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solve using matrices
x+3y-3z=12
3x-y+4z=0
-x+2y-z=1

Put in matrix form (the 4th column is the RHS):
1 3 -3 12
3 -1 4 0
-1 2 -1 1
proceed to transform to echelon form:
1 3 -3 12
0 10 -13 36 (3R1-R2)
0 5 -4 13 (R1+R3)

1 3 -3 12
0 10 -13 36
0 0 5 -10 (2R3-R2)

1 3 -3 12
0 10 -13 36
0 0 1 -2 (R3/5)

Back substitute
1 3 0 6 (3R3+R1)
0 1 0 1 (13R3+R2)/10
0 0 1 -2

1 0 0 3 (R1-3R2)
0 1 0 1
0 0 1 -2

So x=3, y=1, z=-2

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