Stitz-Zeager_College_Algebra_e-book

When traveling downstream the river is helping carl

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Unformatted text preview: trix A below. (Others will be discussed in the Exercises.) 3 1 A = 0 −1 2 1 2 5 4 We found det(A) = −13 by expanding along the first row. Theorem 8.7 guarantees that we will get the same result if we expand along the second row. (Doing so would take advantage of the 0 there.) 3 1 det 0 −1 2 1 2 5 = 0C21 + (−1)C22 + 5C23 4 = (−1)(−1)2+2 det (A22 ) + 5(−1)2+3 det (A23 ) 32 24 = − det 31 21 − 5 det = −((3)(4) − (2)(2)) − 5((3)(1) − (2)(1)) = −8 − 5 = −13 In general, the sign of (−1)i+j in front of the minor in the expansion of the determinant follows an alternating pattern. Below is the pattern for 2 × 2, 3 × 3 and 4 × 4 matrices, and it extends naturally to higher dimensions. +− −+ +−+ − + − +−+ + − + − − + − + + − + − − + − + The reader is cautioned, however, against reading too much into these sign patterns. In the example above, we expanded the 3 × 3 matrix A by its second row and the term which corresponds to the second entry ended up being negative even though the sign attached to the minor is (+). These signs represent only the signs of the (−1)i+j in the formula; the sign of the corresponding entry as well as the minor itself determine the u...
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