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102_solutions7

# 102_solutions7 - MATH 102 SOLUTIONS TO HW#7 Section 4.4...

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SOLUTIONS TO HW #7 Section 4.4, problem 2. a) Since these are symmetric matrices, so are the co- factors. Therefore, we only need to compute the elements C ij for i 6 j . First, computing the the determinants: | M 11 | = 3 , | M 12 | = - 2 , | M 13 | = 1 , | M 22 | = 4 , | M 23 | = - 2 , | M 33 | = 3 . Also, we have that | A | = 4. Thus, we have (from C T = C ): A - 1 = 1 | A | C T = 3 / 4 1 / 2 1 / 4 1 / 2 1 1 / 2 1 / 4 1 / 2 3 / 4 . b) For this matrix we ﬁrst compute the minor determinants: | M 11 | = 2 , | M 12 | = 1 , | M 13 | = 0 , | M 22 | = 2 , | M 23 | = 1 , | M 33 | = 1 . Also, we have that | B | = 1. Thus, we have (from C T = C ): B - 1 = 1 | B | C T = 2 - 1 0 - 1 2 - 1 0 - 1 1 . Section 4.4, problem 14. a) Using Cramer’s rule, we may solve for y by com- puting the determinant: y = 1 ad - bc ± ± ± ± a 1 c 0 ± ± ± ± = - c ad - bc . b) In this case, we solve for y by computing the determinant: y = 1 D ± ± ± ± ± ± a 1 c d 0 f g 0 i ± ± ± ± ± ± = fg - di D . 1

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102_solutions7 - MATH 102 SOLUTIONS TO HW#7 Section 4.4...

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