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