2y - 3z = 2 2x + z = 3 x - y + 3z = 1. +- -+ | 0 2 -3 | 2 | write the system as an augmented matrix | 2 0 1 | 3 | | 1 -1 3 | 1 | +- -+ +- -+ | 1 -1 3 | 1 | interchange first and third row | 2 0 1 | 3 | (to make top left entry non-zero) | 0 2 -3 | 2 | +- -+ +- -+ | 1 -1 3 | 1 | Add -2 times first row to second row | 0 2 -5 | -1 | (to get 0 in first column of row 2) | 0 2 -3 | 2 | +- -+ +- -+ | 1 -1 3 | 1 | Divide second row by 2 | 0 1 -5/2 | -1/2 | (to get a leading 1 in row 2) | 0 2 -3 | 2 | +- -+ +- -+ | 1 0 1/2 | 3/2 | Add second row to first | 0 1 -5/2 | -1/2 | Add -2 times second row to third | 0 0 2 | 1 | (to get 0's in the second column) +- -+ +- -+ | 1 0 1/2 | 3/2 | Divide third row by 2 | 0 1 -5/2 | -1/2 | (to get a leading 1) | 0 0 1 | 1/2 | +- -+ +- -+ | 1 0 0 | 5/4 | Add -1/2 times third row to first row | 0 1 0 | 7/4 | Add 5/2 times third row to second row | 0 0 1 | 1/2 | (to get 0's in third column) +- -+This is now in reduced row-echelon form, so we can read off the answer:
Check that the answer satisfies the initial equations (in case we made arithmatic errors):
All of these check out, so our solution is correct.2y - 3z = 2(7/4) - 3(1/2) = 7/2 - 3/2 = 4/2 = 2
2x + z = 2(5/4) + 1/2 = 5/2 + 1/2 = 6/2 = 3
x - y + 3z = (5/4) - (7/4) + (1/2) = -(2/4) + 3/2 = -(1/2) + 3/2 = 2/2 = 1