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Unformatted text preview: Linear Algebra Homework #4 Solution 1. First we want to show that det for any elementary matrix E. If E is of type 1, interchanging any row (column) of identity matrix I 2 , then 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Hence we get det 1 of type 1 elementary matrix. If E is of type 2, then since the rows are fixed, δ is linear. det k k for every scalar k . If E is of type 3, let E be obtained from I 2 by adding k times row i of I 2 to j, where j ≠ i, and let A be obtained from I 2 by replacing row j and row i. Then the rows of I 2 , E, and A are identical except for row j. Moreover, row j of E is the sum of row j of I 2 and k times row j of A. Since δ is linear, it follows that det 1 k Thus we show that det for any elementary matrix E. Secondly we want to show that . We can consider E is of type 2, multiply a row i k times from I 2 , say 1 k , then det 1 22 21 12 11 22 21 12 11 22 21 12 11 a a a a k a a ka ka a a a a k If E is of type 3, say 1 1 m , ...
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- Spring '11
- Linear Algebra, Det