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Unformatted text preview: Math 216 (Section 50) Written Homework #8 – Solutions Fall 2007 1 ( 5.1–2 ) (a) Since BC = 3 4 5 1 2 3 1 = 12 10 3 9 , AB = 2 3 4 7 3 4 5 1 = 9 11 47 9 , we have A ( BC ) = 2 3 4 7 12 10 3 9 = 33 7 27 103 and ( AB ) C = 9 11 47 9 2 3 1 = 33 7 27 103 . Therefore A ( BC ) = ( AB ) C . (b) We can easily calculate B + C = 3 4 5 1 + 2 3 1 = 3 2 8 and AC = 2 3 4 7 2 3 1 = 9 7 21 1 . Thus A ( B + C ) = 2 3 4 7 3 2 8 = 18 4 68 8 . We already know what AB is from Part (a). So AB + AC = 9 11 47 9 + 9 7 21 1 = 18 4 68 8 . Therefore A ( B + C ) = AB + AC . 2 ( 5.1–7 ) A = A 1 A 2 = 2 1 3 2 1 3 1 2 = 1 2 2 4 Calculation shows AB = 1 2 2 4 2 4 1 2 = 0 0 0 0 , and thus det( AB ) = 0 . On the other hand, det( A ) = 1 2 2 4 = 4 4 = 0 , det( B ) = 2 4 1 2 = 4 4 = 0 , and thus det( A ) · det( B ) = 0 . Therefore det( AB ) = det( A ) · det( B ). 1 3 ( 5.1–9 ) It is straightforward to calculate AB = t 2 t 1 t 3 1 /t 1 t 1 + t 3 t 2 4 t 3 = t 4 t 2 + 6 t 3 t + t 2 4 t 3 + 8 t 4 3 t + t 3 t 4 4 t 2 + t 3 + t 4 , and A = 1 2 3 t 2 1 /t 2 , B = 1 1 6 t 12 t 2 ....
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This homework help was uploaded on 04/02/2008 for the course ENGR 101 taught by Professor Ringenberg during the Fall '07 term at University of Michigan.
 Fall '07
 Ringenberg

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