ism3 - Chapter 3 Determinants Section 3.1, p. 192 2. (a)...

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Chapter 3 Determinants Section 3.1, p. 192 2. (a) even. (b) odd. (c) even. (d) odd. (e) even. (f) even. 4. The number of inversions are: (a) 9, 6. (b) 8, 7. (c) 5, 6. (d) 2, 7. 6. (a) 2. (b) 24. (c) 30. (d) 2. 8. | B | =4; | C | = 8; | D | = 4. 10. det( A ) = det( A T ) = 14. 12. (a) ( λ 1)( λ 2)( λ 3) = λ 3 6 λ 2 +11 λ 6. (b) λ 3 λ . 14. (a) 1, 2, 3. (b) 1, 0, 1. 16. (a) 144. (b) 168. (c) 72. 18. (a) 120. (b) 29. (c) 9. 20. (a) 1 (b) 120. (c) 22. 22. (a) 16. (b) 256. (c) 1 4 . 24. (a) 1. (b) 1. (c) 1. 26. (a) 1. (b) 1. T.1. If j i and j i +1 are interchanged, all inversions between numbers distinct from j i and j i +1 remain unchanged, and all inversions between one of j i , j i +1 and some other number also remain unchanged. If originally j i <j i +1 , then after interchange there is one additional inversion due to j i +1 j i .I f originally j i >j i +1 , then after interchange there is one fewer inversion. Suppose j p and j q are separated by k intervening numbers. Then k interchanges of adjacent numbers will move j p next to j q . One interchange switches j p and j q . Finally, k interchanges of adjacent numbers takes j q back to j p ’s original position. The total number of interchanges is the odd number 2 k +1. T.2. Parallel to proof for the upper triangular case. T.3. cA = ± ca ij ² .By n applications of Theorem 3.5, the result follows.
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60 Chapter 3 T.4. If A is nonsingular, then AA 1 = I n . Therefore det( A ) · det( A 1 ) = det( AA 1 ) = det( I n )=1 . Thus det( A ) ± = 0 and det( A 1 )= 1 det( A ) . T.5. det( AB ) = det( A )det( B ). Thus if det( AB ) = 0, then det A · det B = 0, and either det A =0or det B =0. T.6. det( AB ) = det( A ) · det( B ) = det( B ) · det( A ) = det( BA ). T.7. In the summation det( A ± ( ² ) a 1 j 1 a 2 j 2 ··· a nj n for the definition of det( A ) there is exactly one nonzero term. Thus det( A ) ± T.8. det( A B ) = det( AB ) = det( I n ) = 1. Thus det( A ) ± = 0 and det( B ) ± T.9. (a) [det( A )] 2 = det( A A ) = det( A A 1 ) = det( AA 1 )=1. (b) [det( A )] 2 = det( A A ) = det( A A T ) = det( A A 1 ) = det( AA 1 T.10. det( A 2 ) = [det( A )] 2 = det( A ), so det( A ) is a nonzero root of the equation x 2 x T.11. det( A T B T ) = det( A T B T ) = det( A B T ) = det( A T B ). T.12. ¯ ¯ ¯ ¯ ¯ ¯ a 2 a 1 b 2 b 1 c 2 c 1 ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ a 2 a 1 b 2 a 2 b a 0 c 2 a 2 c a 0 ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ( b a )( b + a ) b a ( c a )( c + a ) c a ¯ ¯ ¯ ¯ =( b a )( c a ) ¯ ¯ ¯ ¯ b + a 1 c + a 1 ¯ ¯ ¯ ¯ b a )( c a )( b c ).
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ism3 - Chapter 3 Determinants Section 3.1, p. 192 2. (a)...

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