Multiply the first equation by -1 and rearrange the system.
2x - y = 1 |*(-1)
3x - y + z = -2
2x - y + z = 3
becomes
3x - y + z = -2
-2x + y = -1
2x - y + z = 3
Then you solve the system like the you solved the first one.
Let A =
1 0 -1
2 1 -2
0 1 1
find A^-1 of the above matrix using inverse matrices and hence solve the bottom 2 systems:
x - z = 3
2x + y - 2z = 0
y + z = 15
2x - y = 1
3x - y + z = -2
2x - y + z = 3
i got A^-1 =
3 -1 1
-2 1 0
2 -1 1
I solved for system 1, how do I solve for system 2 can you please demonstrate.