129a_ans6 - det    2 1 0-3 0 1 1 2   ...

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Spring 2010 Answer of HW6 3.2 #8 Using replacing row operations only to obtain the ref: 1 3 3 - 4 0 1 2 - 4 0 0 0 1 0 0 0 0 , hence the determinant is 0 3.2 #19 14 3.2 #29 Since det B = - 2, det B 5 = ( - 2) 5 = - 32 3.2 #36 Since 0 = det A 4 = (det A ) 4 , det A = 0, hence A is not invertible. 3.2 #40 a. - 2 b. 32 c. - 16 d. 1 e. - 1 3.3 #2 x 1 = det " 6 1 7 2 # det " 4 1 5 2 # = 5 3 , x 2 = det " 4 6 5 7 # det " 4 1 5 2 # = - 2 3 3.3 #5 x 1 = det 7 1 0 - 8 0 1 - 3 1 2 det 2 1 0 - 3 0 1 0 1 2 = 6 4 = 3 2 x 2 = det 2 7 0 - 3 - 8 1 0 - 3 2 det 2 1 0 - 3 0 1 0 1 2 = 16 4 = 4
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x 3 = det 2 1 7 - 3 0 - 8 0 1 - 3
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Unformatted text preview: det    2 1 0-3 0 1 1 2    =-14 4 =-7 2 3.3 #8 For any s , the unique solution is x 1 = 3 s +2 3( s 2 +1) , x 2 = 2 s-9 5( s +1) 3.3 #9 For s 6 =-1, the unique solution is x 1 = 1 3( s +1) , x 2 = 3+4 s 6( s +1) 3.3 #15 the adjugate is    2 2 6-1-9 3    and the inverse is    1 / 3 1 / 3 1-1 / 6-3 / 2 1 / 2   ...
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This note was uploaded on 09/08/2010 for the course MATH 129A at San Jose State University .

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129a_ans6 - det    2 1 0-3 0 1 1 2   ...

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