lecture_9 - MA 265 LECTURE NOTES MONDAY JANUARY 28...

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MA 265 LECTURE NOTES: MONDAY, JANUARY 28 Gauss-Jordan 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 Gauss-Jordan 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 0 - 5 - 10 - 20 0 - 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 0 1 2 4 0 - 7 - 4 2 . 1
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2 MA 265 LECTURE NOTES: MONDAY, JANUARY 28 Let’s eliminate the other entries in the second column: - 2 r 2 + r 1 r 1 7 r 2 + r 3 r 3 = 1 0 - 1 - 2 0 1 2 4 0 0 10 30 .
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