# Student Solutions Manual for Linear Algebra with Applications

## Otto Bretscher, Kyle Burke

Language: English

Pages: 235

ISBN: B01JXOF5BW

Format: PDF / Kindle (mobi) / ePub

Fully bookmarked for all chapters and sections.

This manual contains completely worked-out solutions for all the odd-numbered exercises in the text.

for the output vector x : x = (In − A)−1 b (compare with Exercise 2.3.49). c. The output x required to satisfy a consumer demand b is x = (In − A)−1 b = (In + A + A2 + · · · + Am + · · ·) b = b + Ab + A2 b + · · · + Am b + · · ·. To interpret the terms in this series, keep in mind that whatever output v the industries produce generates an interindustry demand of Av. 60 SSM: Linear Algebra True or False The industries first need to satisfy the consumer demand, b. Producing the output b will

3.1.3, the image of A is the span of the columns of A: im(A) = span 1 1 1 1 , , , 1 2 3 4 . 64 SSM: Linear Algebra Section 3.1 Since any two of these vectors span all of R2 already, we can write im(A) = span 1 1 , 2 1 . 17. By Fact 3.1.3, im(A) = span 2 1 , 4 3 = R2 (the whole plane). 19. Since the four column vectors of A are parallel, we have im(A) = span 1 , a line in −2 R2 . 4 7 3 21. By Fact 3.1.3, im(A) = span 1 , 9 , 2 . 5 6 8 It is hard to tell by

= 0 x3 x1 −t x2 t = , where t x3 −t x4 t .. 1 2 0 2 3 . 0 . 0 0 1 3 2.. 0 7. . 0 0 1 4 −1.. 0 .. 0 0 0 0 1. 0 .. 9. 0 1 2 0 0 . 0 0 1 0 11.. 0 0 0 0 1 −3... 0 .. 0 0 0 0 1. 0 x1 + 2x2 x3 x4 x5 x1 =0 =0 x −→ 3 =0 x4 =0 x5 . 1.. . 0 −1.. . 1 1.. . 0 0.. 0 . 1 0 0.. . 0 1 1 0.. 0 → . 0 0 1 1.. 0 . 0 −1 0 1.. 0 −I 0 0 0 0 0 1 = −x4 = x4 = −x4 is an arbitrary real number. 1 0

3, y = 2. (See Figure 1.9.) Figure 1.9: for Problem 1.3.5 . 7. A unique solution, since there is only one parallelogram with sides along v1 and v2 and one vertex at the tip of v3 . 1 2 3 x 1 9. 4 5 6 y = 4 7 8 9 z 9 11. Undefined since the two vectors do not have the same number of components. 13. 1 2 3 4 29 2 1 7 = + 11 =7 65 4 3 11 or 1 2 3 4 29 1 · 7 + 2 · 11 7 = = 65 3 · 7 + 4 · 11 11 5 6 15. [1 2 3 4] = 5 · 1 + 6 · 2 + 7 · 3 + 4 · 8 = 70 either way. 7

a b c 5 5a + 3b − 9c 2 and 0 d e 3 = 3d − 9e = 0 . 0 0 f −9 −9f 1 21 Chapter 1 SSM: Linear Algebra Clearly, f must equal − 91 . Then, since 3d = 9e, we can choose any non-zero value for the free variable e, and d will be 3e. So, if we choose 1 for e, then d = 3e = 3. Lastly, we . If must resolve 5a + 3b − 9c = 2. Here, b and c are the free variables, and a = 2−3b+9c 5 2−3(1)+9(1) 8 we let b = c = 1. Then, a = = 5. 5 8 1 1 5 So, in our example, A = 0 3 1 0 0 −