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Unformatted text preview: MA 265 LECTURE NOTES: MONDAY, JANUARY 28 GaussJordan Reduction Augmented Matrices. Consider a system of linear equations in the form 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 m 1 x 1 + a m 2 x 2 + ··· + a mn x n = b m 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 m 1 a m 2 ... a mn , x = x 1 x 2 . . . x n and b = x 1 x 2 . . . x n = ⇒ A x = b . Consider the augmented matrix that keeps track of the coefficients to the right of the equal sign: A b = a 11 a 12 ... a 1 n b 1 a 21 a 22 ... a 2 n b 2 . . . . . . . . . . . . . . . a m 1 a m 2 ... a mn b m The method where we place this matrix in (row) echelon form is called Gaussian elimination . The method where we place this matrix in reduced (row) echelon form, is called GaussJordan reduction . For example, if the matrix can be placed in the form I n d , then the unique solution to the system is x = d . Example. Consider the following system of equations x + 2 y + 3 z = 6 3 x + y z = 2 2 x 3 y + 2 z = 14 The augmented matrix for this system is A b = 1 2 3 6 3 1 1 2 2 3 2 14 . We perform Gaussian elimination to find the reduced (row) echelon form. The leading one in the first row exists, so we subtract to eliminate the entries in the first column: 3 r 1 + r 2 → r 2 2 r 1 + r 3 → r 3 = ⇒ 1 2 3 6 5 10 20 7 4 2 . Now divide the second row by 5 to find the “leading one” for that row: ( 1 / 5) r 2 → r 2 = ⇒ 1 2 3 6 1 2 4 7 4 2 ....
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This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.
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