 0 . The only eigenvalue of A is 0. The vector = λv. However A2 = 02 , so every vector is a generalized 01 16. Many examples are possible here. Let A = 0 0 0 0 0 1 v = 1 is not an eigenvector since Av = 0 = λv. 0 0 eigenvector of A corresponding to λ = 0. 0 0 . The only eigenvalue of A is 0. The vector 0 However, A2 = 03 , so every vector is a generalized 17. The characteristic polynomial is det(A − λI ) = det 1−λ −1 1 3−λ = (1 − λ)(3 − λ) + 1 = λ2 − 4λ + 4 = (λ − 2)2 , with roots λ = 2, 2. We have A − 2I = −1 1 −1 1 ∼ 1 −1 0 0 . Because there is only one unpivoted column in this latter matrix, we only have one eigenvector for A. Hence, A is not diagonalizable, and therefore 21 JCF(A) = . 02 To determine the matrix S , we must find a cycle of generalized eigenvectors of length 2. Therefore, it suffices 0 to find a vector v in R2 such that (A − 2I )v = 0. Many choices are possible here. We take v = . Then 1 414 (A − 2I )v = 1 1 . Thus, we have S= 10 11 . 18. The characteristic polynomial is 1−λ 1 1 1−λ 1 = (1 − λ) (1 − λ)2 − 1 = (1 − λ)(λ2 − 2λ) = λ(1 − λ)(λ − 2), det(A − λI ) = det 0 0 1 1−λ with roots λ = 0, 1, 2. Since A is a 3 × 3 matrix with three distinct eigenvalues, it is diagonalizable. Therefore, it’s Jordan canonical form is simply a diagonal matrix with the eigenvalues as its diagonal entries: 000 JCF(A) = 0 1 0 . 002 To determine the invertible matrix S , we must find eigenvectors associated with each eigenvalue. Eigenvalue λ = 0: Consider 111 nullspace(A) = nullspace 0 1 1 , 011 111 and this latter matrix can be row reduced to 0 1 1 . The equations corresponding to the rows of this 000 matrix are x + y + z = 0 and y + z = 0. Setting z = t, then y = −t and x = 0. With t = 1 this gives us the eigenvector (0, −1, 1). Eigenvalue λ = 1: Consider 0 nullspace(A − I ) = nullspace 0 0 11 0 1 . 10 By inspection, we see that z = 0 and y = 0 are required, but x = t is free. Thus, an eigenvector associated with λ = 1 may be chosen as (1, 0, 0). Eigenvalue λ = 2: Consider −1 1 1 1 , nullspace(A − 2I ) = nullspace 0 −1 0 1 −1 1 −1 −1 1 −1 . Setting z = t, we have y = t and x = 2t. Thus, with t = 1 which can be row-reduced to 0 0 0 0 we obtain the eigenvector (2, 1, 1) associated with λ = 2. 415 Placing the eigenvectors obtained as the columns of S (with columns corresponding to the eigenvalues of JCF(A) above), we have 012 S = −1 0 1 . 101 19. We can get the characteristic polynomial by using cofactor expansion along the second column as follows: 5−λ nullspace(A−λI ) = det 1 1 0 4−λ 0 −1 −1 = (4−λ) [(5 − λ)(3 − λ) + 1] = (4−λ)(λ2 −8λ+16) = (4−λ)(λ−4)2 , 3−λ with roots λ = 4, 4, 4. 1 0 −1 We have A − 4I = 1 0 −1 , and so vectors (x, y, z ) in the nullspace of this matrix must satisfy 1 0 −1 x − z = 0. Setting z = t and y s, have x = t. Hence, we obtain two linearly independent eigenvectors = we 0 1 of A corresponding to λ = 4: 0 and 1 . Therefore, JCF(A) contains exactly two Jordan blocks. 0 1 This uniquely determines JCF(A), up to a rearrangement of the Jordan blocks: 4 JCF(A) = 0 0 1 4 0 0 0 . 4 To determine the matrix S , we must seek a generalized eigenvector. It is easy to verify that (A − 4I )2 = 03 , so every nonzero vector v is a generalized eigenvector. We must choose one such that (A − 4I ) = 0 in v 1 order to form a cycle of length 2. There are many choices here, but let us choose v = 0 . Then 0 1 (A − 4I )v = 1 . Notice that this is an eigenvector of A corresponding to λ = 4. To complete the matrix 1 S , we will need a second linearly independent eigenvector. Again, there are a multitude of choices. Let us 0 choose the eigenvector 1 found above. Thus, 0 1 S= 1 1 1 0 0 0 1 . 0 20. We will do cofactor expansion along the first column of the matrix to obtain the characteristic polynomial: 416 4−λ det(A − λI ) = det −1 −1 −4 5 4−λ 2 2 4−λ = (4 − λ)(λ2 − 8λ + 12) + (−4)(4 − λ) − 10 − (−8 − 5(4 − λ)) = (4 − λ)(λ2 − 8λ + 12) + (2 − λ) = (4 − λ)(λ − 2)(λ − 6) + (2 − λ) = (λ − 2) [(4 − λ)(λ − 6) − 1] = (λ − 2)(−λ2 + 10λ − 25) = −(λ − 2)(λ − 5)2 , with eigenvalues λ = 2, 5, 5. Since λ = 5 is a repeated eigenvalue, consider this eigenvalue first. We must consider the nullspace we −1 −4 5 1 1 −2 2 , which can be row-reduced to 0 3 −3 , a matrix that has of the matrix A − 5I = −1 −1 −1 2 −1 00 0 only one unpivoted column, and hence λ = 5 only yields one linearly independent eigenvector. Thus, 200 JCF(A) = 0 5 1 . 005 To determine an invertible matrix S , we first proceed to find a cycle of eigenvectors of length 2 corresponding to λ = 5. Therefore, we must find a vector v in R3 such that (A − 5I )v = 0, but (A − 5I )2 v = 0 (in order 1 0 18 −18 that v is a generalized eigenvector). Note that (A − 5I )2 = , so if we set v = 0 , then 0 9 −9 0 −1 (A − 5I )v = −1 and (A − 5I )2 v = 0. Thus, we obtain the cycle −1 1 −1 −1 , 0 . −1 0 Next, corresponding to = 2, we must find an eigenvector. We need to λ find a nonzero vector (x, y, z ) 2 −4 5 1 −2 −2 0 1 . The middle 2 2 , and this latter matrix can be row-reduced to 0 in nullspace −1 0 0 0 −1 22 row requires that z = 0, and if we set y = t, then x = 2t. Thus, by using t = 1, we obtain the eigenvector (2, 1, 0). Thus, we can form the matrix 2 −1 1 S = 1 −1 0 . 0 −1 0 21. We are given that λ = −5 occurs with multiplicity 2 as a root of the characteristic polynomial of A. To 417 search for corresponding eigenvectors, we consider −1 1 0 1 nullspace(A + 5I ) = nullspace − 1 1 , 2 2 2 1 1 1 −2 2 −2 1 −1 0 0 1 . Since there is only one unpivoted column in this row-echelon and this matrix row-reduces to 0 0 00 form of A, the eigenspace corresponding to λ = −5 is only one-dimensional. Thus, based on the eigenvalues λ = −5, −5, −6, we already know that −5 1 0 0 . JCF(A) = 0 −5 0 0 −6 Next, we seek a cycle of generalized eigenvectors of length 2 corresponding to λ = −5. The cycle takes the form {(A + 5I )v, v}, where v is a vector such that (A + 5I )2 v = 0. We readily compute that 1 −1 1 2 2 2 0 0 . An obvious vector that is killed by (A + 5I )2 (although other choices are (A + 5I )2 = 0 1 1 −2 1 2 2 0 1 also possible) is v = 1 . Then (A + 5I )v = 1 . Hence, we have a cycle of generalized eigenvectors 1 0 corresponding to λ = −5: 0 1 1 , 1 . 0 1 Now consider the eigenspace corresponding to λ = −6. We need only find one eigenvector (x, y, z ) in this eigenspace. To do so, we must compute 010 nullspace(A + 6I ) = nullspace − 1 3 1 , 2 2 2 −1 1 1 2 2 2 1 −3 −1 1 0 . We see that y = 0 and x − 3y − z = 0, which is equivalent and this matrix row-reduces to 0 0 0 0 to x − z = 0. Setting z = t, we have x = t. With t = 1, we obtain the eigenvector (1, 0, 1). Hence, we can form the matrix 101 S = 1 1 0 . 011 22. Because the matrix is upper triangular, the eigenvalues of A appear along the main diagonal: λ = 2, 2, 3. Let us first consider the eigenspace corresponding to λ = 2: We consider 0 −2 14 1 −7 , nullspace(A − 2I ) = nullspace 0 0 0 0 418 0 1 −7 0 . There are two unpivoted columns, so this eigenspace is two-dimensional. which row reduces to 0 0 00 0 Therefore, the matrix A is diagonalizable: 200 JCF(A) = 0 2 0 . 003 Next, we must determine an invertible matrix S such that S −1 AS is the Jordan canonical form just obtained. Using the row-echelon form of A − 2I obtained above, vectors (x, y, z ) in nullspace(A − 2I ) must satisfy y − 7z = 0. Setting z = t, y = 7t, and x = s, we obtain the eigenvectors (1, 0, 0) and (0, 7, 1). Next, consider the eigenspace corresponding to λ = 3. We consider −1 −2 14 0 −7 , nullspace(A − 3I ) = nullspace 0 0 0 −1 1 2 −14 1 . Vectors (x, y, z ) in the nullspace of this matrix must satisfy z = 0 which row reduces to 0 0 00 0 and x + 2y − 14z = 0 or x + 2y = 0. Setting y = t and x = −2t. Hence, setting t = 1 gives the eigenvector (−2, 1, 0). Thus, using the eigenvectors obtained above, we obtain the matrix 1 0 −2 1 . S= 0 7 01 0 23. We use the characteristic polynomial to determine the eigenvalues of A: 7−λ −2 2 4−λ −1 det(A − λI ) = det 0 −1 1 4−λ = (4 − λ) [(7 − λ)(4 − λ) + 2] + (7 − λ − 2) = (4 − λ)(λ2 − 11λ + 30) + (5 − λ) = (4 − λ)(λ − 5)(λ − 6) + (5 − λ) = (5 − λ) [1 − (4 − λ)(6 − λ)] = (5 − λ)(λ − 5)2 = −(λ − 5)3 . Hence, the eigenvalues are λ = 5, 5, 5. Let us consider the eigenspace corresponding to λ = 5. We consider 2 −2 2 nullspace(A − 5I ) = nullspace 0 −1 −1 , −1 1 −1 1 −1 1 1 1 , which contains one unpivoted column. Therefore, the and this latter matrix row-reduces to 0 0 00 eigenspace corresponding to λ = 5 is one-dimensional. Therefore, the Jordan canonical form of A consists 419 of one Jordan block: 5 JCF(A) = 0 0 1 5 0 0 1 . 5 A corresponding invertible matrix S in this case must have columns that consist of one cycle of generalized eigenvectors, which will take the form {(A − 5I )2 v, (A − 5I )v, v}, where v is a generalized eigenvector. Now, we can verify quickly that 2 −2 2 20 4 2 , (A − 5I )3 = 03 . A − 5I = 0 −1 −1 , (A − 5I )2 = 1 0 −1 1 −1 −1 0 −2 The fact that (A − 5I )3 = 03 means that every nonzero vector v is a generalized eigenvector. Hence, we 1 simply choose v such that (A − 5I )2 v = 0. There are many choices. Let us take v = 0 . Then 0 2 2 (A − 5I )v = 0 and (A − 5I )2 v = 1 . Thus, we have the cycle of generalized eigenvectors −1 −1 2 1 2 1 , 0 , 0 . −1 0 −1 Hence, we have 2 21 0 0 . S= 1 −1 −1 0 24. Because the matrix is upper triangular, the eigenvalues of A appear along the main diagonal: λ = −1, −1, −1. Let us consider the eigenspace corresponding to λ = −1. We consider 0 −1 0 0 −2 , nullspace(A + I ) = nullspace 0 0 0 0 and it is straightforward to see that the nullspace here consists precisely of vectors that are multiples of (1, 0, 0). Because only one linearly independent eigenvector was obtained, the Jordan canonical form of this matrix consists of only one Jordan block: −1 1 0 1 . JCF(A) = 0 −1 0 0 −1 A corresponding invertible matrix S in this case must have columns that consist of one cycle of generalized eigenvectors, which will take the form {(A + I )2 v, (A + I )v, v}. Here, we have 0 −1 0 002 0 −2 and (A + I )2 = 0 0 0 and (A + I )3 = 03 . A+I = 0 0 0 0 000 420 Therefore, every nonzero vector is a generalized eigenvector. We wish to choose a vector such that v 0 0 (A + I )2 v = 0. There are many choices, but we will choose v = 0 . Then (A + I )v = −2 and 0 1 2 (A + I )2 v = 0 . Hence, we form the matrix S as follows: 0 2 00 S = 0 −2 0 . 0 01 25. We use the characteristic polynomial to determine the eigenvalues of A: 2−λ −1 0 1 0 3−λ −1 0 det(A − λI ) = det 0 1 1−λ 0 0 −1 0 3−λ = (2 − λ)(3 − λ) [(3 − λ)(1 − λ) + 1] = (2 − λ)(3 − λ)(λ2 − 4λ + 4) = (2 − λ)(3 − λ)(λ − 2)2 , and so the eigenvalues are λ = 2, 2, 2, 3. First, consider the eigenspace corresponding to λ = 2. We consider 0 −1 01 0 1 −1 0 , nullspace(A − 2I ) = nullspace 0 1 −1 0 0 −1 01 0 1 −1 0 0 0 1 −1 . There are two unpivoted columns, and and this latter matrix can be row-reduced to 0 0 0 0 00 0 0 therefore two linearly independent eigenvectors corresponding to λ = 2. Thus, we will obtain two Jordan blocks corresponding to λ = 2, and they necessarily will have size 2 × 2 and 1 × 1. Thus, we are already in a position to write down the Jordan canonical form of A: 2100 0 2 0 0 JCF(A) = 0 0 2 0 . 0003 We continue in order to obtain an invertible matrix S such that S −1 AS is in Jordan canonical form. To this end, we see a generalized eigenvector v such that (A − 2I )v = 0 and (A − 2I )2 v = 0. Note that 0 −2 1 1 0 −1 01 0 0 0 0 1 −1 0 . and (A − 2I )2 = 0 A − 2I = 0 0 0 0 0 1 −1 0 0 −2 1 1 0 −1 01 421 0 0 By inspection, we see that by taking v = −1 (there are many other valid choices, of course), then 1 1 1 (A − 2I )v = and (A − 2I )2 v = 0. We also need a second eigenvector corresponding to λ = 2 that is 1 1 linearly independent from (A − 2I )v just obtained. From the row-echelon form of A − 2I , see that all we s 1 t 0 eigenvectors corresponding to λ = 2 take the form , so for example, we can take . t 0 t 0 Next, we consider the eigenspace corresponding to λ = 3. We consider −1 −1 01 0 0 −1 0 . nullspace(A − 3I ) = nullspace 0 1 −2 0 0 −1 00 Now, if (x, y, z, w) is an eigenvector corresponding to λ = 3, the last three rows of the matrix imply that y = z = 0. Thus, the first row becomes −x + w = 0. Setting w = t, then x = t, so we obtain eigenvectors in the form (t, 0, 0, t). Setting t = 1 gives the eigenvector (1, 0, 0, 1). Thus, we can now form the matrix S such that S −1 AS is the Jordan canonical form we obtained above: 1 011 1 0 0 0 S= 1 −1 0 0 . 