Urgent please help! matrices
posted by Anonymous on .
solve using matrices
x+3y3z=12
3xy+4z=0
x+2yz=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 (3R1R2)
0 5 4 13 (R1+R3)
1 3 3 12
0 10 13 36
0 0 5 10 (2R3R2)
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 (R13R2)
0 1 0 1
0 0 1 2
So x=3, y=1, z=2