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# 5.1 - 1 a a 2 1 b b 2 1 c c 2 → 1 a a 2 b-a b 2-a 2 c-a c...

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Math 2940 Solutions, Fall 2011 Section 5 . 1 2 ) We know A is 3 × 3 and that det ( A ) = - 1. det (2 A ) = 2 3 det ( A ) = - 8, det ( - A ) = ( - 1) 3 det ( A ) = 1, det ( A 2 ) = det ( A ) det ( A ) = 1 and det ( A - 1 ) = 1 /det ( A ) = - 1. 3 ) (a) False. For A = 2 0 0 2 , det ( I + A ) = 9 while 1 + det ( A ) = 5. (b) True. det (( AB ) C ) = det ( AB ) det ( C ) = det ( A ) det ( B ) det ( C ). (c) False. For A = 2 0 0 2 , det (4 A ) = 64 while 4 det ( A ) = 16. (d) False. Let A = 0 0 0 1 and A = a b c d so AB - BA = 0 0 0 1 a b c d - a b c d 0 0 0 1 = 0 - b c 0 which has determinant bc , which is not 0 as long as b, c = 0. 12 ) Assuming ad - bc = 0, the problem is, for 2 × 2 matrices, that det ( αA ) = α 2 det ( A ). 13 ) 1 1 1 1 2 2 1 2 3 1 1 1 0 1 1 0 1 2 1 1 1 0 1 1 0 0 1 so det ( A ) = 1 3 = 1. 1 2 3 2 2 3 3 3 3 1 2 3 0 - 2 - 3 0 - 3 - 6 1 2 3 0 - 2 - 3 0 0 - 3 / 2 so det ( A ) = (1)( - 2)( - 3 / 2) = 3. 18 ) 1 a a 2 1 b b 2 1 c c 2 1 a a 2 0 b - a b
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Unformatted text preview: ) 1 a a 2 1 b b 2 1 c c 2 → 1 a a 2 b-a b 2-a 2 c-a c 2-a 2 → 1 a a 2 b-a b 2-a 2 c 2-a 2-c-a b-a ( b 2-a 2 ) . The last entry becomes c 2-a 2-( c-a )( b + a ) = ( c-a )(( c + a )-( b + a )) = ( c-a )( c-b ) so det ( A ) = ( c-a )( c-b )( b-a ). 24 ) det ( L ) = 1 3 = 1, det ( U ) = (3)(2)(-1) =-6, det ( A ) = det ( L ) det ( U ) =-6. As U-1 L-1 = A-1 , det ( U-1 L-1 ) = 1 /det ( A ) =-1 / 6, and U-1 L-1 A = I so it has determinant 1....
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