Stitz-Zeager_College_Algebra_e-book

Our nal answer is x3 5 x 1 x 2x 1 2 4 6 x2 9 x x

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Unformatted text preview: 13 8 − 13 15 13 3 13 1 13 2 13 8 − 13 1 13 9 13 − 10 13 2 − 13 A− 1 A = 9 13 − 10 13 2 − 13 499 100 Replace R1 with −−−−− − −1− − − → 0 1 0 − 3 R2 + R1 001 7 − 13 15 13 3 13 2 13 8 − 13 1 13 9 13 10 − 13 2 − 13 2 13 8 − 13 1 13 7 − 13 15 13 3 13 . To check our answer, we compute 7 − 13 15 13 3 13 3 1 2 0 −1 2 1 100 5 = 0 1 0 = I3 4 001 and 3 1 AA−1 = 0 −1 2 1 2 5 4 2 13 8 − 13 1 13 9 13 − 10 13 2 − 13 7 − 13 15 13 3 13 100 = 0 1 0 = I3 001 2. Each of the systems in this part has A as its coefficient matrix. The only difference between the systems is the constants which is the matrix B in the associated matrix equation AX = B . We solve each of them using the formula X = A−1 B . (a) X = A−1 B = (b) X = A−1 B = (c) X = A−1 B = 9 13 − 10 13 2 − 13 9 13 − 10 13 2 − 13 9 13 − 10 13 2 − 13 2 13 8 − 13 1 13 2 13 8 − 13 1 13 2 13 8 − 13 1 13 7 − 13 15 13 3 13 7 − 13 15 13 3 13 7 − 13 15 13 3 13 39 = 91 . Our solution is (−39, 91, 26). 117 26 5 4 13 19 9 5 2 = 13 . We get 13 , 19 , 13 . 13 9 5 13 9 1 13 10 9 2 0 = − 13 . We find 13 , − 10...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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