1 101 26. From the characteristic polynomial, we have eigenvalues λ = 2, 2, 4, 4. Let us consider the associated eigenspaces. Corresponding to λ = 2, we seek eigenvectors (x, y, z, w) by computing 0 −4 2 2 −2 −2 1 3 nullspace(A − 2I ) = nullspace −2 −2 1 3 , −2 −6 3 5 2 2 −1 −3 0 2 −1 −1 . Setting w = 2t and z = 2s, we obtain y = s + t and this matrix can be row-reduced to 0 0 0 0 00 0 0 and x = 2t, so we obtain the eigenvectors (2, 1, 0, 2) and (0, 1, 2, 0) corresponding to λ = 2. Next, corresponding to λ = 4, we seek eigenvectors (x, y, z, w) by computing −2 −4 22 −2 −4 1 3 nullspace(A − 4I ) = nullspace −2 −2 −1 3 . −2 −6 33 422 1 2 −1 −1 0 2 −1 −1 . Since there is only one unpivoted column, this eigenspace This matrix can be reduced to 0 0 1 −1 00 0 0 is only one-dimensional, despite λ = 4 occurring with multiplicity 2 as a root of the characteristic equation. Therefore, we must seek a generalized eigenvector v such that (A − 4I )v is an eigenvector. This in turn requires that (A − 4I )2 v = 0. We find that −2 −2 A − 4I = −2 −2 −4 22 −4 1 3 −2 −1 3 −6 33 4 8 −4 4 4 0 (A − 4I )2 = 4 0 4 4 8 −4 and 1 0 Note that the vector v = satisfies (A − 4I )2 v = 0 and (A − 4I )v = 0 1 of generalized eigenvectors corresponding to λ = 4 given by 1 0 10 , . 0 1 1 1 Hence, we can form the matrix 2 1 S= 0 2 0 1 2 0 0 1 1 1 1 0 0 1 2 0 JCF(A) = 0 0 and 27. Since A is upper triangular, the eigenvalues appear along at 0 0 nullspace(A − 2I ) = nullspace 0 0 0 0 2 0 0 −4 −4 . −4 −4 0 1 . Thus, we have the cycle 1 1 0 0 . 1 4 0 0 4 0 the main diagonal: λ = 2, 2, 2, 2, 2. Looking 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 1 0 , we see that the row-echelon form of this matrix will contain two pivots, and therefore, three unpivoted columns. That means that the eigenspace corresponding to λ = 2 is three-dimensional. Therefore, JCF(A) consists of three Jordan blocks. The only list of block sizes for a 5 × 5 matrix with three blocks are (a) 3,1,1 and (b) 2,2,1. In this case, note that (A − 2I ) = 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 1 1 1 1 = 05 , 423 so that it is possible to find a vector v that generates a cycle of generalized eigenvectors of length 3: {(A − 2I )2 v, (A − 2I )v, v}. Thus, JCF(A) contains a Jordan block of size 3 × 3. We conclude that the correct list of block sizes for this matrix is 3,1,1: 21000 0 2 1 0 0 JCF(A) = 0 0 2 0 0 . 0 0 0 2 0 00002 28. Since A is upper triangular, the eigenvalues appear along the main diagonal: λ = 0, 0, 0, 0, 0. Looking at nullspace(A − 0I ) = nullspace(A), we see that eigenvectors (x, y, z, u, v ) corresponding to λ = 0 must satisfy z = u = v = 0 (since the third row gives 6u = 0, the first row gives u + 4v = 0, and the second row gives z + u + v = 0). Thus, we have only two free variables, and thus JCF(A) will consist of two Jordan blocks. The only list of block sizes for a 5 × 5 matrix with two blocks are (a)4,1 and (b) 3,2. In this case, it is easy to verify that A3 = 0, so that the longest possible cycle of generalized eigenvectors {A2 v, Av, v} has length 3. Therefore, case (b) holds: JCF(A) consists of one Jordan block of size 3 × 3 and one Jordan block of size 2 × 2: 01000 0 0 1 0 0 JCF(A) = 0 0 0 0 0 . 0 0 0 0 1 00000 29. Since A is upper triangular, the eigenvalues appear Looking at 0 0 0 0 nullspace(A − I ) = nullspace 0 0 0 0 along the main diagonal: λ = 1, 1, 1, 1, 1, 1, 1, 1. 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 , we see that if (a, b, c, d, e, f, g, h) is an eigenvector of A, then b = d = f = h = 0, and a, c, e, and g are free variables. Thus, we have four linearly independent eigenvectors of A, and hence we expect four Jordan blocks. Now, an easy calculation shows that (A − I )2 = 0, and thus, no Jordan blocks of size greater than 2 × 2 are permissible. Thus, it must be the case that JCF(A) consists of four Jordan blocks, each of which is a 2 × 2 matrix: 11000000 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 JCF(A) = 0 0 0 0 1 1 0 0 . 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 00000001 424 30. NOT SIMILAR. We will compute JCF(A) and JCF(B ). If they are the same (up to a rearrangement of the Jordan blocks), then A and B are similar; otherwise they are not. Both matrices have eigenvalues λ = 6, 6, 6. Consider 1 10 nullspace(A − 6I ) = nullspace −1 −1 0 . 1 00 We see that eigenvectors (x, y, z ) corresponding to λ = 6 must satisfy x + y = 0 and x = 0. Therefore, x = y = 0, while z is a free variable. Since we obtain only one free variable, JCF(A) consists of just one Jordan block corresponding to λ = 6: 610 JCF(A) = 0 6 1 . 006 Next, consider 0 −1 1 0 0 . nullspace(B − 6I ) = nullspace 0 0 00 In this case, eigenvectors (x, y, z ) corresponding to λ = 6 must satisfy −y + z = 0. Therefore, both x and z are free variables, and hence, JCF(B ) consists of two Jordan blocks corresponding to λ = 6: 610 JCF(B ) = 0 6 0 . 006 Since JCF(A) = JCF(B ), we conclude that A and B are not similar. 31. SIMILAR. We will compute JCF(A) and JCF(B ). If they are the same (up to a rearrangement of the Jordan blocks), then A and B are similar; otherwise they are not. Both matrices have eigenvalues λ = 5, 5, 5. (For A, this is easiest to compute by expanding det(A − λI ) along the middle row, and for B , this is easiest to compute by expanding det(B − λI ) along the second column.) In Problem 23, we computed 510 JCF(A) = 0 5 1 . 005 Next, consider −2 −1 −2 1 1 , nullspace(B − 5I ) = nullspace 1 1 0 1 111 and this latter matrix can be row-reduced to 0 1 0 , which has only one unpivoted column. Thus, 000 the eigenspace of B corresponding to λ = 5 is only one-dimensional, and so 510 JCF(B ) = 0 5 1 . 005 Since A and B each had the same Jordan canonical form, they are similar matrices. 425 32. The eigenvalues of A are λ = −1, −1, and the eigenspace corresponding to λ = −1 is only onedimensional. Thus, we seek a generalized eigenvector v of A corresponding to λ = −1 such that {(A + I )v, v} −2 −2 is a cycle of generalized eigenvectors. Note that A + I = and (A + I )2 = 02 . Thus, every 2 2 1 nonzero vector v is a generalized eigenvector of A corresponding to λ = −1. Let us choose v = . Then 0 −2 (A + I )v = . Form the matrices 2 S= −2 2 1 0 and J= −1 1 0 −1 . Via the substitution x = S y, the system x = Ax is transformed into y = J y. The corresponding equations are y1 = −y1 + y2 and y2 = −y2 . The solution to the second equation is y2 (t) = c1 e−t . Substituting this solution into y1 = −y1 + y2 gives y1 + y1 = c1 e−t . This is a first order linear equation with integrating factor I (t) = et . When we multiply the differential equation for y1 (t) by I (t), it becomes (y1 · et ) = c1 . Integrating both sides yields y1 · et = c1 t + c2 . Thus, y1 (t) = c1 te−t + c2 e−t . Thus, we have y(t) = y1 (t) y2 (t) = c1 te−t + c2 e−t c1 e−t . Finally, we solve for x(t): x(t) = S y(t) = = −2 2 1 0 c1 te−t + c2 e−t c1 e−t −2(c1 te−t + c2 e−t ) + c1 e−t 2(c1 te−t + c2 e−t ) = c1 e−t −2t + 1 2t + c2 e−t −2 2 . This is an acceptable answer, or we can write the individual equations comprising the general solution: x1 (t) = −2(c1 te−t + c2 e−t ) + c1 e−t and x2 (t) = 2(c1 te−t + c2 e−t ). 33. The eigenvalues of A are λ = −1, −1, 1. The eigenspace corresponding to λ = −1 is 110 nullspace(A + I ) = nullspace 0 1 1 , 110 426 which is only one-dimensional, spanned by the vector (1, −1, 1). Therefore, we seek a generalized eigenvector v of A corresponding to λ = −1 such that {(A + I )v, v} is a cycle of generalized eigenvectors. Note that 110 121 A + I = 0 1 1 and (A + I )2 = 1 2 1 . 110 121 In order that v be a generalized eigenvector of A corresponding to λ = −1, we should choose v such that 1 (A + I )2 v = 0 and (A + I )v = 0. There are many valid choices; let us choose v = 0 . Then −1 1 (A + I )v = −1 . Hence, we obtain the cycle of generalized eigenvectors corresponding to λ = −1: 1 1 1 −1 , 0 . 1 −1 Next, consider the eigenspace corresponding to λ = 1. For this, we compute −1 1 0 1 . nullspace(A − I ) = nullspace 0 −1 1 1 −2 1 −1 0 1 1 −1 . We find the eigenvector 1 This can be row-reduced to 0 0 0 0 1 Hence, we are ready to form the matrices S and J : −1 1 1 11 0 1 and J = 0 −1 S = −1 0 0 1 −1 1 as a basis for this eigenspace. 0 0 . 1 Via the substitution x = S y, the system x = Ax is transformed into y = J y. The corresponding equations are y1 = −y1 + y2 , y2 = −y2 , y3 = y3 . The third equation has solution y3 (t) = c3 et , the second equation has solution y2 (t) = c2 e−t , and so the first equation becomes y1 + y1 = c2 e−t . This is a first-order linear equation with integrating factor I (t) = et . When we multiply the differential equation for y1 (t) by I (t), it becomes (y1 · et ) = c2 . Integrating both sides yields y1 · et = c2 t + c1 . Thus, y1 (t = c2 te−t + c1 e−t . Thus, we have y1 (t) c2 te−t + c1 e−t . c2 e−t y(t) = y2 (t) = t y3 (t) c3 e 427 Finally, we solve for x(t): 1 11 c2 te−t + c1 e−t 0 1 c2 e−t x(t) = S y(t) = −1 1 −1 1 c3 et c2 te−t + c1 e−t + c2 e−t + c3 et −c2 te−t − c1 e−t + c3 et = c2 te−t + c1 e−t − c2 e−t + c3 et 1 t+1 1 = c1 e−t −1 + c2 e−t −t + c3 et 1 . 1 t−1 1 34. The eigenvalues of A are λ = −2, −2, −2. The eigenspace corresponding to λ = −2 is 0 0 0 nullspace(A + 2I ) = nullspace 1 −1 −1 , −1 1 1 and there are two linearly independent vectors in this nullspace, corresponding to the unpivoted columns of the row-echelon form of this matrix. Therefore, the Jordan canonical form of A is −2 1 0 0 . J = 0 −2 0 0 −2 To form an invertible matrix S such that S −1 AS = J , we must find a cycle of generalized eigenvectors corresponding to λ = −2 of length 2: {(A + 2I )v, v}. Now 0 0 0 A + 2I = 1 −1 −1 and (A + 2I )2 = 03 . −1 1 1 Since (A + 2I )2 = 03 , every nonzero vector in R3 is a generalized eigenvector corresponding to λ = −2. We need only find a nonzero vector v such that (A + 2I )v = 0. There are many valid choices; let us choose 1 0 v = 0 . Then (A + 2I )v = 1 , an eigenvector of A corresponding to λ = −2. We also need a 0 −1 second linearly independent eigenvector corresponding to λ = −2. There are many choices; let us choose 1 1 . Therefore, we can form the matrix 0 01 S= 1 0 −1 0 1 1 . 0 Via the substitution x = S y, the system x = Ax is transformed into y = J y. The corresponding equations are y1 = −2y1 + y2 , y2 = −2y2 , y3 = −2y3 . 428 The third equation has solution y3 (t) = c3 e−2t , the second equation has solution y2 (t) = c2 e−2t , and so the first equation becomes y1 + 2y1 = c2 e−2t . This is a first-order linear equation with integrating factor I (t) = e2t . When we multiply the differential equation for y1 (t) by I (t), it becomes (y1 · e2t ) = c2 . Integrating both sides yields y1 · e2t = c2 t + c1 . Thus, y1 (t) = c2 te−2t + c1 e−2t . Thus, we have y1 (t) c2 te−2t + c1 e−2t . c2 e−2t y(t) = y2 (t) = −2t y3 (t) c3 e Finally, we solve for x(t): 1 c2 te−2t + c1 e−2t 1 c2 e−2t 0 c3 e−2t c2 e−2t + c3 e−2t = c2 te−2t + c1 e−2t + c3 e−2t −(c2 te−2t + c1 e−2t ) 1 1 0 = c1 e−2t 1 + c2 e−2t t + c3 e−2t 1 . 0 −t −1 0 x(t) = S y(t) = 1 −1 1 0 0 35. The eigenvalues of A are λ = 4, 4, 4. The eigenspace corresponding to λ = 4 is 000 nullspace(A − 4I ) = nullspace 1 0 0 , 010 and there is only one eigenvector. Therefore, the Jordan 41 J = 0 4 00 canonical form of A is 0 1 . 4 Next, we need to find an invertible matrix S such that S −1 AS = J . To do this, we must find a cycle of generalized eigenvectors {(A − 4I )2 v, (A − 4I )v, v} of length 3 corresponding to λ = 4. We have 000 000 A − 4I = 1 0 0 and (A − 4I )2 = 0 0 0 and (A − 4I )3 = 03 . 010 100 From (A − 4I )3 = 03 , we know that every nonzero vector is a generalized eigenvector corresponding to 1 λ = 4. We choose v = 0 (any multiple of the chosen vector v would be acceptable as well). Thus, 0 429 0 0 (A − 4I )v = 1 and (A − 4I )2 v = 0 . Thus, we have the cycle of generalized eigenvectors 0 1 0 1 0 0 , 1 , 0 . 1 0 0 Thus, we can form the matrix 0 S= 0 1 0 1 0 1 0 . 0 Via the substitution x = S y, the system x = Ax is transformed into y = J y. The corresponding equations are y1 = 4y1 + y2 , y 2 = 4 y2 + y 3 , y3 = 4y3 . The third equation has solution y3 (t) = c3 e4t , and the second equation becomes y2 − 4y2 = c3 e4t . This is a first-order linear equation with integrating factor I (t) = e−4t . When we multiply the differential equation for y2 (t) by I (t), it becomes (y2 · e−4t ) = c3 . Integrating both sides yields y2 · e−4t = c3 t + c2 . Thus, y2 (t) = c3 te4t + c2 e4t = e4t (c3 t + c2 ). Therefore, the differential equation for y1 (t) becomes y1 − 4y1 = e4t (c3 t + c2 ). This equation is first-order linear with integrating factor I (t) = e−4t . When we multiply the differential equation for y1 (t) by I (t), it becomes (y1 · e−4t ) = c3 t + c2 . Integrating both sides, we obtain y1 · e−4t = c3 t2 + c2 t + c1 . 2 Hence, y1 (t) = e4t c3 t2 + c2 t + c1 . 