06win-m2sols

06win-m2sols - MATH 51 MIDTERM 2 SOLUTIONS March 9 2006 1...

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MATH 51 MIDTERM 2 SOLUTIONS March 9, 2006 1(a). Find the determinant of the matrix C = 1 2 3 - 1 - 1 1 2 1 1 . Solution: for example, 1 2 3 - 1 - 1 1 2 1 1 = 1 2 3 0 1 4 0 - 3 - 5 = 1 2 3 0 1 4 0 0 7 = 7 1(b). Find the inverse of the matrix M = 1 0 - 2 0 1 3 - 2 1 8 . Ans: 1 0 - 2 | 1 0 0 0 1 3 | 0 1 0 - 2 1 8 | 0 0 1 1 0 - 2 | 1 0 0 0 1 3 | 0 1 0 0 1 4 | 2 0 1 1 0 - 2 | 1 0 0 0 1 3 | 0 1 0 0 0 1 | 2 - 1 1 1 0 - 2 | 5 - 2 2 0 1 0 | - 6 4 - 3 0 0 1 | 2 - 1 1 so M - 1 = 5 - 2 2 - 6 4 - 3 2 - 2 1 . 2 . Let v 1 = 1 1 and v 2 = - 2 1 . Let T : R 2 R 2 be the linear transformation such that T ( v 1 ) = 3 v 1 + 2 v 2 (i) T ( v 2 ) = - v 1 - v 2 . (ii) 2(a) . What is the matrix for T with respect to the basis B = { v 1 , v 2 } ? 1
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Solution: The first column of the matrix is [ T ( v 1 )] B , or 3 2 . The second column is [ T ( v 2 )] B , or 1 - 1 . Thus the matrix is B = 3 - 1 2 - 1 . 2(b) . What is the matrix for T with respect to the standard basis { e 1 , e 2 } ? Solution: The change of basis matrix is C = 1 - 2 1 1 (the matrix whose columns are v 1 and v 2 .) The matrix for T in standard coordinates is A = CBC - 1 . Now C - 1 = 1 / 3 2 / 3 - 1 / 3 1 / 3 , so A = 1 - 2 1 1 ‚• 3 - 1 2 - 1 ‚• 1 / 3 2 / 3 - 1 / 3 1 / 3 = - 1 1 5 - 2 ‚• 1 / 3 2 / 3 - 1 / 3 1 / 3 = - 2 / 3 - 1 / 3 7 / 3 8 / 3 3(a) . Find all eigenvalues of the matrix A = 1 4 3 1 1 2 0 0 7 .
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06win-m2sols - MATH 51 MIDTERM 2 SOLUTIONS March 9 2006 1...

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