Linear Algebra Solutions 21

Linear Algebra Solutions 21 - A is non–singular. 17. A =...

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(f) 1 2 3 4 5 6 5 7 9 R 2 R 2 - 4 R 1 R 3 R 3 - 5 R 1 1 2 3 0 - 3 - 6 0 - 3 - 6 R 3 R 3 - R 2 1 2 3 0 - 3 - 6 0 0 0 . Hence A is singular by virtue of the zero row. 15. Suppose that A is non–singular. Then AA - 1 = I n = A - 1 A. Taking transposes throughout gives ( AA - 1 ) t = I t n = ( A - 1 A ) t ( A - 1 ) t A t = I n = A t ( A - 1 ) t , so A t is non–singular and ( A t ) - 1 = ( A - 1 ) t . 16. Let A = a b c d , where ad - bc = 0. Then the equation A 2 - ( a + d ) A + ( ad - bc ) I 2 = 0 reduces to A 2 - ( a + d ) A = 0 and hence A 2 = ( a + d ) A . From the last equation, if A - 1 exists, we deduce that A = ( a + d ) I 2 , or a b c d = a + d 0 0 a + d . Hence a = a + d, b = 0 , c = 0 , d = a + d and a = b = c = d = 0, which contradicts the assumption that
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Unformatted text preview: A is non–singular. 17. A = 1 a b-a 1 c-b-c 1 R 2 → R 2 + aR 1 R 3 → R 3 + bR 1 1 a b 1 + a 2 c + ab ab-c 1 + b 2 R 2 → 1 1+ a 2 R 2 1 a b 1 c + ab 1+ a 2 ab-c 1 + b 2 R 3 → R 3-( ab-c ) R 2 1 a b 1 c + ab 1+ a 2 1 + b 2 + ( c-ab )( c + ab ) 1+ a 2 = B. 24...
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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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