Stitz-Zeager_College_Algebra_e-book

Use any of the methods presented in sections 81

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Unformatted text preview: he problem at hand. From our discussion at the beginning of the section on page 494, we know X = A−1 B = 4 17 3 − 17 3 17 2 17 16 7 = 5 −2 so that our final solution to the system is (x, y ) = (5, −2). As we mentioned, the point of this exercise was not just to solve the system of linear equations, but to develop a general method for finding A−1 . We now take a step back and analyze the foregoing discussion in a more general context. In solving for A−1 , we used two augmented matrices, both of which contained the same entries as A 496 Systems of Equations and Matrices 2 −3 1 3 40 = A 1 0 2 −3 0 3 41 = A 0 1 We also note that the reduced row echelon forms of these augmented matrices can be written as 10 01 10 01 4 17 3 − 17 3 17 2 17 = I2 = I2 x1 x3 x2 x4 where we have identified the entries to the left of the vertical bar as the identity I2 and the entries to the right of the vertical bar as the solutions to our systems. The long and short of the solution process can be summarized as A 1 0 −− − − − − −...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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