Starting with the augmented matrix of the system of equations:
[ 2 -1 1 | -2 ]
[ 3 3 2 | 3 ]
[ 1 1 1 | 2 ]
We can perform row operations to reduce the matrix to row echelon form:
R2 - (3/2)R1 -> R2
[ 2 -1 1 | -2 ]
[ 0 15/2 1/2 | 9 ]
[ 1 1 1 | 2 ]
R3 - (1/2)R1 -> R3
[ 2 -1 1 | -2 ]
[ 0 15/2 1/2 | 9 ]
[ 0 3/2 1/2 | 3 ]
R2 - (1/10)R2 -> R2 and R3 - (1/5)R2 -> R3
[ 2 -1 1 | -2 ]
[ 0 3/2 1/2 | 9/2 ]
[ 0 0 1/3 | 3/2 ]
Finally, we can multiply R3 by 3 and rearrange the rows to obtain the reduced row echelon form:
R3 -> 3R3
[ 2 -1 1 | -2 ]
[ 0 3/2 1/2 | 9/2 ]
[ 0 0 1 | 4.5 ]
R3 -> R3 - (1/2)R2
[ 2 -1 1 | -2 ]
[ 0 3/2 1/2 | 9/2 ]
[ 0 0 1 | 4.5 ]
So the system is equivalent to:
2x - y + z = -2
3/2y + 1/2z = 9/2
z = 4.5
We can solve for y and x in terms of z:
z = 4.5
3/2y + 1/2z = 9/2
3/2y + 1/2(4.5) = 9/2
3/2y = 9/2 - 2.25
3/2y = 1.875
y = 1.25
2x - y + z = -2
2x - 1.25 + 4.5 = -2
2x = -2 - 3.25
2x = -5.25
x = -2.625
Therefore, the solution to the system is:
x = -2.625, y = 1.25, z = 4.5