Linear Algebra Solutions 48

Linear Algebra Solutions 48 - det A = det(( E 1 E m ) E m...

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(c) det E ij ( t ) = 1. Now suppose that B is a non–singular n × n matrix. Then we know that B is a product of elementary row matrices: B = E 1 ··· E m . Consequently we have to prove that det E 1 ··· E m A = det E 1 ··· E m det A. We prove this by induction on m . First the case m = 1. We have to prove det E 1 A = det E 1 det A if E 1 is an elementary row matrix. This follows form Remark 4.0.4: (a) det E ij A = - det A = det E ij det A ; (b) det E i ( t ) A = t det A = det E i ( t )det A ; (c) det E ij ( t ) A = det A = det E ij ( t )det A . Let m 1 and assume the proposition holds for products of m elementary row matrices. Then det E 1 ··· E m E m +1 A = det( E 1 ··· E m )( E m +1 A ) = det( E 1 ··· E m )det( E m +1 A ) = det( E 1 ··· E m )det E m +1
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Unformatted text preview: det A = det(( E 1 E m ) E m +1 )det A and the induction goes through. Hence det BA = det B det A if B is nonsingular. If B is singular, problem 26, Chapter 2.7 tells us that BA is also singlular. However singular matrices have zero determinant, so det B = 0 det BA = 0 , so the equation det BA = det B det A holds trivially in this case. 16. f f f f f f f f a + b + c a + b a a a + b a + b + c a a a a a + b + c a + b a a a + b a + b + c f f f f f f f f 51...
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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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