WrittenHW08-Solutions

# WrittenHW08-Solutions - Math 216(Section 50 1(5.12(a Since...

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Math 216 (Section 50) Written Homework #8 – Solutions Fall 2007 1 ( 5.1–2 ) (a) Since BC = 3 - 4 5 1 0 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 0 2 3 - 1 = - 33 - 7 - 27 103 . Therefore A ( BC ) = ( AB ) C . (b) We can easily calculate B + C = 3 - 4 5 1 + 0 2 3 - 1 = 3 - 2 8 0 and AC = 2 - 3 4 7 0 2 3 - 1 = - 9 7 21 1 . Thus A ( B + C ) = 2 - 3 4 7 3 - 2 8 0 = - 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

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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 . Then ( AB ) = d dt 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 = 1 - 8 t + 18 t 2 1 + 2 t - 12 t 2 + 32 t 3 3 + 3 t 2 - 4 t 3 8 t + 3 t 2 + 4 t 3 , and A B + AB = 1 2 3 t 2 - 1 /t 2 1 - t 1 + t 3 t 2 4 t 3 + t 2 t - 1 t 3 1 /t - 1 1 6 t 12 t 2 = 1 - t + 6 t 2 1 + t + 8 t 3 - 3 + 3 t 2 - 3 t 3 - 4 t + 3 t 2 + 3 t 3 + - 7 t + 12 t 2 t - 12 t 2 + 24 t 3 6 - t 3 12 t + t 3 = 1 - 8 t + 18 t 2 1 + 2 t - 12 t 2
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