Find below a system of linear equations.
4x - y + 2z = 13
x + 2y - 2z = 0
-x + y + z = 5
Solve the system of linear equations using the Gauss Jordan elimination method to produce a reduced row-echelon form.
1. Perform row operations to produce zeros below the diagonal in the first column:
- R2 -> R2 - 1*R1
- R3 -> R3 + 1*R1
This results in the following:
4x - y + 2z = 13
-3y - 4z = -13
3z = 18
2. Solve for z:
z = 6
3. Substitute z back into the third equation and solve for y:
-x + y + 6 = 5
y = 5 + x - 6
y = x - 1
4. Substitute z and y back into the first equation and solve for x:
4x - (x-1) + 2(6) = 13
4x - x + 1 + 12 = 13
3x = 0
x = 0
5. Substitute x back into the second equation and solve for y:
0 + 2y - 2(6) = 0
2y - 12 = 0
2y = 12
y = 6
Therefore, the solution to the system of linear equations is:
x = 0
y = 6
z = 6