2 Thus, we have 4t 2 y1 (t) e c3 t2 + c2 t + c1 y(t) = y2 (t) = e4t (c3 t + c2 ) y3 (t) c3 e4t . 430 Finally, we solve for x(t): 4t 2 1 e c3 t2 + c2 t + c1 0 e4t (c3 t + c2 ) 0 c3 e4t c3 e4t e4t (c3 t + c2 ) 0 x(t) = S y(t) = 0 1 0 1 0 = 2 e4t c3 t2 + c2 t + c1 0 0 1 = c1 e4t 0 + c2 e4t 1 + c3 e4t t . 1 t t2 /2 36. The eigenvalues of A are λ = −3, −3. The eigenspace corresponding to λ = −3 is only one-dimensional, and therefore the Jordan canonical form of A contains one 2 × 2 Jordan block: −3 1 0 −3 J= . Next, we look for a cycle of generalized eigenvectors of the form {(A + 3I )v, v}, where v is a generalized 2 1 −1 eigenvector of A. Since (A +3I )2 = = 02 , every nonzero vector in R2 is a generalized eigenvector. 1 −1 1 1 Let us choose v = . Then (A + 3I )v = . Thus, we form the matrix 0 1 11 10 S= . Via the substitution x = S y, the system x = Ax is transformed into y = J y. The corresponding equations are y1 = −3y1 + y2 and y2 = −3y2 . The second equation has solution y2 (t) = c2 e−3t . Substituting this expression for y2 (t) into the differential equation for y1 (t) yields y1 + 3y1 = c2 e−3t . An integrating factor for this first-order linear differential equation is I (t) = e3t . Multiplying the differential equation for y1 (t) by I (t) gives us (y1 · e3t ) = c2 . Integrating both sides, we obtain y1 · e3t = c2 t + c1 . Thus, y1 (t) = c2 te−3t + c1 e−3t . Thus, y(t) = y1 (t) y2 (t) = c2 te−3t + c1 e−3t c2 e−3t . Finally, we solve for x(t): x(t) = S y(t) = = 1 1 c2 te−3t + c1 e−3t c2 e−3t 1 0 c2 te−3t + c1 e−3t + c2 e−3t c2 te−3t + c1 e−3t = c1 e−3t 1 1 + c2 e−3t t+1 t . 431 Now, we must apply the initial condition: 0 −1 1 1 = x(0) = c1 + c2 1 0 . Therefore, c1 = −1 and c2 = 1. Hence, the unique solution to the given initial-value problem is x(t) = −e−3t 1 1 + e−3t t+1 t . 37. Let J = JCF(A) = JCF(B ). Thus, there exist invertible matrices S and T such that S −1 AS = J and T −1 BT = J. Thus, S −1 AS = T −1 BT, and so B = T S −1 AST −1 = (ST −1 )−1 A(ST −1 ), which implies by definition that A and B are similar matrices. 38. Since the characteristic polynomial has degree 3, we know that A is a 3 × 3 matrix. Moreover, the roots of the characteristic equation has roots λ = 0, 0, 0. Hence, the Jordan canonical form J of A must be one of the three below: 000 010 010 0 0 0 , 0 0 0 , 0 0 1 . 000 000 000 In all three cases, note that J 3 = 03 . Moreover, there exists an invertible matrix S such that S −1 AS = J . Thus, A = SJS −1 , and so A3 = (SJS −1 )3 = SJ 3 S −1 = S 03 S −1 = 03 , which implies that A is nilpotent. 39. (a): Let J be an n × n Jordan block with eigenvalue λ. Then the eigenvalues of J T are λ (with multiplicity n). The matrix J T − λI consists of 1’s on the subdiagonal (the diagonal parallel and directly beneath the main diagonal) and zeros elsewhere. Hence, the null space of J T − λI is one-dimensional (with a free variable corresponding to the right-most column of J T − λI ). Therefore, the Jordan canonical form of J T consists of a single Jordan block, since there is only one linearly independent eigenvector corresponding to the eigenvalue λ. However, a single Jordan block with eigenvalue λ is precisely the matrix J . Therefore, JCF(J T ) = J. (b): Let JCF(A) = J . Then there exists an invertible matrix S such that S −1 AS = J . Transposing both sides, we obtain (S −1 AS )T = J T , or S T AT (S −1 )T = J T , or S T AT (S T )−1 = J T . Hence, the matrix AT is similar to J T . However, by applying part (a) to each block in J T , we find that JCF(J T ) = J . Hence, J T is similar to J . By Problem 26 in Section 5.8, we conclude that AT is similar to J . Hence, AT and J have the same Jordan canonical form. However, since JCF(J ) = J , we deduce that JCF(AT ) = J = JCF(A), as required. Solutions to Section 5.12 432 Problems: 1. NO. Note that T (1, 1) = (2, 0, 0, 1) and T (2, 2) = (4, 0, 0, 4) = 2T (1, 1). Thus, T is not a linear transformation. 2. YES. The function T can be represented by the matrix function T (x) = Ax, where A= 2 −3 0 −1 00 and x is a vector in R3 . Every matrix transformation of the form T (x) = Ax is linear. Since the domain of T has larger dimension than the codomain of T , T cannot be one-to-one. However, since T (0, −1/3, 0) = (1, 0) and T (−1, −2/3, 0) = (0, 1), we see that T is onto. Thus, Rng(T ) = R2 (2-dimensional), and so a basis for Rng(T ) is {(1, 0), (0, 1)}. The kernel of T consists of vectors of the form (0, 0, z ), and hence, a basis for Ker(T ) is {(0, 0, 1)} and Ker(T ) is 1-dimensional. 3. YES. The function T can be represented by the matrix function T (x) = Ax, where A= 0 0 −3 2 −1 5 and x is a vector in R3 . Every matrix transformation of the form T (x) = Ax is linear. Since the domain of T has larger dimension than the codomain of T , T cannot be one-to-one. However, since T (1, 1/3, −1/3) = (1, 0) and T (1, 1, 0) = (0, 1), we see that T is onto. Thus, Rng(T ) = R2 , and so a basis for Rng(T ) is {(1, 0), (0, 1)}, and Rng(T ) is 2-dimensional. The kernel of T consists of vectors of the form (t, 2t, 0), where t ∈ R, and hence, a basis for Ker(T ) is {(1, 2, 0)}. We have that Ker(T ) is 1-dimensional. 4. YES. The function T is a linear transformation, because if g, h ∈ C [0, 1], then T (g + h) = ((g + h)(0), (g + h)(1)) = (g (0) + h(0), g (1) + h(1)) = (g (0), g (1)) + (h(0), h(1)) = T (g ) + T (h), and if c is a scalar, T (cg ) = ((cg )(0), (cg )(1)) = (cg (0), cg (1)) = c(g (0), g (1)) = cT (g ). Note that any function g ∈ C [0, 1] for which g (0) = g (1) = 0 (such as g (x) = x2 − x) belongs to Ker(T ), and hence, T is not one-to-one. However, given (a, b) ∈ R2 , note that g defined by g (x) = a + (b − a)x satisfies T (g ) = (g (0), g (1)) = (a, b), 433 so T is onto. Thus, Rng(T ) = R2 , with basis {(1, 0), (0, 1)} (2-dimensional). Now, Ker(T ) is infinitedimensional. We cannot list a basis for this subspace of C [0, 1]. 5. YES. The function T can be represented by the matrix function T (x) = Ax, where 1/5 A= 1/5 and x is a vector in R2 . Every such matrix transformation is linear. Since the domain of T has larger dimension than the codomain of T , T cannot be one-to-one. However, T (5, 0) = 1, so we see that T is onto. Thus, Rng(T ) = R, a 1-dimensional space with basis {1}. The kernel of T consists of vectors of the form t(1, −1), and hence, a basis for Ker(T ) is {(1, −1)}. We have that Ker(T ) is 1-dimensional. 6. NO. For instance, note that T 1 1 0 1 = (1, 0) T 1 1 0 1 +T 1 0 1 1 + 1 0 1 1 and T 1 0 1 1 = (0, 1), and so = (1, 0) + (0, 1) = (1, 1). However, T 1 1 0 1 =T 2 1 1 2 = (2, 2). Thus, T does not respect addition, and hence, T is not a linear transformation. Similar work could be given to show that T also fails to respect scalar multiplication. 7. YES. We can verify that T respects addition and scalar multiplication as follows: T respects addition: Let a1 + b1 x + c1 x2 and a2 + b2 x + c2 x2 belong to P2 . Then T ((a1 + b1 x + c1 x2 ) + (a2 + b2 x + c2 x2 )) = T ((a1 + a2 ) + (b1 + b2 )x + (c1 + c2 )x2 ) = −(a1 + a2 ) − (b1 + b2 ) 0 3(c1 + c2 ) − (a1 + a2 ) −2(b1 + b2 ) = −a1 − b1 3c1 − a1 0 −2b1 + −a2 − b2 3c2 − a2 0 −2b2 = T (a1 + b1 x + c1 x2 ) + T (a2 + b2 x + c2 x2 ). T respects scalar multiplication: Let a + bx + cx2 belong to P2 and let k be a scalar. Then we have T (k (a + bx + cx2 )) = T ((ka) + (kb)x + (kc)x2 ) = = =k −ka − kb 0 3(kc) − (ka) −2(kb) k (−a − b) 0 k (3c − a) k (−2b) −a − b 3c − a 0 −2b = kT (a + bx + cx2 ). 434 Next, observe that a + bx + cx2 belongs to Ker(T ) if and only if −a − b = 0, 3c − a = 0, and −2b = 0. These equations require that b = 0, a = 0, and c = 0. Thus, Ker(T ) = {0}, which implies that Ker(T ) is 0-dimensional (with basis ∅), and that T is one-to-one. However, since M2 (R) is 4-dimensional and P2 is only 3-dimensional, we see immediately that T cannot be onto. By the Rank-Nullity Theorem, in fact, Rng(T ) must be 3-dimensional, and a basis is given by −1 −1 Basis for Rng(T ) = {T (1), T (x), T (x2 )} = 0 0 −1 0 0 −2 , 0 3 , 0 0 . 8. YES. We can verify that T respects addition and scalar multiplication as follows: T respects addition: Let A, B belong to M2 (R). Then T (A + B ) = (A + B ) + (A + B )T = A + B + AT + B T = (A + AT ) + (B + B T ) = T (A) + T (B ). T respects scalar multiplication: Let A belong to M2 (R), and let k be a scalar. Then T (kA) = (kA) + (kA)T = kA + kAT = k (A + AT ) = kT (A). Thus, T is a linear transformation. Note that if A is any skew-symmetric matrix, then T (A) = A + AT = A + (−A) = 0, so Ker(T ) consists precisely of the 2 × 2 skew-symmetric matrices. These matrices take the 0 −a 0 −1 form , for a constant a, and thus a basis for Ker(T ) is given by , and Ker(T ) is a 0 1 0 1-dimensional. Consequently, T is not one-to-one. Therefore, by Proposition 5.4.13, T also fails to be onto. In fact, by the Rank-Nullity Theorem, Rng(T ) must be 3-dimensional. A typical element of the range of T takes the form T ab cd = ab cd ac bd + = 2a b+c c+b 2d . The characterizing feature of this matrix is that it is symmetric. So Rng(T ) consists of all 2 × 2 symmetric matrices, and hence a basis for Rng(T ) is 1 0 0 0 , 0 1 1 0 , 0 0 0 1 . 9. YES. We can verify that T respects addition and scalar multiplication as follows: T respects addition: Let (a1 , b1 , c1 ) and (a2 , b2 , c2 ) belong to R3 . Then T ((a1 , b1 , c1 ) + (a2 , b2 , c2 )) = T (a1 + a2 , b1 + b2 , c1 + c2 ) = (a1 + a2 )x2 + (2(b1 + b2 ) − (c1 + c2 ))x + (a1 + a2 − 2(b1 + b2 ) + (c1 + c2 )) = [a1 x2 + (2b1 − c1 )x + (a1 − 2b1 + c1 )] + [a2 x2 + (2b2 − c2 )x + (a2 − 2b2 + c2 )] = T ((a1 , b1 , c1 )) + T ((a2 , b2 , c2 )). T respects scalar multiplication: Let (a, b, c) belong to R3 and let k be a scalar. Then T (k (a, b, c)) = T (ka, kb, kc) = (ka)x2 + (2kb − kc)x + (ka − 2kb + kc) = k (ax2 + (2b − c)x + (a − 2b + c)) = kT ((a, b, c)). 435 Thus, T is a linear transformation. Now, (a, b, c) belongs to Ker(T ) if and only if a = 0, 2b − c = 0, and a − 2b + c = 0. These equations collectively require that a = 0 and 2b = c. Setting c = 2t, we find that b = t. Hence, (a, b, c) belongs to Ker(T ) if and only if (a, b, c) has the form (0, t, 2t) = t(0, 1, 2). Hence, {(0, 1, 2)} is a basis for Ker(T ), which is therefore 1-dimensional. Hence, T is not one-to-one. By Proposition 5.4.13, T is also not onto. In fact, the Rank-Nullity Theorem implies that Rng(T ) must be 2-dimensional. It is spanned by {T (1, 0, 0), T (0, 1, 0), T (0, 0, 1)} = {x2 + 1, 2x − 2, −x + 1}, but the last two polynomials are proportional to each other. Omitting the polynomial 2x − 2 (this is an arbitrary choice; we could have omitted −x + 1 instead), we arrive at a basis for Rng(T ): {x2 + 1, −x + 1}. 10. YES. We can verify that T respects addition and scalar multiplication as follows: T respects addition: Let (x1 , x2 , x3 ) and (y1 , y2 , y3 ) be vectors in R3 . Then T ((x1 , x2 , x3 ) + (y1 , y2 , y3 )) = T (x1 + y1 , x2 + y2 , x3 + y3 ) = 0 (x1 + y1 ) − (x2 + y2 ) + (x3 + y3 ) −(x1 + y1 ) + (x2 + y2 ) − (x3 + y3 ) 0 = 0 −x1 + x2 − x3 x1 − x2 + x3 0 + 0 −y1 + y2 − y3 y1 − y2 + y 3 0 = T ((x1 , x2 , x3 )) + T ((y1 , y2 , y3 )). T respects scalar multiplication: Let (x1 , x2 , x3 ) belong to R3 and let k be a scalar. Then T (k (x1 , x2 , x3 )) = T (kx1 , kx2 , kx3 ) = =k 0 (kx1 ) − (kx2 ) + (kx3 ) −(kx1 ) + (kx2 ) − (kx3 ) 0 0 −x1 + x2 − x3 x1 − x2 + x3 0 = kT ((x1 , x2 , x3 )). Thus, T is a linear transformation. Now, (x1 , x2 , x3 ) belongs to Ker(T ) if and only if x1 − x2 + x3 = 0 and −x1 + x2 − x3 = 0. Of course, the latter equation is equivalent to the former, so the kernel of T consists simply of ordered triples (x1 , x2 , x3 ) with x1 − x2 + x3 = 0. Setting x3 = t and x2 = s, we have x1 = s − t, so a typical element of Ker(T ) takes the form (s − t, s, t), where s, t ∈ R. Extracting the free variables, we find a basis for Ker(T ): {(1, 1, 0), (−1, 0, 1)}. Hence, Ker(T ) is 2-dimensional. By the Rank-Nullity Theorem, Rng(T ) must be 1-dimensional. In fact, Rng(T ) consists precisely of the 0 −1 set of 2 × 2 skew-symmetric matrices, with basis . Since M2 (R) is 4-dimensional, T fails to be 1 0 onto. 11. We have T (x, y, z ) = (−x + 8y, 2x − 2y − 5z ). 12. We have T (x, y ) = (−x + 4y, 2y, 3x − 3y, 3x − 3y, 2x − 6y ). 13. We have T (x) = x x T (2) = (−1, 5, 0, −2) = 2 2 x 5x − , , 0, −x . 22 436 ab cd 14. For an arbitrary 2 × 2 matrix ab cd 1 0 =r , if we write 0 1 0 1 +s 1 0 1 0 +t 0 0 1 0 +u 1 0 , we can solve for r, s, t, u to find r = d, t = a − b + c − d, s = c, u = b − c. Thus, T ab cd =T 1 0 d = dT 1 0 0 1 0 1 +c + cT 0 1 1 0 01 10 + (a − b + c − d) + (a − b + c − d)T 1 0 0 0 1 0 + (b − c) 0 0 + (b − c)T 1 0 1 0 1 0 1 0 = d(2, −5) + c(0, −3) + (a − b + c − d)(1, 1) + (b − c)(−6, 2) = (a − 7b + 7c + d, a + b − 4c − 6d). 