Here's one exactly like it. Just put your numbers instead:
2x + y - z = 2
x + y + z = 2
x + 3y + 2z = 1
The object is to get the matrix like this:
where there will be 1's on the diagonal, 0's underneath the
diagonal, and numbers where the @'s are.
Start with:
Swap rows 1 and 2, because it's easier when there's a 1
in the upper left corner.
R1 <-> R2
Multiply row 1 by -2
-2R1->R1
Add Row 1 and Row 2 and put that in place of Row 2
R1+R2->R2
Restore Row 1 both
R1/(-2) -> R1
Multiply Row 1 by -1
Add row 1 to Row 3
Restore row 1
-R1 <-> R1
Multiply row 2 by -1
-R2 <-> R2
Multiply Row by -2
Add row 2 and Row 3 and put that sum in row 3
R2+R3 -> R3
Restore Row 2
R2/(-2) -> R2
Divide row 3 by -5
R3/(-5)( -> R3
Rewrite as a system of equations:
1x + 1y + 1z = 2
0x + 1y + 3z = 2
0x + 0y + 1z = 1
Remove the understood 1's and 0's
x + y + z = 2
y + 3z = 2
z = 1
The bottom equation of the system is already solved for z
z = 1
Substitute in the middle equation of the system for y:
y + 3z = 2
y + 3(1) = 2
y + 3 = 2
y = -1
Substitute in the top equation of the system for z
x + y + z = 2
x + (-1) + (1) = 2
x = 2
(x,y,z) = (2,-1,1)
Edwin
