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Solve the system of linear equations using Gauss- Jordan elimination with back- substitution. Please show all work!
5x + 7y + 4z = 23
4x + 3y + 8z = -6
5x + 6y + 5z = 17
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Solution set:
x1 = 5
x2 = 2
x3 = -4
Your matrix
№ X1 X2 X3 b
1 5 7 4 23
2 4 3 8 -6
3 5 6 5 17
Make the pivot in the 1st column by dividing the 1st row by 5
№ X1 X2 X3 b
1 1 7/5 4/5 23/5
2 4 3 8 -6
3 5 6 5 17
Multiply the 1st row by 4
№ X1 X2 X3 b
1 4 28/5 16/5 92/5
2 4 3 8 -6
3 5 6 5 17
Subtract the 1st row from the 2nd row and restore it
№ X1 X2 X3 b
1 1 7/5 4/5 23/5
2 0 -13/5 24/5 -122/5
3 5 6 5 17
Multiply the 1st row by 5
№ X1 X2 X3 b
1 5 7 4 23
2 0 -13/5 24/5 -122/5
3 5 6 5 17
Subtract the 1st row from the 3rd row and restore it
№ X1 X2 X3 b
1 1 7/5 4/5 23/5
2 0 -13/5 24/5 -122/5
3 0 -1 1 -6
Find the pivot in the 2nd column (inversing the sign in the whole row) and swap the 3rd and the 2nd rows
№ X1 X2 X3 b
1 1 7/5 4/5 23/5
2 0 1 -1 6
3 0 -13/5 24/5 -122/5
Multiply the 2nd row by 7/5
№ X1 X2 X3 b
1 1 7/5 4/5 23/5
2 0 7/5 -7/5 42/5
3 0 -13/5 24/5 -122/5
Subtract the 2nd row from the 1st row and restore it
№ X1 X2 X3 b
1 1 0 11/5 -19/5
2 0 1 -1 6
3 0 -13/5 24/5 -122/5
Multiply the 2nd row by -13/5
№ X1 X2 X3 b
1 1 0 11/5 -19/5
2 0 -13/5 13/5 -78/5
3 0 -13/5 24/5 -122/5
Subtract the 2nd row from the 3rd row and restore it
№ X1 X2 X3 b
1 1 0 11/5 -19/5
2 0 1 -1 6
3 0 0 11/5 -44/5
Make the pivot in the 3rd column by dividing the 3rd row by 11/5
№ X1 X2 X3 b
1 1 0 11/5 -19/5
2 0 1 -1 6
3 0 0 1 -4
Multiply the 3rd row by 11/5
№ X1 X2 X3 b
1 1 0 11/5 -19/5
2 0 1 -1 6
3 0 0 11/5 -44/5
Subtract the 3rd row from the 1st row and restore it
№ X1 X2 X3 b
1 1 0 0 5
2 0 1 -1 6
3 0 0 1 -4
Multiply the 3rd row by -1
№ X1 X2 X3 b
1 1 0 0 5
2 0 1 -1 6
3 0 0 -1 4
Subtract the 3rd row from the 2nd row and restore it
№ X1 X2 X3 b
1 1 0 0 5
2 0 1 0 2
3 0 0 1 -4