WrittenHW08-Solutions

WrittenHW08-Solutions - Math 216 (Section 50) Written...

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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.

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WrittenHW08-Solutions - Math 216 (Section 50) Written...

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