FinalTestReviewSpring2011

255 r 1 0 1 1 1 1 0 1 1 1 255 r 1 1 0 1 1 r 1 1

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Unformatted text preview: = and AT￿ = ￿i i b = , so we get system ti￿ ￿ t2 ￿ 0 14 t i bi 10 i ii i ￿ ￿￿ 40 C 0 = . Hence, 4C = 0, that is, C = 0, and 14D = 10, that is, D = 5/7. 0 14 D 10 Answer: The equation of the line is y = 0 + 5/7t. 0 1 Ex. 2. Evaluate the determinant of the matrix A = 1 1 1 Hint. In the reduction first step add rows 1-4 to row # matrix to lower triangular. Then calculate determinant. ￿ (12 pts) Solution: ￿ ￿ ￿ ￿0 1 1 1 1￿ ￿ 0 1 1 1 1 ￿ −.25￿5 r ￿ ￿ ￿ ￿ ￿1 0 1 1 1￿ ￿ 1 0 1 1 1 ￿ −.25￿5 r ￿ ￿ ￿ ￿ ￿1 1 0 1 1￿ r → ￿ 1 1 0 1 1 ￿ −.25￿5 ￿ ￿ ￿ ￿ ￿1 1 1 0 1￿ ￿ 1 1 1 0 1 ￿ −.25￿5 r ￿ ￿ ￿ ￿ ￿ 1 1 1 1 0 ￿ + ￿4 ￿i ￿4 4 4 4 4￿ r i=1 Answer: |A| = (−1)(−1)(−1)(−1)4 = 4 1 1 0 1 1 1 5. 11 11 01 10 11 In the ￿ ￿ ￿ ￿ ￿ →￿ ￿ ￿ ￿ ￿ 1 1 1 . Show your work! 1 0 second step, transform −1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1 4 4 4 4 0 0 0 0 4 ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ Ex. 3. Use Gram-Schmidt process to find an orthonormal basis for the subspace V of R4 spanned the by vectors 2 1 1 0 , ￿2 = 0 , and ￿3 = −1 . v v ￿1 =...
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