SOLUTION: Use the row echelon method to solve the system of three equations in three unknowns. x - y + 2z = -2 5x + z = 0 x + 2y+z = 4

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Question 257807: Use the row echelon method to solve the system of three equations in three unknowns.
x - y + 2z = -2
5x + z = 0
x + 2y+z = 4
Answer by drk(1908) About Me  (Show Source):
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We are given the following:
x - y + 2z = -2
5x + z = 0
x + 2y+z = 4
--
in row echelon form, we have:
1 . . . . . (-1) . . . . . . .2 . . . . . . .(-2)
5 . . . . . .0 . . . . . . . .1 . . . . . . .0
1 . . . . . . 2 . . . . . . . 1 . . . . . .. 4
We need to get a diagonal matrix so take row 2 - 5 row 1 -> row 2; row 3 - row 1 -> row 3
--
1 . . . . . (-1) . . . . . . .2 . . . . . . .(-2)
0 . . . .. . 5 . . . . . . . .(-9) . . .. . . 10
0 . . . .. . 3 . . . . . .. . . -1 . . . . . . . 6
(1/5) row 2 -> row 2
--
1 . . . . . (-1) . . . . . . .2 . . . . . . .(-2)
0 . . . . . . 1 . . . . . . . .(-9/5) . . 2
0 . . . . . . 1 . . . . . . . .3 . . . . . . . 6
row 1 + row 2 -> row 1; row 3 - row 2 -> row 3
--
1 . . . . . . 0. . . . . . . .(1/5) . . . . 0
0 . . . . . . 1 . . . . . . . .(-9/5) . . . 2
0 . . . . . . 0 . . . . . . . .(22/5) . . . .0
(5/24) row 3 -> row 3
--
1 . . . . . . 0. . . . . . . .(1/5) . . . . 0
0 . . . . . . 1 . . . . . . . .(-9/5) . . . 2
0 . . . . . . 0 . . . . . . . .1 . . . . . . . 0
So,
z = 0
y = 2
x = 0
{0, 2, 0}