Math416_GaussJordan - Remark 2 Let us check that ± 2-1 ²...

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Math 416 - Abstract Linear Algebra Fall 2011, section E1 Preview of Gauss-Jordan elimination ( § 2.2) Consider the linear system ( x 1 - x 2 = 3 4 x 1 + 3 x 2 = 5 whose augmented matrix is ± 1 - 1 3 4 3 5 ² . We solve the system using Gauss-Jordan elimination: ± 1 - 1 3 4 3 5 ² R 2 - 4 R 1 ± 1 - 1 3 0 7 - 7 ² 1 7 R 2 ± 1 - 1 3 0 1 - 1 ² R 1 + R 2 ± 1 0 2 0 1 - 1 ² . The unique solution is -→ x = ± x 1 x 2 ² = ± 2 - 1 ² . Remark 1: The reduced row echelon form of the augmented matrix (last step) is saying x 1 = 2 x 2 = - 1 whence the conclusion.
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Unformatted text preview: Remark 2: Let us check that ± 2-1 ² is indeed a solution, in all three points of view on linear systems. As a system of linear equations: ( (2)-(-1) = 3 4(2) + 3(-1) = 5 . As a matrix-vector equation: ± 1-1 4 3 ²± 2-1 ² = ± 3 5 ² . As a vector equation: 2 ± 1 4 ²-1 ±-1 3 ² = ± 2 8 ² + ± 1-3 ² = ± 3 5 ² . 1...
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This note was uploaded on 12/16/2011 for the course MATH 416 taught by Professor Frankland during the Fall '11 term at University of Illinois at Urbana–Champaign.

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