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Stitz-Zeager_College_Algebra_e-book

# On the other hand when we substitute 0 3 into the

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Unformatted text preview: = y B = 1 A3 = Az = 1 −1 1 A = 1 −1 3 0 −4 z 0 3 0 0 Expanding both det(A) and det (Az ) along the third rows (to take advantage of the 0’s) gives z= −12 6 det (Az ) = = det(A) −10 5 The reader is encouraged to solve this system for x and y similarly and check the answer. Our last application of determinants is to develop an alternative method for ﬁnding the inverse of a matrix.5 Let us consider the 3 × 3 matrix A which we so extensively studied in Section 8.5.1 3 1 2 5 A = 0 −1 2 1 4 We found through a variety of methods that det(A) = −13. To our surprise and delight, its inverse below has a remarkable number of 13’s in the denominators of its entries. This is no coincidence. 9 2 7 13 13 − 13 8 15 A−1 = − 10 − 13 13 13 2 1 3 − 13 13 13 Recall that to ﬁnd A−1 , we are essentially solving the matrix equation AX = I3 , where X = [xij ]3×3 is a 3 × 3 matrix. Because of how matrix multiplication is deﬁned, the ﬁrst column of I3 is the product of A with the ﬁrst column of X , the second column of I3...
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