Stitz-Zeager_College_Algebra_e-book

It amounts to the second property in theorem 86 where

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Unformatted text preview: th 1 R2 + R1 3 3 1 1 0 −4 − − − − − − − −→ 0 0 4 −− − − − − − − − 5 35 0 0 1 24 0 24 At last, we decode to 1 0 1 0 0 0 1 0 −3 −2 3 6 1 19 2 5 35 1 24 24 0 1 0 1 0 −3 1 0 −4 5 1 24 5 0 − 12 0 −1 4 5 1 24 −2 3 3 4 35 24 5 − 12 3 4 35 24 get 5 0 − 12 0 −1 4 5 1 24 5 − 12 3 4 35 24 5 5 x1 − 12 x4 = − 12 Decode from the matrix 1 3 x2 − 4 x4 = −−−−−−− −−−−−−−→ 4 x + 5x = 35 4 3 24 24 5 We have that x4 is free and we assign it the parameter t. We obtain x3 = − 24 t + 35 , x2 = 1 t + 3 , 24 4 4 5 5 5 51 3 5 35 and x1 = 12 t − 12 . Our solution is 12 t − 12 , 4 t + 4 , − 24 t + 24 , t : −∞ < t < ∞ and leave it to the reader to check. 472 Systems of Equations and Matrices Like all good algorithms, putting a matrix in row echelon or reduced row echelon form can easily be programmed into a calculator, and, doubtless, your graphing calculator has such a feature. We use this in our next example....
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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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