04win-m2sols

# 04win-m2sols - MATH 51 MIDTERM 2 SOLUTIONS February 26,...

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MATH 51 MIDTERM 2 SOLUTIONS February 26, 2004 Professors Clingher, Munson, and White 1. Find the inverse of the matrix A = 1 - 1 1 - 1 1 1 1 1 - 1 . Solution: 1 - 1 1 . . . 1 0 0 - 1 1 1 . . . 0 1 0 1 1 - 1 . . . 0 0 1 1 - 1 1 . . . 1 0 0 0 0 2 . . . 1 1 0 0 2 - 2 . . . - 1 0 1 1 - 1 1 . . . 1 0 0 0 0 2 . . . 1 1 0 0 2 0 . . . 0 1 1 1 - 1 1 . . . 1 0 0 0 2 0 . . . 0 1 1 0 0 2 . . . 1 1 0 1 - 1 1 . . . 1 0 0 0 1 0 . . . 0 . 5 . 5 0 0 1 . . . . 5 . 5 0 1 0 0 . . . . 5 0 . 5 0 1 0 . . . 0 . 5 . 5 0 0 1 . . . . 5 . 5 0 so A - 1 = . 5 0 . 5 0 . 5 . 5 . 5 . 5 0 . 2. Suppose A and B are 3x3 matrices, and that det( A ) = 5 and det( B ) = - 2. (a) Find det( AB ). Solution: det( AB ) = det( A )det( B ) = 5( - 2) = - 10 . (b) Find det( A - 1 ). Solution: det( A - 1 ) = (det A ) - 1 = 1 / 5 . (c) Find det(2 A ). Solution: det(2 A ) = 2 3 det( A ) = 8(5) = 40 . 3. Let A = 0 - 1 2 3 . (a) Find the eigenvalues of A . Solution: det( λI - A ) = λ 1 - 2 λ - 3 = λ ( λ - 3) - 1( - 2) = λ 2 - 3 λ + 2 = ( λ - 1)( λ - 2) , so the eigenvalues are 1 and 2 . (b) Find the eigenvalues of A 10 . Solution: 1 10 = 1 and 2 10 (or 1024). 1

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(c) The matrix A = 2 1 1 1 3 - 2 1 - 2 3 has the number 3 as one of its eigenvalues. Find an eigenvector v that has 3 as its associated eigenvalue. Solution: We need a nonzero vector in the nullspace of 3 I - A . We ﬁnd it by Gaussian elimination: 1 - 1 - 1 - 1 0 2 - 1 2 0 1 0 - 2 1 - 1 - 1 - 1 2 0 1 0 - 2 0 - 1 1 0 2 - 2 1 0 - 2 0 1 - 1 0 0 0 So v is in the nullspace of 3 I - A if and only if v 1 - 2 v 3 = 0 and v 2 - v 3 = 0, i.e., if and only if v
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04win-m2sols - MATH 51 MIDTERM 2 SOLUTIONS February 26,...

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