math294_hw1fa06 - Math 294 HW1 Solutions 1.1 Introduction...

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Unformatted text preview: Math 294 - HW1 Solutions 1.1 Introduction to Linear Systems 26. The system reduces to x - 3z y + 2z (k 2 - 4)z = = = 1 1 . k-2 This system has a unique solution if k 2 - 4 = 0, that is, if k = 2. If k = 2, then the last equation is 0 = 0, and there will be infinitely many solutions. If k = -2, then the last equation is 0 = -4, and there will be no solutions. 28. The thermal equilibrium condition requires that T1 = T3 = T2 +400+0+0 . 4 We can rewrite this system as -4T1 + T2 T1 - 4T2 + T3 T2 - 4T3 = = = -200 -200 . -400 T2 +200+0+0 , 4 T2 = T1 +T3 +200+0 , 4 and The solution is (T1 , T2 , T3 ) = (75, 100, 125). 30. Proceeding as in the previous exercise, we obtain the system a+b+c = a + 2b + 4c = a + 3b + 9c = p q r The unique solution is a b c = = = 3p - 3q + r -2.5p + 4q - 1.5r 0.5p - q + 0.5r Only one polynomial of degree 2 goes through the three points, namely, f (t) = 3p - 3q + r + (-2.5p + 4q - 1.5r)t + (0.5p - q + 0.5r)t2 . 1.2 Matrices, Vectors, and Gauss-Jordan Elimination 2. 3 6 4 8 -1 8 -2 3 1 3 R1 1 6 4 3 -1 3 8 -2 8 3 -6R1 +R2 3 1 0 4 3 0 1 8 -3 3 0 -13 This system has no solutions, since the last row represents the equation 0 = -13 x1 10. The system reduces to x2 x3 Let x4 = t. x1 1-t x2 2 + 3t x3 = -3 - 2t x4 t + x4 - 3x4 + 2x4 = = = 1 2 -3 - x1 x2 x3 = = = 1 - x4 2 + 3x4 -3 - 2x4 , where t is an arbitrary real number. x1 + x2 + x3 + x4 x1 + 2x2 + 3x3 + 4x4 x1 + 9x2 + 9x3 + 7x4 0.25x4 1.5x4 2.25x4 = = = 0 0 0 35. We need to solve the system = 0 = 0 , = 0 x1 which reduces to x2 x3 + - + x1 -0.25t x 1.5t The solutions are of the form 2 = x3 -2.25t , where t is an arbitrary real number. x4 t 42. Let x1 , x2 , x3 , and x4 be the traffic volume at the four locations indicated in the figure below. We are told that the number of cars coming into each intersection is the same as the number of cars coming out: x1 + 300 x2 + 300 x3 + x4 + 100 150 + 120 = 320 + x2 = 400 + x3 = 250 = x1 + x4 x1 or x1 - x2 x2 = = = = 20 100 150 270 - x3 x3 + + x4 x4 x1 270 - t x 250 - t The solutions are of the form 2 = x3 150 - t x4 t Since the xi must be positive integers (or zero), t must be an integer with 0 t 150. The lowest possible values are x1 = 120, x2 = 100, x3 = 0, and x4 = 150, while the highest possible values are x1 = 270, x2 = 250, x3 = 150, and x4 = 0 1.3 On the Solutions of Linear Systems; Matrix Algebra 1 0 -1 2 . 4. This matrix has rank 2 since its rref is 0 1 0 0 0 1 0 24. By fact 1.3.4, rref(A) = 0 0 0 1 0 0 0 0 1 0 0 0 . 0 1 0 1 0 0 0 0 . 1 0 1 0 26. From Exercise 3d, we know that rank(A) = 3, so that rref(A) = 0 0 Since all variables are leading, the system Ax = c cannot have infinitely many solutions, but it could have a unique solution (for example, if c = b) or no solutions at all (compare with Example 3c). 48. The fact that x1 is a solution of Ax = b means that Ax1 = b. a. A (x1 + xh ) = Ax1 + Axh = b + 0 = b b. A (x2 - x1 ) = Ax2 - Ax1 = b - b = 0 c. Parts (a) and (b) show that the solutions of Ax = b are exactly the vectors of the form x1 + xh , where xh is a solution of Ax = 0; indeed if x2 is a solution of Ax = b, we can write x2 = x1 + (x2 - x1 ), and x2 - x1 will be a solution of Ax = 0, by part (b). Geometrically, the vectors of the form x1 + xh are those whose tips are on the line L in the figure below; the line L runs through the tip of x1 and is parallel to the given line consisting of the solutions of Ax = 0. a 0 1 2 b 0 0 0 60. We need c = c1 3 + c2 4 + c3 5 = d 0 0 6 that a, c, and d can be any value, while b must equal zero. c2 + 2c3 0 3c1 + 4c2 + 5c3 . From this we see 6c3 ...
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