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Stitz-Zeager_College_Algebra_e-book

# Substituting these values into e 1 gives z 1 when x 1

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Unformatted text preview: = det = = = = For a generic 2 × 2 matrix A = 4 −3 2 1 4 det (A11 ) − (−3) det (A12 ) 4 det([1]) + 3 det([2]) 4(1) + 3(2) 10 ab cd we get ab cd det(A) = det = a det (A11 ) − b det (A12 ) = a det ([d]) − b det ([c]) = ad − bc This formula is worth remembering Equation 8.1. For a 2 × 2 matrix, det ab cd = ad − bc 3 1 2 5 we obtain Applying Deﬁnition 8.13 to the 3 × 3 matrix A = 0 −1 2 1 4 3 1 2 5 det(A) = det 0 −1 2 1 4 = 3 det (A11 ) − 1 det (A12 ) + 2 det (A13 ) = 3 det −1 5 14 − det 05 24 + 2 det 0 −1 2 1 = 3((−1)(4) − (5)(1)) − ((0)(4) − (5)(2)) + 2((0)(1) − (−1)(2)) = 3(−9) − (−10) + 2(2) = −13 To evaluate the determinant of a 4 × 4 matrix, we would have to evaluate the determinants of four 3 × 3 matrices, each of which involves the ﬁnding the determinants of three 2 × 2 matrices. As you can see, our method of evaluating determinants quickly gets out of hand and many of you may be reaching for the calculator. There is some mathematical machinery which can assist us in calculat...
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