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assign9_practice_soln

# assign9_practice_soln - Math 136 Practice Problems 9 213 1...

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Math 136 Practice Problems # 9 1. Let A = 2 1 3 3 a - 1 - 2 1 4 . Assuming that A is invertible, find A - 1 by the cofactor method. Solution: We have cof( A ) = 4 a + 1 - 10 3 + 2 a - 1 14 - 4 - 1 - 3 a 11 2 a - 3 , so (cof( A )) T = 4 a + 1 - 1 - 1 - 3 a - 10 14 11 3 + 2 a - 4 2 a - 3 Moreover, by definition of the determinant we have the determinant of A is the dot product of the i -th row of A with the i -th column of (cof( A )) T . Hence we get det A = 3( - 1) + a (14) + ( - 1)( - 4) = 1 + 14 a Hence, A - 1 = 1 det A (cof( A )) T = 1 1+14 a 4 a + 1 - 1 - 1 - 3 a - 10 14 11 3 + 2 a - 4 2 a - 3 . 2. Use Cramer’s Rule to solve the following systems. a) 3 x 1 - x 2 = 2 4 x 1 + 7 x 2 = 5 Solution: The coefficient matrix is A = 3 - 1 4 7 , so det A = 21 + 4 = 25. Hence, x 1 = 1 25 2 - 1 5 7 = 19 25 x 2 = 1 25 3 2 4 5 = 7 25 Thus, the solution is ~x = 19 / 25 7 / 25 . 1

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2 b) 2 x 1 + 3 x 2 + x 3 = 1 x 1 + x 2 - x 3 = - 1 - 2 x 1 + 2 x 3 = 1 Solution: The coefficient matrix is A = 2 3 1 1 1 - 1 - 2 0 2 , so det A = 6. Hence, x 1 = 1 6 1 3 1 - 1 1 - 1 1 0 2 = 4 6 x 2 = 1 6 2 1 1 1 - 1 - 1 - 2 1 2 = - 3 6 x 3 = 1 6 2 3 1 1 1 - 1 - 2 0 1 = 7 6 Hence, the solution is ~x = 2 / 3 - 1 / 2 7 / 6 . c) 2 x 1 + x 2 = 1 3 x 1 + 7 x 2 = - 2 Solution: The coefficient matrix is A = 2 1 3 7 , so det A = 14 - 3 = 11. Hence, x 1 = 1 11 1 1 - 2 7 = 9 11 x 1 = 1 11 2 1 3 - 2 = 7 11 Thus, the solution is
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