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Unformatted text preview: Assignment 10 Math 417 Winter 2009 Due April 10 6.1#10. Using the method discussed in class. det 1 1 1 1 2 3 1 3 6 = det 1 1 1 0 1 2 0 2 5 R 2 R 2 R 1 R 3 R 3 R 1 = det 1 1 1 0 1 2 0 0 1 = 1 ( R 3 R 3 2 R 2 ) 6.1#20. Using the method discussed in class. det 1 k 1 1 k + 1 k + 2 1 k + 2 2 k + 4 = det 1 1 1 0 1 k + 1 0 2 2 k + 3 R 2 R 2 R 1 R 3 R 3 R 1 = det 1 1 1 0 1 k + 1 0 0 1 = 1 ( R 3 R 3 2 R 2 ) 6.1#28. First note that A I 3 = 5 7 11 0 3 13 0 0 2  1 0 0 0 1 0 0 0 1 = 5 7 11 3 13 2 is upper triangular for all values of . Thus, det( A I 3 ) = (5 )(3 )(2 ). This determinant is 0 precisely when = 2 , 3 , 5. These are the only values of for which A I 3 fails to be invertible. 1 6.1#42. In this case, cofactor expansion (Theorem 6.2.10) works well: det 1 1 0 1 1 0 2 0 7 0 2 3 4 5 0 0 0 0 3 0 3 4 5 2 6...
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 Spring '09
 eveet
 Math, Linear Algebra, Algebra

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