Question 395394
Here is the system of equations: 
3a+b-c=0
2a+3b-5c=1
a-2b+3c=-4 
My augmented matrix is: 
3 1 -1 0
2 3 -5 1
1 -2 3 -4 

I swapped R3 and R1->R1 
1 -2 3 -4
2 3 -5 1
3 1 -1 0 
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Add -2 times R1 to R2
Add -3 times R1 to R3
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1 -2 3 -4
0 7 -11 9
0 7 -10 12
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Subtract R2 from R3
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1 -2 3 -4
0 7 -11 9
0 0  1  3
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z = 3
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Substitute into R2 to solve for "y":
7y -11*3 = 9
7y = 42
y = 6
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Substitute into$1 to solve for "x":
x -2*6+3*3 = -4
x -12+9 = -4
x-3= -4
x = -1
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Solution: (-1,6,3)
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Cheers,
Stan H.