Week8 - v satisfying the equation A h-2 I v = Row reducing the augmented matrix we find 1 4 3 0 3 h-4 3 h-4 0 8-4 h 8-4 h which row reduces to 1 4

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Assignment 8 Applied Linear Algebra Math 232(Fall 2008) Additional question: Show that h - 2 is an eigenvalue of the matrix A = h - 1 4 3 2 - h 2 2 h - 2 0 0 . Find an eigenvector corresponding to this eigenvalue. A1. We must show that A - ( h - 2) I is not invertible. Notice that A - ( h - 2) I = 1 4 3 2 - h 4 - h 2 h - 2 0 2 - h . We now compute the determinant using cofactor expansion along the third row to see that the determinant is ( h - 2) ± ± ± ± 4 3 4 - h 2 ± ± ± ± + (2 - h ) ± ± ± ± 1 4 2 - h 4 - h ± ± ± ± . Using the formula for the determinant of a 2 × 2 matrix, we see that the determinant is ( h - 2)(8 - 12 + 3 h ) - ( h - 2)(4 - h - 8 + 4 h ) = ( h - 2)(3 h - 4) - ( h - 2)(3 h - 4) = 0 Thus the determinant is zero and so h - 2 is an eigenvalue of A . To find a corresponding eigenvector, we wish to find a non-trivial vector
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Unformatted text preview: v satisfying the equation ( A-( h-2) I ) v = . Row reducing the augmented matrix we find 1 4 3 0 3 h-4 3 h-4 0 8-4 h 8-4 h which row reduces to 1 4 3 0 1 1 0 0 0 . 1 The solution to this system consists of all vectors of the form t 1-1 1 where t is a free variable. Thus 1-1 1 is an eigenvector corresponding to the eigenvalue h-2. 2 Assignment 8 Applied Linear Algebra Math 232 (Fall 2008) Section 5.1 Section 5.2 Section 5.3...
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This note was uploaded on 06/14/2011 for the course MATH 232 taught by Professor Russel during the Fall '10 term at Simon Fraser.

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Week8 - v satisfying the equation A h-2 I v = Row reducing the augmented matrix we find 1 4 3 0 3 h-4 3 h-4 0 8-4 h 8-4 h which row reduces to 1 4

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