15. For an arbitrary element ax2 + bx + c in P2 , if we write ax2 + bx + c = r(x2 − x − 3) + s(2x + 5) + 6t = rx2 + (−r + 2s)x + (−3r + 5s + 6t), we can solve for r, s, t to find r = a, s = 1 (a + b), and t = 2 1 12 a − 5 12 b + 1 c. Thus, 6 1 5 1 a(x2 − x − 3) + (a + b)(2x + 5) + a − b + c 2 2 2 1 5 c 1 a− b+ T (6) = aT (x2 − x − 3) + (a + b)T (2x + 5) + 2 12 12 6 1 1 5 c −2 1 0 1 12 6 =a + a− b+ + (a + b) 2 −2 6 18 −4 −1 2 12 12 6 T (ax2 + bx + c) = T = −a − 5b + 2c 2a − 2b + c − 5 a − 3 b + c − 1 a − 17 b + 3c 2 2 2 2 . 16. Since dim[P5 ] = 6 and dim[M2 (R)] = 4 = dim[Rng(T )] (since T is onto), the Rank-Nullity Theorem gives dim[Ker(T )] = 6 − 4 = 2. 17. Since T is one-to-one, dim[Ker(T )] = 0, so the Rank-Nullity Theorem gives dim[Rng(T )] = dim[M2×3 (R)] = 6. 18. Since A is lower triangular, its eigenvalues lie along the main diagonal, λ1 = 3 and λ2 = −1. Since A is 2 × 2 with two distinct eigenvalues, A is diagonalizable. To get an invertible matrix S such that S −1 AS = D, we need to find an eigenvector associated with each eigenvector: Eigenvalue λ1 = 3: To get an eigenvector, we consider nullspace(A − 3I ) = nullspace 0 16 0 −4 , 437 1 4 and we see that one possible eigenvector is . Eigenvalue λ2 = −1: To get an eigenvector, we consider nullspace(A + I ) = nullspace 0 1 and we see that one possible eigenvector is 40 16 0 , 3 0 0 −1 . . Putting the above results together, we form S= 1 4 0 1 and D= 19. To compute the eigenvalues, we find the characteristic equation det(A − λI ) = det 13 − λ 25 −9 −17 − λ = (13 − λ)(−17 − λ) + 225 = λ2 + 4λ + 4, and the roots of this equation are λ = −2, −2. Eigenvalue λ = 2: We compute nullspace(A − 2I ) = nullspace 15 −9 25 −15 , but since there is only one linearly independent solution to the corresponding system (one free variable), the eigenvalue λ = 2 does not have two linearly independent solutions. Hence, A is not diagonalizable. 20. To compute the eigenvalues, we find the characteristic equation −4 − λ 3 0 = (−1 − λ) [(−4 − λ)(5 − λ) + 18] 5−λ 0 det(A − λI ) = det −6 3 −3 −1 − λ = (−1 − λ)(λ2 − λ − 2) = −(λ + 1)2 (λ − 2), so the eigenvalues are λ1 = −1 and λ2 = 2. Eigenvalue λ1 = −1: To get eigenvectors, we consider −3 30 1 −1 0 6 0 ∼ 0 0 0 . nullspace(A + I ) = nullspace −6 3 −3 0 0 00 There are two free variables, z = t and y = s. From the first equation x = s. Thus, two linearly independent eigenvectors can be obtained corresponding to λ1 = −1: 1 0 1 and 0 . 0 1 438 Eigenvalue λ2 = 2: To get an eigenvector, we consider −6 3 0 1 −1 −1 3 0 ∼ 0 1 2 . nullspace(A − 2I ) = nullspace −6 3 −3 −3 0 0 0 We let z = t. Then y = −2t from the middle line, and x = −t from the top line. Thus, an eigenvector corresponding to λ2 = 2 may be chosen as −1 −2 . 1 Putting the above results together, we −1 S = −2 1 form 1 1 0 0 0 1 and 2 0 0 0 . D = 0 −1 0 0 −1 21. To compute the eigenvalues, we find the characteristic equation 1−λ 1 0 = (−2 − λ) [(1 − λ)(5 − λ) + 4] 5−λ 0 det(A − λI ) = det −4 17 −11 −2 − λ = (−2 − λ)(λ2 − 6λ + 9) = (−2 − λ)(λ − 3)2 , so the eigenvalues are λ1 = 3 and λ2 = −2. Eigenvalue λ1 = 3: To get eigenvectors, we consider −2 1 0 −2 1 0 1 −3 −5 1 −3 −5 2 0 ∼ 17 −11 −5 ∼ −2 1 0 ∼ 0 1 2 . nullspace(A−3I ) = nullspace −4 17 −11 −5 0 0 0 0 0 0 0 0 0 The latter matrix contains only one unpivoted column, so that only one linearly independent eigenvector can be obtained. However, λ1 = 3 occurs with multiplicity 2 as a root of the characteristic equation for the matrix. Therefore, the matrix is not diagonalizable. 22. We are given that the only eigenvalue of A is λ = 2. Eigenvalue λ = 2: To get eigenvectors, we consider −3 −1 3 1 0 −1 1 0 −1 2 −4 ∼ −3 −1 3 ∼ 0 1 0 . nullspace(A − 2I ) = nullspace 4 −1 0 1 4 2 −4 02 0 We see that only one unpivoted column will occur in a row-echelon form of A − 2I , and thus, only one linearly independent eigenvector can be obtained. Since the eigenvalue λ = 2 occurs with multiplicity 3 as a root of the characteristic equation for the matrix, the matrix is not diagonalizable. 23. We are given that the eigenvalues of A are λ1 = 4 and λ2 = −1. 439 Eigenvalue λ1 = 4: We consider 5 5 −5 0 . nullspace(A − 4I ) = nullspace 0 −5 10 5 −10 The middle row tells us that nullspace vectors (x, y, z ) must have y = 0. From this information, the first and last rows of the matrix tell us the same thing: x = z . Thus, an eigenvector corresponding to λ1 = 4 may be chosen as 1 0 . 1 Eigenvalue λ2 = −1: We consider 10 5 −5 10 5 −5 0 ∼ 0 0 0 . nullspace(A + I ) = nullspace 0 0 10 5 −5 00 0 From the first row, a vector (x, y, z ) in the nullspace must satisfy 10x + 5y − 5z = 0. Setting z = t and y = s, 1 we get x = 2 t − 1 s. Hence, the eigenvectors corresponding to λ2 = −1 take the form ( 1 t − 1 s, s, t), and so 2 2 2 a basis for this eigenspace is −1/2 1/2 0 , 1 . 1 0 Putting the above results together, 1 S= 0 1 we form 1/2 −1/2 0 1 1 0 and 4 0 0 0 . D = 0 −1 0 0 −1 28. We will compute the dimension of each of the two eigenspaces associated For λ1 = 1, we compute as follows: 4 8 16 1 0 8 ∼ 0 nullspace(A − I ) = nullspace 4 −4 −4 −12 0 with the matrix A. 2 1 0 4 1 , 0 which has only one unpivoted column. Thus, this eigenspace is 1-dimensional. For λ2 = −3, we compute as follows: 8 8 16 11 4 8 ∼ 0 0 nullspace(A + 3I ) = nullspace 4 −4 −4 −8 00 2 0 , 0 which has two unpivoted columns. Thus, this eigenspace is 2-dimensional. Between the two eigenspaces, we have a complete set of linearly independent eigenvectors. Hence, the matrix A in this case is diagonalizable. Therefore, A is diagonalizable, and we may take 1 0 0 0 . J = 0 −3 0 0 −3 440 (Of course, the eigenvalues of A may be listed in any order along the main diagonal of the Jordan canonical form, thus yielding other valid Jordan canonical forms for A.) 29. We will compute the dimension of each of the two eigenspaces associated with the matrix A. For λ1 = −1, we compute as follows: 3 nullspace(A + I ) = nullspace 2 −1 1 1 10 1 2 −2 ∼ 0 1 −2 , 0 −1 00 0 which has one unpivoted column. Thus, this eigenspace is 1-dimensional. For λ2 = 3, we compute as follows: −1 1 1 1 −1 −1 1 6 , nullspace(A − 3I ) = nullspace 2 −2 −2 ∼ 0 −1 0 −5 0 0 0 which has one unpivoted column. Thus, this eigenspace is also one-dimensional. Since we have only generated two linearly independent eigenvectors from the eigenvalues of A, we know that A is not diagonalizable, and hence, the Jordan canonical form of A is not a diagonal matrix. We must have one 1 × 1 Jordan block and one 2 × 2 Jordan block. To determine which eigenvalue corresponds to the 1 × 1 block and which corresponds to the 2 × 2 block, we must determine the multiplicity of the eigenvalues as roots of the characteristic equation of A. A short calculation shows that λ1 = −1 occurs with multiplicity 2, while λ2 = 3 occurs with multiplicity 1. Thus, the Jordan canonical form of A is −1 10 J = 0 −1 0 . 0 03 30. There are 3 different possible Jordan canonical forms, up to a rearrangement of the Jordan blocks: Case 1: −1 0 00 0 −1 0 0 . J = 0 0 −1 0 0 0 02 In this case, the matrix has four linearly independent eigenvectors, and because all Jordan blocks have size 1 × 1, the maximum length of a cycle of generalized eigenvectors for this matrix is 1. Case 2: −1 1 00 0 −1 0 0 . J = 0 0 −1 0 0 0 02 In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = −1 and one corresponding to λ = 2). There is a Jordan block of size 2 × 2, and so a cycle of generalized eigenvectors can have a maximum length of 2 in this case. 441 Case 3: −1 1 0 0 −1 1 J = 0 0 −1 0 0 0 0 0 . 0 2 In this case, the matrix has two linearly independent eigenvectors (one corresponding to λ = −1 and one corresponding to λ = 2). There is a Jordan block of size 3 × 3, and so a cycle of generalized eigenvectors can have a maximum length of 3 in this case. 31. There are 7 different possible Jordan canonical forms, up to a rearrangement of the Jordan blocks: Case 1: J = 4 0 0 0 0 0 4 0 0 0 0 0 4 0 0 0 0 0 4 0 0 0 0 0 4 . In this case, the matrix has five linearly independent eigenvectors, and because all Jordan blocks have size 1 × 1, the maximum length of a cycle of generalized eigenvectors for this matrix is 1. Case 2: J = 4 0 0 0 0 1 4 0 0 0 0 0 4 0 0 0 0 0 4 0 0 0 0 0 4 . In this case, the matrix has four linearly independent eigenvectors, and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 3: J = 4 0 0 0 0 1 4 0 0 0 0 0 4 0 0 0 0 1 4 0 0 0 0 0 4 . In this case, the matrix has three linearly independent eigenvectors, and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 4: J = 4 0 0 0 0 1 4 0 0 0 0 1 4 0 0 0 0 0 4 0 0 0 0 0 4 . In this case, the matrix has three linearly independent eigenvectors, and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. 442 Case 5: J = 4 0 0 0 0 1 4 0 0 0 0 1 4 0 0 0 0 0 4 0 0 0 0 1 4 . In this case, the matrix has two linearly independent eigenvectors, and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. Case 6: J = 4 0 0 0 0 1 4 0 0 0 0 1 4 0 0 0 0 1 4 0 0 0 0 0 4 . In this case, the matrix has two linearly independent eigenvectors, and because the largest Jordan block is of size 4 × 4, the maximum length of a cycle of generalized eigenvectors for this matrix is 4. Case 7: J = 4 0 0 0 0 1 4 0 0 0 0 1 4 0 0 0 0 1 4 0 0 0 0 1 4 . In this case, the matrix has only one linearly independent eigenvector, and because the largest Jordan block is of size 5 × 5, the maximum length of a cycle of generalized eigenvectors for this matrix is 5. 32. There are 10 different possible Jordan canonical forms, up to a rearrangement of the Jordan blocks: Case 1: J = 6 0 0 0 0 0 0 6 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 −3 0 0 0 −3 . In this case, the matrix has six linearly independent eigenvectors, and because all Jordan blocks have size 1 × 1, the maximum length of a cycle of generalized eigenvectors for this matrix is 1. Case 2: J = 6 0 0 0 0 0 1 6 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 −3 0 0 0 −3 . In this case, the matrix has five linearly independent eigenvectors (three corresponding to λ = 6 and two corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. 443 Case 3: J = 6 0 0 0 0 0 1 6 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 1 0 0 6 0 0 0 −3 0 0 0 −3 . In this case, the matrix has four linearly independent eigenvectors (two corresponding to λ = 6 and two corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 4: J = 6 0 0 0 0 0 1 6 0 0 0 0 0 1 6 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 −3 0 0 0 −3 . In this case, the matrix has four linearly independent eigenvectors (two corresponding to λ = 6 and two corresponding to λ = −3), and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. Case 5: J = 6 0 0 0 0 0 1 6 0 0 0 0 0 1 6 0 0 0 0 0 0 0 0 0 1 0 0 6 0 0 0 −3 0 0 0 −3 . In this case, the matrix has three linearly independent eigenvectors (one corresponding to λ = 6 and two corresponding to λ = −3), and because the largest Jordan block is of size 4 × 4, the maximum length of a cycle of generalized eigenvectors for this matrix is 4. Case 6: J = 6 0 0 0 0 0 0 6 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 −3 1 0 0 −3 . In this case, the matrix has five linearly independent eigenvectors (four corresponding to λ = 6 and one corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. 444 Case 7: J = 6 0 0 0 0 0 1 6 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 −3 1 0 0 −3 . In this case, the matrix has four linearly independent eigenvectors (three corresponding to λ = 6 and one corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 8: J = 6 0 0 0 0 0 1 6 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 1 0 0 6 0 0 0 −3 1 0 0 −3 . In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = 6 and one corresponding to λ = −3), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 9: J = 6 0 0 0 0 0 1 6 0 0 0 0 0 1 6 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 −3 1 0 0 −3 . In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = 6 and one corresponding to λ = −3), and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. Case 10: J = 6 0 0 0 0 0 1 6 0 0 0 0 0 1 6 0 0 0 0 0 0 0 0 0 1 0 0 6 0 0 0 −3 1 0 0 −3 . In this case, the matrix has two linearly independent eigenvectors (one corresponding to λ = 6 and one corresponding to λ = −3), and because the largest Jordan block is of size 4 × 4, the maximum length of a cycle of generalized eigenvectors for this matrix is 4. 