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Unformatted text preview: Quiz 5.3 Using the information given below, find a matrix P so that D = P 1 AP is a diagonal matrix A = 6 2 7 1 1 1 4 2 5 Note that you do not have to compute P 1 . The characteristic polynomial for A is det( A xI ) = 2 + x + 2 x 2 x 3 . x = 1 is a root of the characteristic polynomial. The characteristic polynomial divided by ( x 1) is x 2 + x + 2 . SOLUTION: From the information we know that det( A xI ) = ( x 1)( x 2 + x + 2). Further factoring tells us that 0 = det( A xI ) = ( x 1)( x 2 + x +2) = ( x 1)( x 2 x 2) = ( x 1)( x 2)( x +1) x = 1 , 1 , 2 For each of these eigenvalues we must find an eigenvector which will provide a column for P . x = 1 A I = 5 2 7 1 1 4 2 6 R 1 R 2 =  1 1 5 2 7 4 2 6 R 1  R 1 = 1 1 5 2 7 4 2 6 ( R 2 ,R 3) ( R 2 5 R 1 ,R 3 4 R 1) = 1 1 2 2 2 2 R 3...
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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

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