Posted by **Anonymous** on Wednesday, December 7, 2011 at 10:18pm.

solve using matrices

x+3y-3z=12

3x-y+4z=0

-x+2y-z=1

- Urgent please help! matrices -
**MathMate**, Wednesday, December 7, 2011 at 10:53pm
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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