33. There are 15 different possible Jordan canonical forms, up to a rearrangement of Jordan blocks: 445 Case 1: J = 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 0 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has seven linearly independent eigenvectors (four corresponding to λ = 2 and three corresponding to λ = −4), and because all Jordan blocks are size 1 × 1, the maximum length of a cycle of generalized eigenvectors for this matrix is 1. Case 2: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 0 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has six linearly independent eigenvectors (three corresponding to λ = 2 and three corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 3: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 −4 0 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has five linearly independent eigenvectors (two corresponding to λ = 2 and three corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 4: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 0 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has five linearly independent eigenvectors (two corresponding to λ = 2 and three corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. 446 Case 5: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 −4 0 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has four linearly independent eigenvectors (one corresponding to λ = 2 and three corresponding to λ = −4), and because the largest Jordan block is of size 4 × 4, the maximum length of a cycle of generalized eigenvectors for this matrix is 4. Case 6: J = 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has six linearly independent eigenvectors (four corresponding to λ = 2 and two corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 7: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has five linearly independent eigenvectors (three corresponding to λ = 2 and two corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. Case 8: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has four linearly independent eigenvectors (two corresponding to λ = 2 and two corresponding to λ = −4), and because the largest Jordan block is of size 2 × 2, the maximum length of a cycle of generalized eigenvectors for this matrix is 2. 447 Case 9: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has four linearly independent eigenvectors (two corresponding to λ = 2 and two corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. Case 10: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 0 0 0 0 −4 . In this case, the matrix has three linearly independent eigenvectors (one corresponding to λ = 2 and two corresponding to λ = −4), and because the largest Jordan block is of size 4 × 4, the maximum length of a cycle of generalized eigenvectors for this matrix is 4. Case 11: J = 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 1 0 0 0 −4 . In this case, the matrix has five linearly independent eigenvectors (four corresponding to λ = 2 and one corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. Case 12: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 1 0 0 0 −4 . In this case, the matrix has four linearly independent eigenvectors (three corresponding to λ = 2 and one corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. 448 Case 13: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 1 0 0 0 −4 . In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = 2 and one corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. Case 14: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 1 0 0 0 −4 . In this case, the matrix has three linearly independent eigenvectors (two corresponding to λ = 2 and one corresponding to λ = −4), and because the largest Jordan block is of size 3 × 3, the maximum length of a cycle of generalized eigenvectors for this matrix is 3. Case 15: J = 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 −4 1 0 0 0 −4 1 0 0 0 −4 . In this case, the matrix has two linearly independent eigenvectors (one corresponding to λ = 2 and one corresponding to λ = −4), and because the largest Jordan block is of size 4 × 4, the maximum length of a cycle of generalized eigenvectors for this matrix is 4. 34. FALSE. For instance, if A = the matrix A − B = 0 −1 1 0 1 0 1 1 and B = 1 1 0 1 , then we have eigenvalues λA = λB = 1, but is invertible, and hence, zero is not an eigenvalue of A − B . This can also be verified directly. 11 , then A2 = B 2 = I2 , but the matrices A and B are 01 not similar. (Otherwise, there would exist an invertible matrix S such that S −1 AS = B . But since A = I2 this reduces to I2 = B , which is clearly not the case. Thus, no such invertible matrix S exists.) 35. FALSE. For instance, if A = I2 and B = 36. To see that T1 + T2 is a linear transformation, we must verify that it respects addition and scalar multiplication: 449 T1 + T2 respects addition: Let v1 and v2 belong to V . Then we have (T1 + T2 )(v1 + v2 ) = T1 (v1 + v2 ) + T2 (v1 + v2 ) = [T1 (v1 ) + T1 (v2 )] + [T2 (v1 ) + T2 (v2 )] = [T1 (v1 ) + T2 (v1 )] + [T1 (v2 ) + T2 (v2 )] = (T1 + T2 )(v1 ) + (T1 + T2 )(v2 ), where we have used the linearity of T1 and T2 individually in the second step. T1 + T2 respects scalar multiplication: Let v belong to V and let k be a scalar. Then we have (T1 + T2 )(k v) = T1 (k v) + T2 (k v) = kT1 (v) + kT2 (v) = k [T1 (v) + T2 (v)] = k (T1 + T2 )(v), as required. There is no particular relationship between Ker(T1 ), Ker(T2 ), and Ker(T1 + T2 ). 37. FALSE. For instance, consider T1 : R → R defined by T1 (x) = x, and consider T2 : R → R defined by T2 (x) = −x. Both T1 and T2 are linear transformations, and both of them are onto. However, (T1 + T2 )(x) = T1 (x) + T2 (x) = x + (−x) = 0, so Rng(T1 + T2 ) = {0}, which implies that T1 + T2 is not onto. 38. FALSE. For instance, consider T1 : R → R defined by T1 (x) = x, and consider T2 : R → R defined by T2 (x) = −x. Both T1 and T2 are linear transformations, and both of them are one-to-one. However, (T1 + T2 )(x) = T1 (x) + T2 (x) = x + (−x) = 0, so Ker(T1 + T2 ) = R, which implies that T1 + T2 is not one-to-one. 39. Assume that c1 T (v1 + c2 T (v2 ) + · · · + cn T (vn ) = 0. We wish to show that c1 = c2 = · · · = cn = 0. To do this, use the linearity of T to rewrite the above equation as T (c1 v1 + c2 v2 + · · · + cn vn ) = 0. Now, since Ker(T ) = {0}, we conclude that c1 v1 + c2 v2 + · · · + cn vn = 0. Since {v1 , v2 , . . . , vn } is a linearly independent set, we conclude that c1 = c2 = · · · = cn = 0, as required. 40. Assume that V1 ∼ V2 and V2 ∼ V3 . Then there exist isomorphisms T1 : V1 → V2 and T2 : V2 → V3 . = = Since the composition of two linear transformations is a linear transformation (Theorem 5.4.2), we have a linear transformation T2 T1 : V1 → V3 . Moreover, since both T1 and T2 are one-to-one and onto, T2 T1 is also one-to-one and onto (see Problem 39 in Section 5.4). Thus, T2 T1 : V1 → V3 is an isomorphism. Hence, V1 ∼ V3 , as required. = 41. We have Ai v = λ1 v 450 for each i = 1, 2, . . . , k . Thus, (A1 A2 . . . Ak )v = (A1 A2 . . . Ak−1 )(Ak v) = (A1 A2 . . . Ak−1 )(λk v) = λk (A1 A2 . . . Ak−1 )v = λk (A1 A2 . . . Ak−2 )(Ak−1 v) = λk (A1 A2 . . . Ak−2 )(λk−1 v) = λk−1 λk (A1 A2 . . . Ak−2 )v . . . = λ2 λ3 . . . λk (A1 v) = λ2 λ3 . . . λk (λ1 v) = (λ1 λ2 . . . λk )v, which shows that v is an eigenvector of A1 A2 . . . Ak with corresponding eigenvalue λ1 λ2 . . . λk . 42. We first show that T is a linear transformation: T respects addition: Let A and B belong to Mn (R). Then T (A + B ) = S −1 (A + B )S = S −1 AS + S −1 BS = T (A) + T (B ), and so T respects addition. T respects scalar multiplication: Let A belong to Mn (R), and let k be a scalar. Then T (kA) = S −1 (kA)S = k (S −1 AS ) = kT (A), and so T respects scalar multiplication. Next, we verify that T is both one-to-one and onto (of course, in view of Proposition 5.4.13, it is only necessary to confirm one of these two properties, but we will nonetheless verify them both): T is one-to-one: Assume that T (A) = 0n . That is, S −1 AS = 0n . Left multiplying by S and right multiplying by S −1 on both sides of this equation yields A = S 0n S −1 = 0n . Hence, Ker(T ) = {0n }, and so T is one-toone. T is onto: Let B be an arbitrary matrix in Mn (R). Then T (SBS −1 ) = S −1 (SBS −1 )S = (S −1 S )B (S −1 S ) = In BIn = B, and hence, B belongs to Rng(T ). Since B was an arbitrary element of Mn (R), we conclude that Rng(T ) = Mn (R). That is, T is onto. Solutions to Section 6.1 True-False Review: 1. TRUE. This is essentially the statement of Theorem 6.1.3. 2. FALSE. As stated in Theorem 6.1.5, if there is any point x0 in I such that W [y1 , y2 , . . . , yn ](x0 ) = 0, then {y1 , y2 , . . . , yn } is linearly dependent on I . 3. FALSE. Many counterexamples are possible. Note that (xD − Dx)(x) = xD(x) − D(x2 ) = x − 2x = (−1)(x). Therefore, xD − Dx = −1. Setting L1 = x and L2 = D, we therefore see that L1 L2 = L2 L1 in this example. 4. TRUE. By assumption, L1 (y1 + y2 ) = L1 (y1 ) + L1 (y2 ) and L1 (cy ) = cL1 (y ) for all functions y, y1 , y2 . Likewise, L2 (y1 + y2 ) = L2 (y1 ) + L2 (y2 ) and L2 (cy ) = cL2 (y ) for all functions y, y1 , y2 . Therefore, (L1 +L2 )(y1 +y2 ) = L1 (y1 +y2 )+L2 (y1 +y2 ) = (L1 (y1 )+L1 (y2 ))+(L2 (y1 )+L2 (y2 )) = (L1 (y1 )+L2 (y1 ))+(L1 (y2 )+L2 (y2 )) = (L1 451 and ((L1 + L2 )(cy ) = L1 (cy ) + L2 (cy ) = cL1 (y ) + cL2 (y ) = c(L1 (y ) + L2 (y )) = c(L1 + L2 )(y ). Therefore, L1 + L2 is a linear differential operator. 5. TRUE. By assumption L(y1 + y2 ) = L(y1 ) + L(y2 ) and L(ky ) = kL(y ) for all scalars k . Therefore, for all constants c, we have (cL)(y1 + y2 ) = cL(y1 + y2 ) = c(L(y1 ) + L(y2 )) = cL(y1 ) + cL(y2 ) = (cL)(y1 ) + (cL)(y2 ) and (cL)(ky ) = c(L(ky )) = c(k (L(y ))) = k (cL)(y ). Therefore, cL is a linear differential operator. 6. TRUE. We have L(yp + u) = L(yp ) + L(u) = F + 0 = F. 7. TRUE. We have L(y1 + y2 ) = L(y1 ) + L(y2 ) = F1 + F2 . Solutions to Section 6.2 True-False Review: 1. FALSE. Even if the auxiliary polynomial fails to have n distinct roots, the differential equation still has n linearly independent solutions. For example, if L = D2 + 2D + 1, then the differential equation Ly = 0 has auxiliary polynomial with (repeated) roots r = −1, −1. Yet we do have two linearly independent solutions y1 (x) = e−x and y2 (x) = xe−x to the differential equation. 2. FALSE. Theorem 6.2.1 only applies to polynomial differential operators. However, in general, many counterexamples can be given. For example, note that (xD − Dx)(x) = xD(x) − D(x2 ) = x − 2x = (−1)(x). Therefore, xD − Dx = −1. Setting L1 = x and L2 = D, we therefore see that L1 L2 = L2 L1 in this example. 3. TRUE. This is really just the statement that a polynomial of degree n always has n roots, with multiplicities counted. 4. TRUE. Since 0 is a root of multiplicity four, each term of the polynomial differential operator must contain a factor of D4 , so that any polynomial of degree three or less becomes zero after taking four (or more) derivatives. Therefore, for a homogeneous differential equation of this type, a polynomial of degree three or less must be a solution. 5. FALSE. Note that r = 0 is a root of the auxiliary polynomial, but only of multiplicity 1. The expression c1 + c2 x in the solution reflects r = 0 as a root of multiplicity 2. 6. TRUE. The roots of the auxiliary polynomial are r = −3, −3, 5i, −5i. The portion of the solution corresponding to the repeated root r = −3 is c1 e−3x + c2 xe−3x , and the portion of the solution corresponding to the complex conjugate pair r = ±5i is c3 cos 5x + c4 sin 5x. 7. TRUE. The roots of the auxiliary polynomial are r = 2 ± i, 2 ± i. The terms corresponding to the first pair 2 ± i are c1 e2x cos x and c2 e2x sin x, and the repeated root gives two more terms: c3 xe2x cos x and c4 xe2x sin x. 452 8. FALSE. Many counterexamples can be given. For instance, if P (D) = D − 1, then the general solution is y (x) = cex . However, the differential equation (P (D))2 y = (D − 1)2 y has auxiliary equation with roots r = 1, 1 and general solution z (x) = c1 ex + c2 xex = xy (x). Solutions to Section 6.3 True-False Review: 1. FALSE. Under the given assumptions, we have A1 (D)F1 (x) = 0 and A2 (D)F2 (x) = 0. However, this means that (A1 (D)+A2 (D))(F1 (x)+F2 (x)) = A1 (D)F1 (x)+A1 (D)F2 (x)+A2 (D)F1 (x)+A2 (D)F2 (x) = A1 (D)F2 (x)+A2 (D)F1 (x), which is not necessarily zero. As a specific example, if A1 (D) = D − 1 and A2 (D) = D − 2, then A1 (D) annihilates F1 (x) = ex and A2 (D) annihilates F2 (x) = e2x . However, A1 (D) + A2 (D) = 2D − 3 does not annihilate ex + e2x . 2. FALSE. The annihilator of F (x) in this case is A(D) = Dk+1 , since it takes k + 1 derivatives in order to annihilate xk . 3. TRUE. We apply rule 1 in this section with k = 1, or we can compute directly that (D − a)2 annihilates xeax . 