895 9.2 Matrix Solution of Linear Systems EXAMPLE 2 Using the Gauss-Jordan Method Solve the system. x - y + 5z = -6 3x + 3y - z = 10 x + 3y + 2z = 5 SOLUTION £ 1 3 1 -1 3 3 5 -1 2 3 -6 10 5§ Write the augmented matrix. There is already a 1 in the first row, first column. Introduce 0 in the second row of the first column by multiplying each element in the first row by -3 and adding the result to the corresponding element in the second row. £1 0 1 -1 6 3 5 -16 2 3 -6 28 5§ -3R1 + R2 To change the third element in the first column to 0, multiply each element of the first row by -1. Add the result to the corresponding element of the third row. £1 0 0 -1 6 4 5 -16 -3 3 -6 28 11§ -1R1 + R3 Use the same procedure to transform the second and third columns. Obtain 1 in the appropriate position of each column by multiplying the elements of the row by the reciprocal of the number in that position. £1 0 0 -1 1 4 5 -8 3 -3 3 -6 14 3 11§ 1 6 R2 D1 0 0 0 1 4 7 3 -8 3 -3 4 -4 3 14 3 11T R2 + R1 D1 0 0 0 1 0 7 3 -8 3 23 3 4 -4 3 14 3 -23 3 T4R2 + R3 D1 0 0 0 1 0 7 3 -8 3 1 4 -4 3 14 3 -1T 3 23 R3 £1 0 0 0 1 0 0 -8 3 1 3 1 14 3 -1§ -7 3 R3 + R1 £1 0 0 0 1 0 0 0 1 3 1 2 -1§ 8 3 R3 + R2 0 6 0 4 0 1 1 0 0 0 0 0 1 1 0 0 0 1 0 Reduced row- echelon form
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