Linear Algebra Solutions 46

Linear Algebra Solutions 46 - + x n a 2 n a 22 a 2 n . . ....

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12. Let A be a real 3 × 3 matrix satisfying A t A = I 3 . Then (i) A t ( A - I 3 ) = A t A - A t = I 3 - A t = - ( A t - I 3 ) = - ( A t - I t 3 ) = - ( A - I 3 ) t . Taking determinants of both sides then gives det A t det( A - I 3 ) = det( - ( A - I 3 ) t ) det A det( A - I 3 ) = ( - 1) 3 det( A - I 3 ) t = - det( A - I 3 ) (1) . (ii) Also det AA t = det I 3 , so det A t det A = 1 = (det A ) 2 . Hence det A = ± 1. (iii) Suppose that det A = 1. Then equation (1) gives det( A - I 3 ) = - det( A - I 3 ) , so (1 + 1)det( A - I 3 ) = 0 and hence det( A - I 3 ) = 0. 13. Suppose that column 1 is a linear combination of the remaining columns: A * 1 = x 2 A * 2 + ··· + x n A * n . Then det A = f f f f f f f f f x 2 a 12 + ··· + x n a 1 n a 12 ··· a 1 n x 2 a 22 +
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Unformatted text preview: + x n a 2 n a 22 a 2 n . . . . . . . . . . . . x 2 a n 2 + + x n a nn a n 2 a nn f f f f f f f f f . Now det A is unchanged in value if we perform the operation C 1 C 1-x 2 C 2--x n C n : det A = f f f f f f f f f a 12 a 1 n a 22 a 2 n . . . . . . . . . . . . a n 2 a nn f f f f f f f f f = 0 . 49...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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