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Unformatted text preview: Math 341 Homework #8 (3.2) #11. Combine the methods of row reduction and cofactor expansion to com pute the determinant. 2 5 3 1 3 1 3 6 4 9 4 10 4 1 Solution: Here we go: 2 5 3 1 3 1 3 6 4 9 4 10 4 1 2 R 1 + R 4 = 2 5 3 1 3 1 3 6 0 4 9 2 1 = ( 1) 1+2 5 3 1 3 6 4 9 2 1 2 R 1 + R 2 = 5 3 1 3 2 3 2 1 = 5 3 2 3 2 1 = 5 3 ( 2 6) = 120 (3.2) #31. Show that if A is invertible, then det ( A 1 ) = 1 det( A ) . Solution: Since A is invertible, AA 1 = I (by the definition of invertible, or the Invertible Matrix Theorem). Hence, det ( AA 1 ) = det( I ) = 1. But by Theorem 6, det ( AA 1 ) = det( A )det ( A 1 ) , so det( A )det ( A 1 ) = 1. Solving for det ( A 1 ) , we get det ( A 1 ) = 1 det( A ) . (3.2) #32. Find a formula for det( rA ) when A is an n n matrix. Solution: By Theorem 3c, if we multiply a row of A by a scalar we multiply the determi nant by that scalar. In rA , weve multiplied each of the n rows by r , so weve multiplied the determinant by r n times; that is, weve multiplied the determinant by n factors of r . Hence, det( rA ) = r n det( A ). (3.2) #33. Let A and B be square matrices. Show that even though AB and BA may not be equal, it is always true that det( AB ) = det( BA )....
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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada.
 Spring '08
 Johnson,J
 Linear Algebra, Algebra, Determinant

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