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Unformatted text preview: Purdue University MA 353: Linear Algebra II with Applications Homework 7, due Mar. 5, Some Solutions sec. 2.4 #4: Let A and B be n × n invertible matrices. Prove that AB is invertible and ( AB ) 1 = B 1 A 1 . Proof. Recall that AB is invertible iff there exists an n × n matrix C such that C ( AB ) = ( AB ) C = I n . And if this is the case C = ( AB ) 1 . So let C = B 1 A 1 . Then C ( AB ) = ( B 1 A 1 )( AB ) = B 1 ( A 1 A ) B = B 1 B = I n , and also ( AB ) C = ( AB )( B 1 A 1 ) = A ( BB 1 ) A 1 = AA 1 = I n . Hence AB is invertible and ( AB ) 1 = B 1 A 1 . #5: Let A be invertible. Prove that A t is invertible and ( A t ) 1 = ( A 1 ) t . Proof. The argument is similar to the above problem. Namely, consider ( A t )( A 1 ) t = ( A 1 A ) t = ( I n ) t = I n , and also ( A 1 ) t ( A t ) = ( AA 1 ) t = ( I n ) t = I n . #12: Prove Theorem 2.21, namely φ β : V → F n given by φ ( v ) = [ v ] β is an isomor phism, where β = { v 1 ,v 2 ,...,v n } is an order basis....
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 Spring '08
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 Linear Algebra, Algebra, Matrices, Vector Space, v0

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