4. FALSE. Some functions cannot be annihilated by a polynomial differential operator. Only those of the forms listed in 1-4 can be annihilated. For example, F (x) = ln x does not have an annihilator. 5. FALSE. For instance, if F (x) = x, A1 (D) = A2 (D) = D, then although A1 (D)A2 (D)F (x) = 0, neither A1 (D) nor A2 (D) annihilates F (x) = x. 6. FALSE. The annihilator of F (x) = 3 − 5x is D2 , but since r = 0 already occurs twice as a root of the auxiliary equation, the appropriate trial solution here is yp (x) = A0 x2 + A0 x3 . 7. FALSE. The annihilator of F (x) = x4 is D5 , but since r = 0 already occurs three times as a root of the auxiliary equation, the appropriate trial solution here is yp (x) = A0 x3 + A1 x4 + A2 x5 + A3 x6 + A4 x7 . 8. TRUE. The annihilator of F (x) = cos x is D2 + 1, but since r = ±i already occurs once as a complex conjugate pair of roots of the auxiliary equation, the appropriate trial solution is not yp (x) = A0 cos x + B0 sin x; we must multiply by a factor of x to occur for the fact that r = ±i is a pair of roots of the auxiliary equation. Solutions to Section 6.4 True-False Review: 1. TRUE. Solutions to Section 6.5 True-False Review: 1. TRUE. This is reflected in the negative sign appearing in Hooke’s Law. The spring force Fs is given by Fs = −kL0 , where k > 0 and L0 is the displacement of the spring from its equilibrium position. The spring force acts in a direction opposite to that of the displacement of the mass. 2. FALSE. The circular frequency is the square root of k/m: ω0 = k . m 453 3. TRUE. The frequency of oscillation, denoted f in the text, and the period of oscillation, denoted T in the text, are inverses of one another: f T = 1. For instance, if a system undergoes f = 3 oscillations in one 1 second, then each oscillation takes one-third of a second, so T = 3 . 4. TRUE. This is mathematically seen from Equation (6.5.14) in which the exponential factor e−ct/2m decays to zero for large t. Therefore, y (t) becomes smaller and smaller as t increases. This is depicted in Figure 6.5.5. 5. FALSE. In the cases of critical damping and overdamping, the system cannot oscillate. In fact, the system passes through the equilibrium position at most once in these cases. 6. TRUE. The frequency of the driving term is denoted by ω , while the circular frequency of the spring-mass system is denoted by ω0 . Case 1(b) in this section describes resonance, the situation in which ω = ω0 . 7. TRUE. Air resistance tends to dampen the motion of the spring-mass system, since it acts in a direction opposite to that of the motion of the spring. This is reflected in Equation (6.5.4) by the negative sign appearing in the formula for the damping force Fd . 8. FALSE. From the formula T = 2π m k for the period of oscillation, we see that for a larger mass, the period is larger (i.e. longer). 9. TRUE. We see in the section on free oscillations of a mechanical system, we see that the resulting motion of the mass is given by (6.5.11), (6.5.12), or (6.5.13), and in all of these cases, the amplitude of this system is bounded. Only in the case of forced oscillations with resonance can the amplitude increase without bound. Solutions to Section 6.6 True-False Review: 1. TRUE. The differential equation governing the situation in which no driving electromotive force is present is (6.6.3), and its solutions are given in (6.6.4). In all cases, q (t) → 0 as t → ∞. Since the charge decays to zero, the rate of change of charge, which is the current in the circuit, eventually decreases to zero as well. 2. TRUE. For the given constants, it is true that R2 < 4L/C , since R2 = 16 and 4L/C = 272. 3. TRUE. The amplitude of the steady-state current is given by Equation (6.6.6), which is maximum when 1 ω 2 = ω . Substituting ω 2 = LC , it follows that the amplitude of the steady-state current will be a maximum when 1 . ω = ωmax = √ LC 4. TRUE. From the form of the amplitude of the steady-state current given in Equation (6.6.6), we see that the amplitude A is directly proportional to the amplitude E0 of the external driving force. 5. FALSE. The current i(t) in the circuit is the derivative of the charge q (t) on the capacitor, given in the solution to Example 6.6.1: q (t) = A0 e−Rt/2L cos(µt − φ) + E0 cos(ωt − η ). H The derivative of this is not inversely proportional to R. 6. FALSE. The charge on the capacitor decays over time if there is no driving force. 454 Solutions to Section 6.7 True-False Review: 1. TRUE. This is essentially the statement of the variation-of-parameters method (Theorem 6.7.1). 2. FALSE. The solutions y1 , y2 , . . . , yn must form a linearly independent set of solutions to the associated homogeneous differential equation (6.7.18). 3. FALSE. The requirement on the functions u1 , u2 , . . . , un , according to Theorem 6.7.6, is that they satisfy Equations (6.7.22). Because these equations involve the derivatives of u1 , u2 , . . . , un only, constants of integration can be arbitrarily chosen in solving (6.7.22), and therefore, addition of constants to the functions u1 , u2 , . . . , un again yields valid solutions to (6.7.22). Solutions to Section 6.8 True-False Review: 1. FALSE. First of all, the given equation only addresses the case of a second-order Cauchy-Euler equation. Secondly, the term containing y must contain a factor of x (see Equation 6.8.1): x2 y + a1 xy + a2 y = 0. 2. TRUE. The indicial equation (6.8.2) in this case is r(r − 1) − 2r − 18 = r2 − 3r − 18 = 0, with real and distinct roots r = 6 and r = −3. Hence, y1 (x) = x6 and y2 (x) = x−3 are two linearly independent solutions to this Cauchy-Euler equation. 3. FALSE. The indicial equation (6.8.2) in this case is r(r − 1) + 9r + 16 = r2 + 8r + 16 = (r + 4)2 = 0, and so we have only one real root, r = 4. Therefore, the only solutions to this Cauchy-Euler equation of the form y = xr take the form y = cx−4 . Therefore, only one linearly independent solution of the form y = xr to the Cauchy-Euler equation has been obtained. 4. TRUE. The indicial equation (6.8.2) in this case is r(r − 1) + 6r + 6 = r2 + 5r + 6 = (r + 2)(r + 3) = 0, with roots r = −2 and r = −3. Therefore, the general solution to this Cauchy-Euler equation is y (x) = c1 x−2 + c2 x−3 = c1 c2 + 3. x2 x Therefore, as x → +∞, y (x) for all values of the constants c1 and c2 . 5. TRUE. A solution obtained by the method in this section containing the function ln x implies a repeated root to the indicial equation. In fact, such a solution takes the form y2 (x) = xr1 lnx. Therefore, if y (x) = ln x/x is a solution, we conclude that r1 = r2 = −1 is the only root of the indicial equation r(r − 1)+ a1 r + a2 = − 0. From Case 2 in this section, we see that −1 = − a12 1 , and so a1 = 3. Moreover, the discriminant of Equation (6.8.3) must be zero: (a1 − 1)2 − 4a2 = 0. That is, 22 − 4a2 = 0. Therefore, a2 = 1. Therefore, the Cauchy-Euler equation in question, with a1 = 3 and a2 = 1, must be as given. 455 6. FALSE. The indicial equation (6.8.2) in this case is r(r − 1) − 5r − 7 = r2 − 6r − 7 = (r − 7)(r + 1) = 0, with roots r = 7 and r = −1. Therefore, the general solution to this Cauchy-Euler equation is y (x) = c1 x7 + c2 x−1 , and this is not an oscillatory function of x for any choice of constants c1 and c2 . Solutions to Section 7.1 True-False Review: 1. TRUE. If we make the substitution x → x1 and y → x2 , then in terms of Definition 7.1.1, we have two equations with a11 (t) = t2 ,a12 (t) = −t, a21 (t) = sin t, a22 (t) = 5, and b1 (t) = b2 (t) = 0. 2. TRUE. If we make the substitution x → x1 and y → x2 , then in terms of Definition 7.1.1, we have two equations with a11 (t) = t4 , a12 (t) = −et , a21 (t) = t2 + 3, a22 (t) = 0, and b1 (t) = 4 and b2 (t) = −t2 . 3. FALSE. The term txy on the right-hand side of the formula for x is non-linear because the unknown functions x and y are multiplied together, and this is prohibited in the form of a first-order linear system of differential equations. 4. FALSE. The term ey x on the right-hand side of the formula for y is non-linear because the known function y appears as an exponent. This is prohibited in the form of a first-order linear system of differential equations. 5. TRUE. If we consider Equation (7.1.16) with n = 3, then the text describes how the substitution x1 = x, x2 = x , x3 = x enables us to replace (7.1.16) with the equivalent first-order linear system x1 = x2 , x2 = x3 , x3 = −a3 (t)x1 − a2 (t)x2 − a1 (t)x1 + F (t). 6. FALSE. The technique for converting a first-order linear system of two differential equations to a secondorder linear differential equation only applies in the case where the coefficients aij (t) of the system (7.1.1) are constants. 7. FALSE. As indicated in Definition 7.1.6, an initial-value problem consists of auxiliary conditions all of which are applied at the same time t0 . In this system, the value of x is specified at t = 0 while the value of y is specified at t = 1. 8. TRUE. This has the required form in Definition 7.1.6. Note that both x and y have initial-values specified at t0 = 0. 9. FALSE. There is no initial-value specified for the function y (t). The value y (2) would be required in order for the to be an initial-value problem. 10. FALSE. As indicated in Definition 7.1.6, an initial-value problem consists of auxiliary conditions all of which are applied at the same time t0 . In this system, the value of x is specified at t = 3 while the value of y is specified at t = −3. Solutions to Section 7.2 True-False Review: 1. TRUE. If the unknown function x(t) is a column n-vector function, then A(t) must contain n columns in order to be able to compute A(t)x(t). And since x (t) is also a column n-vector function, then A(t)x(t) 456 must have n rows, and therefore, A(t) must also have n rows. Therefore, A(t) contains the same number of rows and columns. 2. FALSE. If two columns of a matrix are interchanged, the determinant changes sign: det([x1 (t), x2 (t)]) = −det([x2 (t), x1 (t)]). 3. FALSE. Many counterexamples are possible. For instance, if x1 = 1 1 , x2 = 1 t , x3 = 1 t2 , then if we set c1 x1 (t) + c2 x2 (t) + c2 x3 (t) = 0, then we arrive at the equations c1 + c2 + c3 = 0 and c1 + tc2 + t2 c3 = 0. Setting t = 0, we conclude that c1 = 0. Next, setting t = −1, we have −c2 + c3 = 0 = c2 + c3 , and this requires c2 = c3 = 0. Therefore c1 = c2 = c3 = 0. Hence, {x1 , x2 , x3 } is linearly independent. 4. FALSE. Theorem 7.2.4 asserts that the Wronskian of these vector functions must be nonzero for some t0 ∈ I , not for all values in I . 5. FALSE. To the contrary, an n × n linear system x (t) = Ax(t) always has n linearly independent solutions, regardless of any condition on the determinant of the matrix A. This is explained in the paragraph preceding Example 7.2.7 and is proved in Theorem 7.3.2 in the next section. 6. TRUE. This is actually described in the preceding section with Equation (7.1.16) and following. We can replace the fourth-order linear differential equation for x(t) by making the change of variables x1 = x, x2 = x , x3 = x , and x4 = x . In so doing, we obtain the equivalent first-order linear system x1 = x2 , x2 = x3 , x3 = x4 , x4 = −a4 (t)x1 − a3 (t)x2 − a2 (t)x3 − a1 (t)x4 + F (t). 7. FALSE. We have (x0 (t) + b(t)) = x0 (t) + b (t) = A(t)x0 (t) + b (t), and this in general is not the same as A(t)(x0 (t) + b(t)) + b(t). Solutions to Section 7.3 True-False Review: 1. FALSE. With b(t) = 0, note that x(t) = 0 is not a solution to the system x (t) = A(t)x(t) + b(t), and lacking the zero vector, the solution set to the differential equation cannot form a vector space. 2. TRUE. Since any fundamental matrix X (t) = [x1 , x2 , . . . , xn ] is comprised of linearly independent columns, the Invertible Matrix Theorem guarantees that the fundamental matrix is invertible under all circumstances. 3. TRUE. More than this, a fundamental solution set for x (t) = A(t)x(t) is in fact a basis for the space of solutions to the linear system. In particular, it spans the space of all solutions to x = Ax. 457 4. TRUE. This is essentially the content of Theorem 7.3.6. The general solution to the homogeneous vector differential equation x (t) = A(t)x(t) is xc (t) = c1 x1 + c2 x2 + · · · + cn xn , and therefore the general solution to x (t) = A(t)x(t) + b(t) is of the form x(t) = xc (t) + xp (t). Solutions to Section 7.4 True-False Review: 1. TRUE. If x(t) = eλt v, then x (t) = λeλt v = eλt (λv) = eλt (Av) = A(eλt v) = Ax(t). This calculation appears prior to Theorem 7.4.1. 