Linear Algebra by otto brestscher chapter 05 solutions

Linear Algebra with Applications (3rd Edition)

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SSM: Linear Algebra Section 5.1 Chapter 5 5.1 1. v = 7 2 + 11 2 = 49 + 121 = 170 13 . 04 3. v = 2 2 + 3 2 + 4 2 + 5 2 = 4 + 9 + 16 + 25 = 54 7 . 35 5. θ = arccos u · v u v = arccos 2+6+12 14 29 0 . 122 (radians) 7. Use the fact that u · v = u v cos θ , so that the angle is acute if u · v > 0, and obtuse if u · v < 0. Since u · v = 10 12 = 2, the angle is obtuse. 9. Since u · v = 3 4 + 5 3 = 1, the angle is acute (see Exercise 7). 11. a. θ n = arccos u · v u v = arccos 1 n θ 2 = arccos 1 2 = π 4 (= 45 ) θ 3 = arccos 1 3 0 . 955 (radians) a 4 = arccos 1 2 = π 3 (= 60 ) b. Since y = arccos( x ) is a continuous function, lim n →∞ θ n = arccos lim n →∞ 1 n = arccos(0) = π 2 (= 90 ) 13. Figure 5.1 shows that F 2 + F 3 = 2 cos ( θ 2 ) F 2 = 20 cos ( θ 2 ) . It is required that F 2 + F 3 = 16, so that 20 cos ( θ 2 ) = 16, or θ = 2 arccos(0 . 8) 74 . Figure 5.1: for Problem 5.1.13 . 15. The subspace consists of all vectors x in R 4 such that 121
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Chapter 5 SSM: Linear Algebra x · v = x 1 x 2 x 3 x 4 · 1 2 3 4 = x 1 + 2 x 2 + 3 x 3 + 4 x 4 = 0. These are vectors of the form 2 r 3 s 4 t r s t = r 2 1 0 0 + s 3 0 1 0 + t 4 0 0 1 . The three vectors to the right form a basis. 17. The orthogonal complement W of W consists of the vectors x in R 4 such that x 1 x 2 x 3 x 4 · 1 2 3 4 = 0 and x 1 x 2 x 3 x 4 · 5 6 7 8 = 0. Finding these vectors amounts to solving the system x 1 + 2 x 2 + 3 x 3 + 4 x 4 = 0 5 x 1 + 6 x 2 + 7 x 3 + 8 x 4 = 0 . The solutions are of the form x 1 x 2 x 3 x 4 = s + 2 t 2 s 3 t s t = s 1 2 1 0 + t 2 3 0 1 . The two vectors to the right form a basis of W . 19. See Figure 5.2. 21. Call the three given vectors v 1 , v 2 , and v 3 . Since v 2 is required to be a unit vector, we must have b = g = 0. Now v 1 · v 2 = d must be zero, so that d = 0. Likewise, v 2 · v 3 = e must be zero, so that e = 0. Since v 3 must be a unit vector, we have v 3 2 = c 2 + 1 4 = 1, so that c = ± 3 2 . Since we are asked to find just one solution, let us pick c = 3 2 . The condition v 1 · v 3 = 0 now implies that 3 2 a + 1 2 f = 0, or f = 3 a . Finally, it is required that v 1 2 = a 2 + f 2 = a 2 + 3 a 2 = 4 a 2 = 1, so that a = ± 1 2 . Let us pick a = 1 2 , so that f = 3 2 . 122
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SSM: Linear Algebra Section 5.1 Figure 5.2: for Problem 5.1.19 . Summary: v 1 = 1 2 0 3 2 , v 2 = 0 1 0 , v 3 = 3 2 0 1 2 There are other solutions; some components will have different signs. 23. We will follow the hint. Let v be a vector in V . Then v · x = 0 for all x in V . Since ( V ) contains all vectors y such that y · x = 0, v is in ( V ) . So V is a subspace of ( V ) .
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