week09 - Math 43, Fall 2007 B. Dodson Week 9: Monday:...

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Unformatted text preview: Math 43, Fall 2007 B. Dodson Week 9: Monday: Finish Suggested Hw8, start on Hw9 material 1. Determinants 2. Properties of Dets 3. Eigenvalues and Eigenvectors (2-by-2 in 4.1; then n-by-n in 4.3) We compute det 2 1 5 4 2 3 9 5 1 using the (first) row expansion (by minors): det ( A ) = 2 fl fl fl fl 2 3 5 1 fl fl fl fl- 1 fl fl fl fl 4 3 9 1 fl fl fl fl + 5 fl fl fl fl 4 2 9 5 fl fl fl fl = 2(2- 15)- (4- 27) + 5(20- 18) = 2(- 13)- (- 23) + 5(2) =- 26 + 33 = 7 . Problem Reduce A = 2 1 3 5 3 1 2 4 1 4 3 5 2 5 3 to an upper triagular matrix and use the reduction to find det ( A ) . Solution: A - 1 1 2 3 3 1 2 4 1 4 3 5 2 5 3 ( r 1 r 1- r 2 ) - 1 1 2 3 3 7 11 5 12 15 7 15 18 ( r 2 r 2 + 3 r 1 , r 3 r 3 + 4 r 1 , r 4 r 4 + 5 r 1 ) 2 - 1 1 2 3 3 7 11 5 12 15 1 1- 4 ( r 4 r 4- 2 r 2 ) - 1 1 2 3 1 1- 4 5 12 15 3 7 11 ( r 2 r 4 ) - 1 1 2 3 1 1- 4 7 35 4 23 ( r 3 r 3- 5 r 2 , r 4 r 4- 3 r 2 ) - 1 1 2 3 1 1- 4- 1- 11 4 23 ( r 3 r 3- 2 r 4 ) - 1 1 2 3 1 1- 4- 1- 11- 21 = U ( r 4 r 4 + 4 r 3 ). Now we have det U = (- 1)(1)(- 1)(- 21) =- 21 , and det A = (- 1) det U = 21 , since all of the 3rd elementary operations change the determinant by a factor of (1) so no change - there are no 2nd EROs and exactly one 1st ERO, with each 1st ERO changing the determinant by a factor of (- 1) . Notice that, for the problem of finding the value of the determinant, we do not need a reduced row echelon or 1s on the diagonal, and stop at a triangular form. 3 3. Eigenvalues and Eigenvectors A vector v = 0 in R n is an eigenvector with eigenvalue of an n-by- n matrix A if Av = v....
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week09 - Math 43, Fall 2007 B. Dodson Week 9: Monday:...

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