2. TRUE. The two real-valued solutions are derived in the work preceding Example 7.4.6. They are x1 (t) = eat (cos btr − sin bts) and x2 (t) = eat (sin btr + cos bts), where r and s are the real and imaginary parts of an eigenvector v = r + is corresponding to λ = a + bi. 00 01 and B = , then 00 00 the matrices A and B have the same characteristic equation: λ2 = 0. However, whereas any constant vector c1 c2 function x(t) = is a solution to x = Ax, we have B x = and x = 0. Therefore, unless c2 = 0, c2 0 x is not a solution to the system x = B x. Hence, the systems x = Ax and x = B x have differing solution sets. 3. FALSE. Many counterexamples can be given. For instance, if A = 4. TRUE. Assume that the eigenvalue/eigenvector pairs are (λ1 , v1 ), (λ2 , v2 ), . . . , (λn , vn ). In this case, the general solution to both linear systems is the same: x(t) = c1 eλ1 t v1 + c2 eλ2 t v2 + · · · + cn eλn t vn . 5. TRUE. The two real-valued solutions forming a basis for the set of solutions to x = Ax in this case are x1 (t) = eat (cos bt r − sin bt s) and x2 (t) = eat (sin bt r + cos bt s). The general solution is a linear combination of these, and since a > 0, eat → ∞ as t → ∞. Since x1 and x2 cannot cancel out in a linear combination (since they are linearly independent), eat remains as a factor in any particular solution to the vector differential equation, and as this factor tends to ∞, x(t) → ∞ as t → ∞. 6. ?????????????????????????????????? Solutions to Section 7.5 True-False Review: 1. FALSE. Every n × n linear system of differential equations has n linearly independent solutions (Theorem 7.3.2), regardless of whether A is defective or nondefective. 2. TRUE. This fact is part of the information contained in Theorem 7.5.4. 3. FALSE. The number of linearly independent solutions to x = Ax corresponding to λ is equal to the algebraic multiplicity of the eigenvalue λ as a root of the characteristic equation for A, not the dimension of the eigenspace Eλ . 458 4. FALSE. We have Ax = A(eλt v1 ) = eλt (Av1 ) = eλt (v0 + λv1 ) = eλt v0 + λeλt v1 = eλt v0 + x (t). Therefore, Ax = x . Solutions to Section 7.6 True-False Review: 1. FALSE. The particular solution, given by Equation (7.6.4), depends on a fundamental matrix X (t) for the nonhomogeneous linear system, and the columns of any fundamental matrix consist of linearly independent solutions to the corresponding homogeneous vector differential equation x = Ax. 2. FALSE. The vector function u(t) is not arbitrary; it must satisfy the equation X (t)u (t) = b(t), according to Theorem 7.6.1. 3. TRUE. By definition, we have X = [x1 , x2 , . . . , xn ], so X = [x1 , x2 , . . . , xn ] = [Ax1 , Ax2 , . . . , Axn ] = AX. 4. FALSE. If we assume that xp is a solution to x = Ax + b, then note that (c · xp ) = c · xp = c · (Axp + b), whereas A(c · xp ) + b = c · (Axp ) + b. The right-hand sides of the above expressions are not equal. Therefore, c · xp is not a solution to x = Ax + b in general. 5. FALSE. A formula for the particular solution is given in Theorem 7.6.1: t xp (t) = X (t) X −1 (s)b(s)ds. Since an arbitrary integration constant can be applied to this expression, we have an arbitrary number of different choices for the particular solution xp (t). 6. TRUE. From X u = b, we use the invertibility of X to obtain u = X −1 b. Therefore, u is obtained by integrating X −1 b: t u(t) = X −1 (s)b(s)ds. Solutions to Section 7.7 True-False Review: 1. FALSE. In order to solve a coupled spring-mass system with two masses and two springs as a first-order linear system, a 4 × 4 linear system is required. More precisely, if the masses are m1 and m2 , and the spring constants are k1 and k2 , then the first-order linear system for the spring-mass system is x = Ax where 0 1 0 0 k2 1 − (k1 + k2 ) 0 0 m1 . A = m1 0 0 0 1 k2 k 0 − m22 0 m2 459 2. TRUE. Hooke’s Law states that the force due to a spring on a mass is proportional to the displacement of the mass from equilibrium, and oriented in the direction opposite to the displacement: F = −kx. The units of force are Newtons, and the units of displacement are meters. Therefore, in solving for the spring constant k , we have k = − F , with units of Newtons on top of the fraction and units of meters on the bottom x of the fraction. 3. FALSE. The amount of chemical in a tank of solution is measured in grams in the metric system, but the concentration is measured in grams/L, since concentration is computed as the amount of substance per unit of volume. 4. FALSE. The correct relation is that rin + r12 − r21 = 0, where r12 and r21 are the rates of flow from tank 1 to tank 2, and from tank 2 to tank 1, respectively. Example 7.7.2 shows quite clearly that r12 and r21 need not be the same. 5. FALSE. Although this is a closed system, chemicals still pass between the two tanks. So, for example, if tank 1 contained 100 grams of chemical and tank 2 contained no chemical, then as time passes and fluid flows between the two tanks, some of the chemical will move to tank 2. The amount of chemical in each tank changes over time. Solutions to Section 7.8 True-False Review: 1. TRUE. This is precisely the content of Equation (7.8.2). 2. TRUE. If we denote the fundamental matrix for the linear system x = Ax by X (t) = [x1 , x2 , . . . , xn ], then from Equations (7.8.4) and (7.8.7), we see that eAt = X (t)X −1 (0) = X0 (t). 3. TRUE. For each generalized eigenvector v, eAt v is a solution to x = Ax. If A is an n × n matrix, then A has n linearly independent generalized eigenvectors, and therefore, the general solution to x = Ax can be formulated as linear combinations of n linearly independent solutions of the form eAt v. As usual, application of the initial condition yields the unique solution to the given initial-value problem. 4. TRUE. Since the transition matrix for x = Ax is a fundamental matrix for the system, its columns consist of linearly independent vector functions, and hence, this matrix is invertible. 5. TRUE. This follows immediately from Equations (7.8.4) and (7.8.6). 6. TRUE. If c1 eAt v1 + c2 eAt v2 + · · · + cn eAt vn = 0, then eAt [c1 v1 + c2 v2 + · · · + cn vn ] = 0. Since eAt is invertible for all matrices A, we conclude that c1 v1 + c2 v2 + · · · + cn vn = 0. Since {v1 , v2 , . . . , vn } is linearly independent, c1 = c2 = · · · = cn = 0, and hence, {eAt v1 , eAt v2 , . . . , eAt vn } is linearly independent. 460 Solutions to Section 7.9 True-False Review: 1. FALSE. The trajectories need not approach the equilibrium point as t → ∞. For instance, Figures 7.9.4 and 7.9.8 show equilibrium points for which not all solution trajectories approach the origin as t → ∞. 2. TRUE. This is well-illustrated in Figure 7.9.7. All trajectories in this case are closed curves, and the equilibrium point is a stable center. 3. TRUE. This is case (b), described below Theorem 7.9.3. 4. ????????????? Solutions to Section 7.10 True-False Review: 1. TRUE. This is just the definition of the Jacobian matrix given in the text. 2. FALSE. Although the conclusion is valid in most cases, Table 7.10.1 shows that in the case of pure imaginary eigenvalues for the linear system, the behavior of the linear approximation can diverge from that of the given nonlinear system. 3. TRUE. From Equations (7.10.2) and (7.10.3), we find that the Jacobian of the system is J (x, y ) = a − by cy −bx cx − d , a 0 . Since a, d > 0, we have one positive and one negative eigenvalue, and therefore, 0 −d the equilibrium point (0, 0) is a saddle point. so that J (0, 0) = 4. TRUE. The linear system corresponding to the Van der Pol Equation is given in (7.10.5), and the only equilibrium point of this system occurs at (0, 0). 5. FALSE. The equilibrium point is only an unstable spiral if µ < 2. In this equation, µ = 3, in which case the equilibrium point is an unstable node, as discussed in the text. Solutions to Section 8.1 True-False Review: 1. FALSE. This function has a discontinuity at every integer n = 0, 1, and therefore, it has infinitely many discontinuities and cannot be piecewise continuous. 2. FALSE. This function has a discontinuity at every multiple of π , and therefore, it has infinitely many discontinuities and cannot be piecewise continuous. 3. TRUE. If f has discontinuities at a1 , a2 , . . . , ar in I and g has discontinuities at b1 , b2 , . . . , bs in I , then f + g has at most r + s discontinuities in I . Therefore, we can divide I into finitely many subintervals so that f + g is continuous on each subinterval and f approaches a finite limit as the endpoints of each subinterval are approached from within. Hence, from Definition 8.1.4, f is piecewise continuous on the interval I . 4. TRUE. By definition, we can divide [a, b] into finitely many subintervals so that (1) and (2) in Definition 8.1.4 are satisfied, and likewise, we can divide [b, c] into finitely many subintervals so that (1) and (2) in 461 Definition 8.1.4 are satisfied. Taking all of the subintervals so obtained, we have divided [a, c] into finitely many subintervals so that (1) and (2) in Definition 8.1.4 are satisfied. 5. FALSE. The lower limit of the integral defining the Laplace transform must be 0, not 1 (see Definition 8.1.1). 1 6. TRUE. The formula L[eat ] = s−a only applies for s > a, for otherwise the improper integral defining L[eat ] fails to converge. In this case a = 1, so we must restrict s so that s > 1. 7. FALSE. We have L[2 cos 3t] = 2L[cos 3t], and L[cos 3t] = s2s provided that s > 0, according to +9 Equation (8.1.4). So this Laplace transform is defined for s > 0, not just s > 3. 8. FALSE. For instance, if we take f (t) = et , then L[f ] = L[1/f ] = 1/L[f ] in this case. 9. FALSE. For instance, if we take f (t) = et , then L[f ] = 1 s−1 , 1 s−1 , but L[1/f ] = L[e−t ] = but L[f 2 ] = L[e2t ] = 1 s−2 1 s+1 . = Obviously, 1 s−1 2 . Solutions to Section 8.2 True-False Review: 1. TRUE. If f is of exponential order, then by Definition 8.2.1, we have |f (t)| ≤ M eαt (t > 0) for some constants M and α. But then |g (t)| < f (t) ≤ |f (t)| ≤ M eαt for all t > 0, and so g is also of exponential order. 2. TRUE. Suppose that |f (t)| ≤ M1 eα1 t and |g (t)| ≤ M2 eα2 t for t > 0, for some constants M1 , M2 , α1 , α2 . Then |(f + g )(t)| = |f (t) + g (t)| ≤ |f (t)| + |g (t)| ≤ M1 eα1 t + M2 eα2 t . If we set M = max{M1 , M2 } and α = max{α1 , α2 }, then the last expression on the right above is ≤ 2M eαt for all t > 0. Therefore, f + g is of exponential order. 3. FALSE. Looking at the Comparison Test for Improper Integrals, we should conclude from the assump∞ ∞ tions that if 0 H (t)dt converges, then so does 0 G(t)dt. As a specific counterexample, we could take ∞ ∞ G(t) = 0 and H (t) = t. Then although G(t) ≤ H (t) for all t ≥ 0 and 0 G(t)dt = 0 converges, 0 H (t)dt does not converge. 4. TRUE. Assume that L−1 [F ](t) = f (t) and L−1 [G](t) = g (t). Then we have L−1 [F + G](t) = (f + g )(t) = f (t) + g (t) = L−1 [F ](t) + L−1 [G](t). Therefore, L−1 [F + G] = L−1 [F ] + L−1 [G]. Moreover, we have L−1 [cF ](t) = (cf )(t) = cf (t) = cL−1 [F ](t), and therefore, L−1 [cF ] = cL−1 [F ]. Hence, L−1 is a linear transformation. 5. FALSE. From the formulas preceding Example 8.2.6, we see that the inverse Laplace transform of is f (t) = cos 3t. s s2 +9 6. FALSE. From the formulas preceding Example 8.2.6, we see that the inverse Laplace transform of is f (t) = e−3t . 1 s+3 Solutions to Section 8.3 True-False Review: 462 1. FALSE. The period of f is required to be the smallest positive real number T such that f (t + T ) = f (t) for all t ≥ 0. There can be only one smallest positive real number T , so the period, if it exists, is uniquely determined. 2. TRUE. The functions f and g are simply horizontally shifted from one another. Therefore, if f has period T , then f (t + T ) = f (t), and so then g (t + T ) = f (t + T + c) = f (t + c) = g (t). Thus, g also has period T . 3. FALSE. The function f (t) = cos(2t) has period π , not π/2. 4. FALSE. For all T > 0, we have f (t + T ) = sin((t + T )2 ) = sin(t2 ), so f is not periodic. 5. FALSE. Even a continuous function need not be periodic. For instance, the function f (t) = sin(t2 ) in the previous item is continuous, hence piecewise continuous, but not periodic. 6. FALSE. If f (t) = cos t and g (t) = sin t, then f and g are periodic with period m = n = 2π . However, (f + g )(t) = cos t + sin t is periodic with period 2π , not period 4π . 7. FALSE. If f (t) = cos t and g (t) = sin t, then f and g are periodic with period m = n = 2π . However, (f g )(t) = cos t sin t is periodic with period 2π , not period (2π )2 . 8. FALSE. Many examples are possible. The function f given in Example 8.3.4 is one. More simply, we simply note that L[sin t] = s21 , and F (s) = s21 is not a periodic function. +1 +1 Solutions to Section 8.4 True-False Review: 1. FALSE. The Laplace transform of f may not exist unless we assume additionally that f is of exponential order. 