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College Algebra Exam Review 353

# College Algebra Exam Review 353 - R is a ﬁeld One can...

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8.3. MULTILINEAR MAPS AND DETERMINANTS 363 Lemma 8.3.10. Let ' W M n ±! N be an alternating mulilinear map. For any x 1 ;:::;x n 2 M , any pair of indices i ¤ j , and any r 2 R , '.x 1 ;:::;x i ± 1 ;x i C rx j ;x i C 1 ;:::;x n / D '.x 1 ;:::;x n /: Proof. Using the linearity of ' in the i —th variable, and the alternating property, '.x 1 ;:::;x i ± 1 ;x i C rx j ;x i C 1 ;:::;x n / D '.x 1 ;:::;x n / C r '.x 1 ;:::;x j ;:::;x j ;:::;x n / D '.x 1 ;:::;x n /: n Proposition 8.3.11. Let A and B be n –by– n matrices over R . (a) If B is obtained from A by interchanging two rows or columns, then det .B/ D ± det .A/ . (b) If B is obtained from A by multiplying one row or column of A by r 2 R , then det .B/ D r det .A/ . (c) If B is obtained from A by adding a multiple of one column (resp. row) to another column (resp. row), then det .B/ D det .A/ . Proof. Part (a) follows from the skew-symmetry of the determinant, part (b) from multilinearity, and part (c) from the previous lemma. n It is exceedingly inefﬁcient to compute determinants by a formula in- volving summation over all permuations. The previous proposition pro- vides an efﬁcient method of computing determinants, when
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Unformatted text preview: R is a ﬁeld. One can reduce a given matrix A to triangular form by elementary row operations: interchanging two rows or adding a multiple of one row to another row. Operations of the ﬁrst type change the sign of the determi-nant while operations of the second type leave the determinant unchanged. If B is an upper triangular matrix obtained from A in this manner, then det .A/ D . ± 1/ k det .B/ , where k is the number of row interchanges per-formed in the reduction. But det .B/ is the product of the diagonal entries of B , by part (c) of Corollary 8.3.9 . The same method works for matrices over an integral domain, as one can work in the ﬁeld of fractions; of course, the determinant in the ﬁeld of fractions is the same as the determinant in the integral domain....
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