ENG
Chapter 8

# Chapter 8 - 8 Matrices Exercises 8.1 1 2 4 6 8 1 2 3 2 7...

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8 Matrices Exercises 8.1 1. 2 × 4 2. 3 × 2 3. 3 × 3 4. 1 × 3 5. 3 × 4 6. 8 × 1 7. Not equal 8. Not equal 9. Not equal 10. Not equal 11. Solving x = y 2, y = 3 x 2 we obtain x = 2, y = 4. 12. Solving x 2 = 9, y = 4 x we obtain x = 3, y = 12 and x = 3, t = 12. 13. c 23 = 2(0) 3( 3) = 9; c 12 = 2(3) 3( 2) = 12 14. c 23 = 2(1) 3(0) = 2; c 12 = 2( 1) 3(0) = 2 15. (a) A + B = 4 2 5 + 6 6 + 8 9 10 = 2 11 2 1 (b) B A = 2 4 6 5 8 + 6 10 9 = 6 1 14 19 (c) 2 A + 3 B = 8 10 12 18 + 6 18 24 30 = 2 28 12 12 16. (a) A B = 2 3 0 + 1 4 0 1 2 7 + 4 3 + 2 = 5 1 4 1 11 5 (b) B A = 3 + 2 1 0 0 4 2 1 4 7 2 3 = 5 1 4 1 11 5 (c) 2( A + B ) = 2 1 1 4 3 3 1 = 2 2 8 6 6 2 17. (a) AB = 2 9 12 6 5 + 12 30 + 8 = 11 6 17 22 (b) BA = 2 30 3 + 24 6 10 9 + 8 = 32 27 4 1 (c) A 2 = 4 + 15 6 12 10 20 15 + 16 = 19 18 30 31 (d) B 2 = 1 + 18 6 + 12 3 + 6 18 + 4 = 19 6 3 22 18. (a) AB = 4 + 4 6 12 3 + 8 20 + 10 30 30 15 + 20 32 + 12 48 36 24 + 24 = 0 6 5 10 0 5 20 12 0 313

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Exercises 8.1 (b) BA = 4 + 30 24 16 + 60 36 1 15 + 16 4 30 + 24 = 2 8 2 2 19. (a) BC = 9 24 3 8 (b) A ( BC ) = 1 2 2 4 9 24 3 8 = 3 8 6 16 (c) C ( BA ) = 0 2 3 4 0 0 0 0 = 0 0 0 0 (d) A ( B + C ) = 1 2 2 4 6 5 5 5 = 4 5 8 10 20. (a) AB = [5 6 7] 3 4 1 = ( 16) (b) BA = 3 4 1 [5 6 7] = 15 18 21 20 24 28 5 6 7 (c) ( BA ) C = 15 18 21 20 24 28 5 6 7 1 2 4 0 1 1 3 2 1 = 78 54 99 104 72 132 26 18 33 (d) Since AB is 1 × 1 and C is 3 × 3 the product ( AB ) C is not defined. 21. (a) A T A = [4 8 10] 4 8 10 = (180) (b) B T B = 2 4 5 [2 4 5] = 4 8 10 8 16 20 10 20 25 (c) A + B T = 4 8 10 + 2 4 5 = 6 12 5 22. (a) A + B T = 1 2 2 4 + 2 5 3 7 = 1 7 5 11 (b) 2 A T B T = 2 4 4 8 2 5 3 7 = 4 1 1 1 (c) A T ( A B ) = 1 2 2 4 3 1 3 3 = 3 7 6 14 23. (a) ( AB ) T = 7 10 38 75 T = 7 38 10 75 314
Exercises 8.1 (b) B T A T = 5 2 10 5 3 8 4 1 = 7 38 10 75 24. (a) A T + B = 5 4 9 6 + 3 11 7 2 = 2 7 2 8 (b) 2 A + B T = 10 18 8 12 + 3 7 11 2 = 7 11 3 14 25. 4 8 4 16 + 6 9 = 14 1 26. 6 3 3 + 5 5 15 + 6 8 10 = 5 10 22 27. 19 18 19 20 = 38 2 28. 7 17 6 + 1 1 4 2 8 6 = 10 10 4 29. 4 × 5 30. 3 × 2 31. A T = 2 3 4 2 ; ( A T ) T = 2 4 3 2 = A 32. ( A + B ) T = 6 6 14 10 = A T + B T 33. ( AB ) T = 16 40 8 20 T = 16 8 40 20 ; B T A T = 4 2 10 5 2 3 4 2 = 16 8 40 20 34. (6 A ) T = 12 18 24 12 = 6 A T 35. B = AA T = 2 1 6 3 2 5 2 6 2 1 3 5 = 5 15 9

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• Fall '07
• Delahanty
• Linear Algebra, Pallavolo Modena, Det, Howard Staunton, A-1

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