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

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Unformatted text preview: 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- 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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lecture_9 - MA 265 LECTURE NOTES: MONDAY, JANUARY 28...

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