2. TRUE. This is illustrated in the examples throughout this section. The procedure is outlined above Figure 8.4.1 in the text and indicates that the initial condition is imposed at the outset of the solution. 3. TRUE. If we leave the initial conditions as arbitrary constants, the solution technique presented in this section will result in the general solution to the differential equation. 4. FALSE. The expression for Y (s) is affected by the initial conditions, because the initial conditions arise in the formulas for the Laplace transforms of the derivatives of the unknown function y . Solutions to Section 8.5 True-False Review: 1. TRUE. This follows at once from the first formula in the First Shifting Theorem, with a replaced by −a. 2. FALSE. For instance, if f (t) = et , then f (t − 1) = et−1 = et − 1. 3. TRUE. The formula for f (t) is obtained from the formula for f (t + 2) by replacing t + 2 with t (and hence t + 3 with t + 1 and t with t − 2). 4. FALSE. If we take f (x) = x and g (x) = x − 3, for example, then 1 f (t)dt = 0 1 2 1 and 0 5 g (t)dt = − , 2 463 and so 1 1 g (t)dt − 3. f (t)dt = 0 0 5. FALSE. The correct formula is L[e−t sin 2t] = 2 . (s + 1)2 + 4 6. TRUE. The Laplace transform of f (t) = t3 is F (s) = 6 a = 2 gives L[e2t t3 ] = (s−2)4 . 3! s4 = 6 s4 , and so the First Shifting Theorem with 7. TRUE. The inverse Laplace transform of F (s) = s2s is f (t) = cos 3t, and so the First Shifting Theorem +9 with a = −4 gives the indicated inverse Laplace transform. 8. FALSE. The correct formula is L−1 3 1 = e−t sin 6t. (s + 1)2 + 36 2 Solutions to Section 8.6 True-False Review: 1. FALSE. The given function is not well-defined at t = a. The unit step function is 0 for 0 ≤ t < a, not 0 ≤ t ≤ a. 2. FALSE. As Figure 8.6.2 shows, the value of ua (t) − ub (t) at t = b is 0, not 1. 3. FALSE. For values of t with a < t < b, we have ua (t) = 1 and ub (t) = 0, so the given inequality does not hold for such t. 4. FALSE. The given function is precisely the one sketched in Figure 8.6.2, which is ua (t) − ub (t), not ub (t) − ua (t). Solutions to Section 8.7 True-False Review: 1. FALSE. According to Equation (8.7.2), the inverse Laplace transform of eas F (s) is u−a (t)f (t + a). Note that this requires a < 0. 2. TRUE. This is Corollary 8.7.2. 3. TRUE. This is an immediate consequence of the Second Shifting Theorem. 4. FALSE. According to the Second Shifting Theorem, we have L[u2 (t) cos 4(t − 2)] = not L[u2 (t) cos 4t] = se−2s , s2 + 16 se−2s . s2 + 16 464 5. FALSE. We have L[u3 (t)et ] = L[u3 (t)e(t+3)−3 ] = e−3s L[et+3 ] = e−3s e3 1 e3 = 3s . s−1 e (s − 1) 6. ????? 7. FALSE. The correct formula is L−1 1 = u2 (t). se2s Solutions to Section 8.8 True-False Review: 1. TRUE. We have ∞ I= F (t)dt, −∞ where F (t) is the magnitude of the applied force at time t. 2. TRUE. A unit impulse force acts instantaneously on an object and delivers a unit impulse. 3. FALSE. The correct formula is L[δ (t − a)] = e−as . 4. TRUE. An excellent illustration of this is given in Example 8.8.3. A spring-mass system that receives an instantaneous blow experiences an acceleration, which is the second derivative of the position. 5. FALSE. The initial conditions are unrelated to the instantaneous blow. They are the pre-blow position and velocity of the mass, and these are not related in any way to the instantaneous blow. Solutions to Section 8.9 True-False Review: t 1. TRUE. We have (f ∗ g )(t) = 0 f (t − τ )g (τ )dτ and (g ∗ f )(t) = becomes the same as the first one via the u-substitution u = t − τ . t 0 g (t − τ )f (τ )dτ . The latter integral 2. TRUE. Since both f and g are positive functions, the expression f (t − τ )g (τ ) is positive for all t, τ . Thus, if we increase the interval of integration [0, t] by increasing t, the value of (f ∗ g )(t) also increases. 3. FALSE. The Convolution Theorem states that L[f ∗ g ] = L[f ]L[g ]. 4. TRUE. This is exactly the form of the equation in (8.9.3). 5. FALSE. For instance, if f is identically zero, then f ∗ g = f ∗ h = 0 for all functions g and h. 6. TRUE. This is expressed mathematically in Equation (8.9.2). 7. TRUE. This follows from the fact that constants can be factored out of integrals, and a(f ∗ g ), (af ) ∗ g , and f ∗ (ag ) can be expressed as integrals. In short, integration is linear. Solutions to Section 9.1 True-False Review: 465 (x) 1. TRUE. Theorem 9.1.7 addresses this for a rational function f (x) = p(x) . The radius of convergence of q the power series representation of f around the point x0 is the distance from x0 to the nearest root of q (x). 2. TRUE. We have ∞ ∞ (an + bn )xn = n=0 ∞ an xn + n=0 bn xn , n=0 and if each of the latter sums converges at x1 , say to f and g respectively, then ∞ (an + bn )xn = f + g. n=0 3. FALSE. It could be the case, for example, that bn = −an for all n = 0, 1, 2, . . . . Then ∞ (an + bn )xn = 0, n=0 but the individual summations need not converge. 4. FALSE. This is not necessarily the case, as Example 9.1.3 illustrates. In that example, the radius of convergence is R = 3, but the series diverges at the endpoints x = ±3. The endpoints must be considered separately in general. 5. TRUE. This is part of the statement in Theorem 9.1.6. 6. FALSE. The Taylor series expansion of an infinitely differentiable function f need not converge for all real values of x in the domain. It is only guaranteed to converge for x in the domain that lie within the interval of convergence. 7. TRUE. A polynomial is a rational function p(x)/q (x) where q (x) = 1. Since q (x) = 1 has no roots, Theorem 9.1.7 guarantees that the power series representation of p(x)/q (x) about any point x0 has an infinite radius of convergence. 8. ???????????? 9. FALSE. According to the algebra of power series, the coefficient cn of xn is n cn = ak bn−k . k=0 10. TRUE. This is the definition of a Maclaurin series. Solutions to Section 9.2 True-False Review: 1. TRUE. Since a polynomial is analytic at every x0 in R, it follows from Definition 9.2.1 that every point of R is an ordinary point. 2. FALSE. According to Theorem 9.2.4 with x0 = −3 and R = 2, the radius of convergence of the power series representation of the solution to this differential equation is at least 2. 466 3. FALSE. The nearest singularity to x = 2 occurs in p with root x = 1. Hence, R = 1 is the radius of convergence of p(x) = x21 1 about x = 2. Therefore, by Theorem 9.2.4, the radius of convergence of a power − series solution to this differential equation is at least 1 (not at least 2). 4. TRUE. This is seen directly by computing y (x0 ) and y (x0 ) for the series solution ∞ an (x − x0 )n , y (x) = n=0 as explained in the statement of Theorem 9.2.4. 5. TRUE. The radii of convergence of the power series expansions of p and q are both positive, and thus, Theorem 9.2.4 implies that the general solution to the differential equation can be represented as a power series with a positive radius of convergence. 6. FALSE. The two linearly independent solutions to y + (2 − 4x2 )y − 8xy = 0 found in Example 9.2.7 both contain common powers of x. 7. FALSE. For example, the power series y1 (x) found in Example 9.2.6 is a solution to (9.2.10), but xy1 (x) is not a solution. 8. FALSE. The value of a0 only determines the values a0 , a2 , a4 , . . . , but there is no information about the terms a1 , a3 , a5 , . . . unless a1 is also specified. 9. TRUE. The value of ak depends on ak−1 and ak−3 , so once a0 , a1 , and a2 are specified, we can find a3 , a4 , a5 , . . . uniquely from the recurrence relation. 10. FALSE. If the recurrence relation cannot be solved, it simply means that lengthy brute force calculations may be required to determine the solution, not that the solution does not exist. Solutions to Section 9.3 True-False Review: 1. FALSE. Theorem 9.2.4 guarantees that R = 1 is a lower bound on the radius of convergence of the power series solutions to Legendre’s equation about x = 0. The radius of convergence could be greater than 1. 2. TRUE. Relative to the inner product 1 p, q = p(x)q (x)dx, −1 {P0 , P1 , . . . , PN } is an orthogonal set (Theorem 9.3.4), and since each Pi is of different degree, ranging from 0 to N , {P0 , P1 , . . . , PN } is linearly independent. 3. TRUE. If α is a positive even integer or a negative odd integer or zero, then eventually the coefficients of (9.3.3) become zero, whereas if α is a negative even integer or a positive odd integer, then eventually the coefficients of (9.3.4) become zero. Therefore, for any integer α, either (9.3.3) or (9.3.4) contains only finitely many nonzero terms. 4. TRUE. The Legendre polynomials correspond to polynomial solutions to (9.3.3) and (9.3.4), and we see directly from their form that these solutions contain terms with all odd powers of x or with all even powers of x. 467 Solutions to Section 9.4 True-False Review: 1. FALSE. It is required that (x − x0 )2 Q(x) be analytic at x = x0 , not that (x − x0 )Q(x) be analytic at x = x0 . 2. FALSE. It is necessary to assume that r1 and r2 are distinct and do not differ by an integer in order to draw this conclusion. 3. TRUE. This is well-illustrated by Examples 9.4.5 and 9.4.6. The values of a0 , a1 , a2 , . . . in the Frobenius series solution are obtained by directly substituting it into both sides of the differential equation and matching up the coefficients of x resulting on each side of the differential equation. 4. TRUE. Example 9.4.2 is a fine illustration of this. 1 5. TRUE. For instance, the only singular point of y + x2 y + y = 0 is x0 = 0, and it is an irregular singular point. Solutions to Section 9.5 True-False Review: 1. FALSE. The indicial equation (9.5.4) in this case reads r(r − 1) − r + 1 = 0, or r2 − 2r + 1 = 0, and the roots of this equation are r = 1, 1, so the roots are not distinct. √ √ 2. TRUE. The indicial equation (9.5.4) in this case reads r(r − 1)+(1 − 2 5)r + 19 = 0, or r2 − 2 5r + 19 = 0. 4 4 √ The quadratic formula quickly yields the roots r = 5 ± 1 , which are distinct and differ by 1. 2 3. FALSE. The indicial equation (9.5.4) in this case reads r(r − 1) + 9r + 25 = 0, or r2 + 8r + 25 = 0. The quadratic formula quickly yields the roots r = −4 ± 3i, which do not differ by an integer. 7 7 4. TRUE. The indicial equation (9.5.4) in this case reads r(r − 1) + 4r − 4 = 0, or r2 + 3r − 4 = 0. The 7 1 quadratic formula quickly yields the roots r = − 2 and r = 2 , which are distinct and differ by 4. 5. TRUE. The indicial equation (9.5.4) in this case reads r(r − 1) = 0, with roots r = 0 and r = 1. They differ by 1. 6. FALSE. As indicated in Theorem 9.5.1, if the roots of the indicial equation do not differ by an integer, then two linearly independent solutions to x2 y + xp(x)y + q (x)y = 0 can be found as in Equations (9.5.13) and (9.5.14). Solutions to Section 9.6 True-False Review: 1. FALSE. The requirement that guarantees the existence of two linearly independent Frobenius series solutions is that 2p is not an integer, not that p is a positive noninteger. 2. TRUE. Since the roots of the indicial equation are r = ±p, it follows from the general Frobenius theory that, provided 2p is not equal to an integer, we can get two linearly independent Frobenius series solutions to Bessel’s equation of order p, written as Jp and J−p in (9.6.18). The variation appearing in (9.6.20) also applies when p is not a positive integer, which holds in particular if 2p is not an integer. 3. FALSE. The gamma function is not defined for p = 0, −1, −2, . . . . 468 4. TRUE. From Figure 9.6.2, the values of p such that Γ(p) < 0 occur in the intervals (−1, 0), (−3, −2), (−5, −4), and so on. In these intervals, the greatest integer less than or equal to p are −1, −3, −5, . . . , which are the odd, negative integers. 5. FALSE. Although it is possible via Property 3 in Equation (9.6.27) to express Jp (x) in terms of Jp−1 (x) and Jp−2 (x) as Jp (x) = 2x−1 (p − 1)Jp−1 (x) − Jp−2 (x), the expression on the right-hand side is not a linear combination. 6. FALSE. The correct formula for Jp (x) can be computed directly from Equation (9.6.22). A valid expression for Jp (x) is also given in Property 4 (Equation (9.6.28)). 7. FALSE. From Equation (9.6.31), we see that Jp (λn x) and Jp (λm x) are orthogonal on (0, 1) relative to the weight function w(x) = x: 1 xJp (λm x)Jp (λn x)dx = 0. 0 ...
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This note was uploaded on 02/17/2012 for the course MA 262 taught by Professor Ber during the Spring '08 term at Purdue University-West Lafayette.

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