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lecture_15- Cramer's Rule Part 3

# lecture_15- Cramer's Rule Part 3 - MA 265 LECTURE NOTES...

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Unformatted text preview: MA 265 LECTURE NOTES: MONDAY, FEBRUARY 11 Cramer’s Rule Let A be an n × n matrix. We have seen that if det( A ) negationslash = 0, then A is nonsingular. In fact, its inverse is the n × n matrix A- 1 = 1 det( A ) adj( A ) = bracketleftbigg A ij det( A ) bracketrightbigg T = bracketleftbigg A ji det( A ) bracketrightbigg . We give a brief example. Consider a 2 × 2 matrix: A = bracketleftbigg a b c d bracketrightbigg . We have the adjoint adj A = bracketleftbigg d − c − b a bracketrightbigg T = bracketleftbigg d − b − c a bracketrightbigg = ⇒ A- 1 = d ad − bc − b ad − bc − c ad − bc a ad − bc . Linear Systems. Let’s return to a system of linear equations: a 11 x 1 + a 12 x 2 + ··· + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + ··· + a 2 n x n = b 2 . . . . . . . . . . . . a n 1 x 1 + a n 2 x 2 + ··· + a nn x n = b n where the number of equations is the same as the number of unknowns. Recall that we can express this system as a product of matrices: A = a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . a n 1 a n 2 . . . a nn , x = x 1 x 2 . . . x n and b = b 1 b 2 . . . b n = ⇒ A x = b . Hence if det( A ) negationslash = 0, then A is nonsingular, so that the unique solution so this system is x = A- 1 b . Cramer’s Rule. Now that we have an explicit formula for A- 1 , we use this to work out an explicit formula for x . Substituting in the expressions above: x = A- 1 b = 1 det( A ) · adj( A ) · b = 1 det( A ) A 11 A 21 ··· A n 1 A 12 A 22 ··· A n 2 . . . . . . . . . . . . A 1 n A 2 n ··· A nn b 1 b 2 . . . b n = 1 det( A ) A 11 b 1 + A 21 b 2 + ··· + A n 1 b n A 12 b 1 + A 22 b 2 + ··· + A n 2 b n ....
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lecture_15- Cramer's Rule Part 3 - MA 265 LECTURE NOTES...

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