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. k ~v k = 7 2 +11 2 = 49 + 121 = 170 13 . 04 3. k ~v k = 2 2 +3 2 +4 2 +5 2 = 4+9+16+25= 54 7 . 35 5. θ = arccos ~u · ~v k ~u k ~v k = arccos 2+6+12 14 29 0 . 122 (radians) 7. Use the fact that ~u · ~v = k ~u k ~v k 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 k ~u k ~v k = 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 ³ lim n →∞ 1 n ´ = arccos(0) = π 2 (= 90 ) 13. Figure 5.1 shows that k ~ F 2 + ~ F 3 k = 2 cos ( θ 2 ) k ~ F 2 k =20cos ( θ 2 ) . It is required that k ~ F 2 + ~ F 3 k = 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 =0and x 1 x 2 x 3 x 4 · 5 6 7 8 =0. Finding these vectors amounts to solving the system ¯ ¯ ¯ ¯ x 1 x 2 x 3 x 4 5 x 1 +6 x 2 +7 x 3 +8 x 4 ¯ ¯ ¯ ¯ . The solutions are of the form x 1 x 2 x 3 x 4 = s 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 . Now ~v 1 · ~v 2 = d must be zero, so that d Likewise, ~v 2 · ~v 3 = e must be zero, so that e . Since ~v 3 must be a unit vector, we have k ~v 3 k 2 = c 2 + 1 4 =1 ,sothat c = ± 3 2 . Since we are asked to ±nd 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 k ~v 1 k 2 = a 2 + f 2 = a 2 a 2 =4 a 2 =1,sothat a = ± 1 2 . Let us pick a = 1 2 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 3 = 3 2 0 1 2 There are other solutions; some components will have di±erent 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 ) .S o V is a subspace of ( V ) .
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Linear Algebra by otto brestscher chapter 05 solutions -...

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