SOLUTION: Solve System of Equation using Matrix notation (Eq.1) 2x-y+3z = 4 (Eq.2) 3x+2y+z = -1 (Eq.3) -x+3y+2z = 5

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Question 662424: Solve System of Equation using Matrix notation
(Eq.1) 2x-y+3z = 4
(Eq.2) 3x+2y+z = -1
(Eq.3) -x+3y+2z = 5
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
2	-1	3	4
3	2	1	-1
-1	3	2	5



1	-0.5	1.5	2	0.5*R1
3	2	1	-1
-1	3	2	5



1	-0.5	1.5	2
0	3.5	-3.5	-7	R2 - 3*R1
-1	3	2	5



1	-0.5	1.5	2
0	3.5	-3.5	-7
0	2.5	3.5	7	R3 + R1



1	-0.5	1.5	2
0	1	-1	-2	0.285714*R2
0	2.5	3.5	7



1	-0.5	1.5	2
0	1	-1	-2
0	0	6	12	R3 - 2.5*R2



1	-0.5	1.5	2
0	1	-1	-2
0	0	1	2	(1/6)*R3

The Matrix is now in reduced echelon form (ref) and it is upper triangular 
1	-0.5	1.5	2
0	1	0	0	R2 + R3
0	0	1	2



1	-0.5	0	-1	R1 - 1.5*R3
0	1	0	0
0	0	1	2



1	0	0	-1	R1 + 0.5*R2
0	1	0	0
0	0	1	2

The Matrix is now in reduced row echelon form (rref)

So the answers are x = -1, y = 0 and z = 2

The answer as an ordered triple is (-1, 0, 2)