ISM-Chp_6

# Linear Algebra and Its Applications (3rd Edition)

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335 6.1 SOLUTIONS Notes : The first half of this section is computational and is easily learned. The second half concerns the concepts of orthogonality and orthogonal complements, which are essential for later work. Theorem 3 is an important general fact, but is needed only for Supplementary Exercise 13 at the end of the chapter and in Section 7.4. The optional material on angles is not used later. Exercises 27–31 concern facts used later. 1 . Since 1 2 ± ªº «» ¬¼ u and 4 , 6 v 22 (1 ) 2 5 ² ± ³ uu , v ² u = 4(–1) + 6(2) = 8, and 8 . 5 ² ² vu 2 . Since 3 1 5 ± ± w and 6 2, 3 ± x 222 3( 1 )( 5 )3 5 ² ³± ww , x ² w = 6(3) + (–2)(–1) + 3(–5) = 5, and 51 . 35 7 ² ² xw 3 . Since 3 1, 5 ± ± w 1 5 5 , and 3/35 1 1/35 . 1/7 ± ² ± w 4 . Since 1 , 2 ± u and 1/5 1 . 2/5 ± ² u 5 . Since 1 2 ± u and 4 , 6 v u ² v = (–1)(4) + 2(6) = 8, 465 2 , ² ³ vv and 48 / 1 3 2 . 61 2 / 1 3 13 ªº ª º ² §· ¨¸ «» « » ² ©¹ ¬¼ ¬ ¼ uv v 6 . Since 6 2 3 ± x and 3 5 ± ± w x ² w = 6(3) + (–2)(–1) + 3(–5) = 5, 2 6( 2 )34 9 , ³± ³ xx and 63 0 / 4 9 5 21 0 / 4 9 . 49 31 5 / 4 9 ªºª º ² «»« » ± ± » ² » ¬¼¬ ¼ x

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336 CHAPTER 6 • Orthogonality and Least Squares 7 . Since 3 1, 5 ªº «» ± ± ¬¼ w 222 || || 3 ( 1) ( 5) 35. ² ³ ± ³ ± ww w 8 . Since 6 2, 3 ± x 22 2 || || 6 ( 2) 3 49 7. ² ³ ±³ xx x 9 . A unit vector in the direction of the given vector is 30 30 3/5 11 40 40 4/5 50 ( 30) 40 ±± ± ª º « » ¬ ¼ ±³ 10 . A unit vector in the direction of the given vector is 2 6/ 61 66 44 4 / 6 1 61 (6 ) 4 (3 ) 33 36 1 ± ±³³ ± ± 11 . A unit vector in the direction of the given vector is 2 7/ 69 7/4 1/2 2/ 69 69/16 (7/ 4) (1/2) 1 4/ 69 ³³ 12 . A unit vector in the direction of the given vector is 8/3 3 / 5 100/9 (8/3) 2 ³ 13 . Since 10 3 ± x and 1 , 5 ± ± y 2 || || [10 ( 1)] [ 3 ( 5)] 125 ± ± ± ± ± xy and dist ( , ) 125 5 5. 14 . Since 0 5 2 ± u and 4 8 ± ± z 2 2 || || [0 ( 4)] [ 5 ( 1)] [2 8] 68 ± ³±±± ³ ± uz and dist ( , ) 68 2 17. 15 . Since a ² b = 8(–2) + (–5)( –3) = –1 0, a and b are not orthogonal. 16 . Since u ² v ± = 12(2) + (3)( –3) + (–5)(3) = 0, u and v are orthogonal. 17 . Since u ² v = 3(–4) + 2(1) + (–5)( –2) + 0(6) = 0, u and v are orthogonal. 18 . Since y ² z ± = (–3)(1) + 7(–8) + 4(15) + 0(–7) = 1 0, y and z are not orthogonal. 19 . a . True. See the definition of || v ||. b . True. See Theorem 1(c). c . True. See the discussion of Figure 5.
6.1 • Solutions 337 d . False. Counterexample: 11 . 00 ªº «» ¬¼ e . True. See the box following Example 6. 20 . a . True. See Example 1 and Theorem 1(a). b . False. The absolute value sign is missing. See the box before Example 2. c . True. See the defintion of orthogonal complement. d . True. See the Pythagorean Theorem. e . True. See Theorem 3. 21 . Theorem 1(b): () ( ) TT T ±² ± ± ± ²± ² uvw uvw u vwuwvwuwvw The second and third equalities used Theorems 3(b) and 2(c), respectively, from Section 2.1. Theorem 1(c): ( ) ( ) cc ² ² uv uv uv uv The second and third equalities used Theorems 3(c) and 2(d), respectively, from Section 2.1.

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## This document was uploaded on 05/26/2010.

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ISM-Chp_6 - 6.1 SOLUTIONS Notes: The first half of this...

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