4x + y + 2z = 13
x + 2y - 2z = 0
-x +y + z = 5
Solve the system of linear equation using the Gauss Jordan elimination method to produce a reduced row-echelon form.
1) [4 1 2 | 13]
[1 2 -2 | 0]
[-1 1 1 | 5]
2) Swap row 1 and row 2
[1 2 -2 | 0]
[4 1 2 | 13]
[-1 1 1 | 5]
3) Add -4*row1 to row2
[1 2 -2 | 0]
[0 -7 10 | 13]
[-1 1 1 | 5]
4) Add row1 to row3
[1 2 -2 | 0]
[0 -7 10 | 13]
[0 3 -1 | 5]
5) Multiply row2 by -1/7
[1 2 -2 | 0]
[0 1 -10/7 | -13/7]
[0 3 -1 | 5]
6) Add -2*row2 to row1
[1 0 16/7 | 26/7]
[0 1 -10/7 | -13/7]
[0 3 -1 | 5]
7) Add -3*row2 to row3
[1 0 16/7 | 26/7]
[0 1 -10/7 | -13/7]
[0 0 17/7 | 32/7]
8) Multiply row3 by 7/17
[1 0 16/7 | 26/7]
[0 1 -10/7 | -13/7]
[0 0 1 | 32/17]
The reduced row-echelon form of the system is:
x = 16/7
y = -10/7
z = 32/17