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Stitz-Zeager_College_Algebra_e-book

# Rewriting this in standard form we get y 2 x 4 and we

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Unformatted text preview: is the product of A with the second column of X and the third column of I3 is the product of A with the third column of X .6 In other words, we are solving three equations x11 1 x12 0 x13 0 A x21 = 0 A x22 = 1 A x23 = 0 x31 0 x32 0 x33 1 We can solve each of these systems using Cramer’s Rule. Focusing on the ﬁrst system, we have 1 1 2 312 3 1 1 5 A2 = 0 0 5 A3 = 0 −1 0 A1 = 0 −1 0 1 4 204 2 1 0 5 We are developing a method in the forthcoming discussion. As with the discussion in Section 8.4 when we developed the ﬁrst algorithm to ﬁnd matrix inverses, we ask that you indulge us. 6 The reader is encouraged to stop and think this through. 8.5 Determinants and Cramer’s Rule 515 If we expand det (A1 ) along the ﬁrst row, we get det (A1 ) = det −1 5 14 = det −1 5 14 − det 05 04 + 2 det 0 −1 0 1 Amazingly, this is none other than the C11 cofactor of A. The reader is invited to check this, as well as the claims that det (A2 ) = C12 and det (A3 ) = C13 .7 (To see this, though it seems unnatural to do so, expand along the ﬁr...
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