MATH311ansHW4

MATH311ansHW4 - Using the cofactor expansion with respect...

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Math 311, Homework 4 partial answers and solutions 2.5.5. det - 1 0 1 2 0 1 2 - 1 1 2 - 1 0 2 - 1 0 1 = det - 1 0 1 2 0 1 2 - 1 0 2 0 2 0 - 1 2 5 = - det 1 2 - 1 2 0 2 - 1 2 5 = - det 1 2 - 1 0 - 4 4 0 4 4 = - det ± - 4 4 4 4 ² = - ( - 16 - 16) = 32 . 2.5.6. Answer: 12 . 2.5.8. Answer: AB = - 2 0 2 6 - 3 - 9 0 12 20 , BA = - 2 0 4 4 - 3 - 12 0 9 20 . The determinants are calculated as usual. 2.5.9. If A is invertible, then A - 1 is defined, and det( A ) det( A - 1 ) = det( AA - 1 ) = det I = 1 . It follows that det A 6 = 0 , and det A - 1 = (det A ) - 1 . 2.5.11. Using the cofactor expansion with respect to the first column, the determinant of the given matrix is equal to det - 1 5 6 0 3 - 1 0 0 4 .
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Unformatted text preview: Using the cofactor expansion with respect to the rst column of this matrix, the determinant is (-1) det 3-1 4 . Proceeding in this way, we see that the determinant of this matrix is the product of its diagonal entries, (-1) 3 4 =-12 . 2.5.17. Cofactor expansion with respect to the rst column gives det A = 0 . Thus the matrix is not invertible. 2.5.18. Answer: 1-1 0-1 1 1 . 1...
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This note was uploaded on 04/29/2008 for the course MATH 311 taught by Professor Anshelvich during the Spring '08 term at Texas A&M.

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