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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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