chapter4 - Chapter 4 Solutions to §4.1 1 a σ(2 1 3 4...

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Unformatted text preview: Chapter 4 Solutions to §4.1 1. a) σ (2, 1, 3, 4) = (–1) N(2, 1, 3, 4) = (–1) 1 = –1, odd. b) σ (1, 3, 2, 4) = (–1) N(1, 3, 2, 4) = (–1) 1 = –1, odd. c) σ (1, 4, 3, 5, 2) = (–1) N(1, 4, 3, 5, 2) = (–1) 4 = 1, even. d) σ (5, 4, 3, 2, 1) = (–1) N(5, 4, 3, 2, 1) = (–1) 10 = 1, even. 3. det(A) = 1 –1 2 3 = 1 ⋅ 3 – (–1)2 = 5. 5. det(A) = 1 –1 2 3 6 2 –1 = 1 ⋅ 3(–1) + (–1)6 ⋅ 0 + 0 ⋅ 2 ⋅ 2 – 0 ⋅ 3 ⋅ 0 – 6 ⋅ 2 ⋅ 1 – (–1)(–1)2 = –17. 7. π π 2 2 2 π = 2 π 2 – 2 π 2 = (2 – 2) π 2 . 9. 3 2 6 2 1 –1 –1 1 4 = (3)(1)(4) + (2)(–1)(–1) + (6)(2)(1) – (–1)(1)(6) – (1)(–1)(3) – (4)(2)(2) = 19. 11. π 1/2 e 2 e –1 67 1/2 1/30 2001 π π 2 π 3 = (π 1/2 )(1/30)( π 3 ) + (e 2 )(2001)( π ) + (e –1 )(67 1/2 )( π 2 ) – ( π )(1/30)(e –1 ) – ( π 2 )(2001)( π 1/2 ) – ( π 3 )(67 1/2 )(e 2 ) ≈ 9601.882. 116 Sec. 4.1 The Definition of a Determinant 117 13. y 1 ´´´ – y 1 ´´ + 4y 1 ´ – 4y 1 = 8sin 2x + 4cos 2x – 8sin 2x – 4cos 2x = 0, y 2 ´´´ – y 2 ´´ + 4y 2 ´ – 4y 2 = –8cos 2x + 4sin 2x + 8cos 2x – 4sin 2x = 0, y 3 ´´´ – y 3 ´´ + 4y 3 ´ – 4y 3 = e x – e x + 4e x – 4e x = 0. y 1 y 2 y 3 y 1 ´ y 2 ´ y 3 ´ y 1 ´´ y 2 ´´ y 3 ´´ = cos 2x sin 2x e x –2sin 2x 2cos 2x e x –4cos 2x –4sin 2x e x = 2e x cos 2 2x – 4e x sin 2xcos 2x + 8e x sin 2 2x + 8e x cos 2 2x + 4e x sin 2xcos 2x + 2e x sin 2 2x = 10e x . 15 . a) S 4 = {1, 2, 3, 4} N(1, 2, 3, 4) = 0, σ (1, 2, 3, 4) = 1; N(1, 2, 4, 3) = 1, σ (1, 2, 4, 3) = –1; N(1, 3, 2, 4) = 1, σ (1, 3, 2, 4) = –1; N(1, 3, 4, 2) = 2, σ (1, 3, 4, 2) = 1; N(1, 4, 2, 3) = 2, σ (1, 4, 2, 3) = 1; N(1, 4, 3, 2) = 3, σ (1, 4, 3, 2) = –1; N(2, 1, 3, 4) = 1, σ (2, 1, 3, 4) = –1; N(2, 1, 4, 3) = 2, σ (2, 1, 4, 3) = 1; N(2, 3, 1, 4) = 2, σ (2, 3, 1, 4) = 1; N(2, 3, 4, 1) = 3, σ (2, 3, 4, 1) = –1; N(2, 4, 1, 3) = 3, σ (2, 4, 1, 3) = –1; N(2, 4, 3, 1) = 4, σ (2, 4, 3, 1) = 1; N(3, 1, 2, 4) = 2, σ (3,1,2,4) = 1; N(3, 1, 4, 2) = 3, σ (3, 1, 4, 2) = –1; N(3, 2, 1, 4) = 3, σ (3,2,1,4) = –1; N(3, 2, 4, 1) = 4, σ (3, 2, 4, 1) = 1; N(3, 4, 1, 2) = 4, σ (3, 4, 1, 2) = 1; N(3, 4, 2, 1) = 5, σ (3, 4, 2, 1) = –1; N(4, 1, 2, 3) = 3, σ (4, 1, 2, 3) = –1; N(4, 1, 3, 2) = 4, σ (4, 1, 3, 2) = 1; N(4, 2, 1, 3) = 4, σ (4, 2, 1, 3) = 1; N(4, 2, 3, 1) = 5, σ (4, 2, 3, 1) = –1; N(4, 3, 1, 2) = 5, σ (4, 3, 1, 2) = –1; N(4, 3, 2, 1) = 6, σ (4, 3, 2, 1) = 1; det(A) = a 11 a 22 a 33 a 44 – a 11 a 22 a 34 a 43 – a 11 a 23 a 32 a 44 + a 11 a 23 a 34 a 42 + a 11 a 24 a 32 a 43 – a 11 a 24 a 33 a 42 – a 12 a 21 a 33 a 44 + a 12 a 21 a 34 a 43 + a 12 a 23 a 31 a 44 – a 12 a 23 a 34 a 41 – a 12 a 24 a 31 a 43 + a...
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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chapter4 - Chapter 4 Solutions to §4.1 1 a σ(2 1 3 4...

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