chapter3 - Chapter 3 Solutions to 3.1 1. a31 = 0, a24 = 1,...

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83 Chapter 3 Solutions to §3.1 1. a 31 = 0, a 24 = –1, a 14 = 2, a 32 = 2, a 21 = 7, a 34 = –4. 3. 2 1 –1 0 4 –2 , 2 × 3. 5. 1 –3 –2 3 6 0 2 7 4 –4 –1 5 , 4 × 3. 7. Tr(A) = 1 + 3 = 4. 10. Row vectors: [1 –1], [3 5]. Column vectors: 1 3 , –1 5 . 11. Row vectors: [1 3 –4], [ ] –1 –2 5 , [ ] 2 6 7 . Column vectors: 1 –1 2 , 3 –2 6 , –4 5 7 . 13. A = 1 2 3 4 5 1 , 3 x 2 column vectors: 1 3 5 , 2 4 1 . 15. A = [a 1 , a 2 , . .., a p ] has p columns and each column q-vector has q rows so the resulting matrix has dimensions q by p. 17. 2 3 1 2 0 5 6 2 0 0 3 5 0 0 0 1 . 19. 3 0 0 0 2 0 0 0 5 .
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84 Ch. 3 Matrices and Systems of Linear Algebraic Equations 21. Let A be a symmetric upper triangular matrix. Then all elements below the main diagonal are zeros. Consequently, since A is symmetric, all elements above the main diagonal must also before zero. Hence A is a diagonal matrix. Solutions to §3.2 1. 2A = 2 4 –2 6 10 4 . –3B = –6 3 –9 –3 –12 –15 . A 2B = –3 4 –7 1 –3 –8 . 3A + 4B = 3 1 2 –1 3 5 2 + 4 2 –1 3 1 4 5 = 3 6 –3 9 15 6 + 8 –4 12 4 16 20 = 11 2 9 13 31 26 . 3. AB = 5 10 –3 27 22 3 , BC = 9 8 –6 , DC = [ ] 10 , DB = [ ] 6 14 –4 , CD = 2 –2 3 –2 2 –3 4 –4 6 . CA and AD cannot be computed. 5. AB = 3 + 2i 2 – 4i 5 + i 3i – 1 i – 1 2i + 3 4 – 3i 1 + i = (3 + 2i)(i – 1) + (2 – 4i)(4 – 3i) (3 + 2i)(2i + 3) + (2 – 4i)(1 + i) (5 + i)(i – 1) + (3i – 1)(4 – 3i) (5 + i)(2i + 3) + (3i – 1)(1 + i) = –9 – 21i 11 + 10i –1 + 19i 9 + 15i . 7. ABC = (AB)C = ( 1 –1 2 3 –2 3 4 6 3 2 1 5 4 –3 –1 6 ) C = 7 9 7 35 –3 2 1 –4 = –12 –22 14 –126 .
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Sec. 3.2 Matrix Algebra 85 CAB = (CA)B = ( –3 2 1 –4 1 –1 2 3 –2 3 4 6 ) B = –7 9 2 3 9 –13 –14 –21 3 2 1 5 4 –3 –1 6 = –7 43 –21 –131 . 9. A c = 1 3 –5 4 6 –2 = 6 1 –5 + (–2) 3 4 = 0 –38 .
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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 University-West Lafayette.

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chapter3 - Chapter 3 Solutions to 3.1 1. a31 = 0, a24 = 1,...

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