SOLUTION: Solve the system of equations using matrices. Use Gaussian elimination with back-substitution. x + y + z = -5 x - y + 3z = -1 4x + y + z = -2

Answer.  x = 1       
         y = -4     
         z = -2     

Your matrix

	X1	X2	X3	b
1	1	1	1	-5
2	1	-1	3	-1
3	4	1	1	-2

Find the pivot in the 1st column in the 1st row

	X1	X2	X3	b
1	1	1	1	-5
2	1	-1	3	-1
3	4	1	1	-2

Subtract the 1st row from the 2nd row

	X1	X2	X3	b
1	1	1	1	-5
2	0	-2	2	4
3	4	1	1	-2

Multiply the 1st row by 4

	X1	X2	X3	b
1	4	4	4	-20
2	0	-2	2	4
3	4	1	1	-2

Subtract the 1st row from the 3rd row and restore it

	X1	X2	X3	b
1	1	1	1	-5
2	0	-2	2	4
3	0	-3	-3	18

Make the pivot in the 2nd column by dividing the 2nd row by -2

	X1	X2	X3	b
1	1	1	1	-5
2	0	1	-1	-2
3	0	-3	-3	18

Subtract the 2nd row from the 1st row

X1	X2	X3	b
1	1	0	2	-3
2	0	1	-1	-2
3	0	-3	-3	18

Multiply the 2nd row by -3

	X1	X2	X3	b
1	1	0	2	-3
2	0	-3	3	6
3	0	-3	-3	18

Subtract the 2nd row from the 3rd row and restore it

	X1	X2	X3	b
1	1	0	2	-3
2	0	1	-1	-2
3	0	0	-6	12

Make the pivot in the 3rd column by dividing the 3rd row by -6

	X1	X2	X3	b
1	1	0	2	-3
2	0	1	-1	-2
3	0	0	1	-2

Multiply the 3rd row by 2

	X1	X2	X3	b
1	1	0	2	-3
2	0	1	-1	-2
3	0	0	2	-4

Subtract the 3rd row from the 1st row and restore it

	X1	X2	X3	b
1	1	0	0	1
2	0	1	-1	-2
3	0	0	1	-2

Multiply the 3rd row by -1

	X1	X2	X3	b
1	1	0	0	1
2	0	1	-1	-2
3	0	0	-1	2

Subtract the 3rd row from the 2nd row and restore it

	X1	X2	X3	b
1	1	0	0	1
2	0	1	0	-4
3	0	0	1	-2

Solution set:

x1 = 1      
x2 = -4   
x3 = -2