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

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Sure! To solve the system of equations using matrices and Gaussian elimination with back-substitution, we need to represent the system of equations as an augmented matrix.

The augmented matrix for the given system of equations is:

[1 1 1 | -5]
[1 -1 3 | -1]
[4 1 1 | -2]

We can start by performing Gaussian elimination on this augmented matrix to transform it into row-echelon form. The goal is to turn the matrix into an upper triangular form.

Step 1: Swap R1 and R2
[1 -1 3 | -1]
[1 1 1 | -5]
[4 1 1 | -2]

Step 2: Subtract R1 from R2
[1 -1 3 | -1]
[0 2 -2 | -4]
[4 1 1 | -2]

Step 3: Subtract 4 times R1 from R3
[1 -1 3 | -1]
[0 2 -2 | -4]
[0 5 -11 | 2]

Step 4: Subtract 2.5 times R2 from R3
[1 -1 3 | -1]
[0 2 -2 | -4]
[0 0 -6.5 | 12]

Step 5: Divide R3 by -6.5
[1 -1 3 | -1]
[0 2 -2 | -4]
[0 0 1 | -12/6.5]

Step 6: Subtract 3 times R3 from R1
[1 -1 0 | -1 + 12/6.5]
[0 2 -2 | -4]
[0 0 1 | -12/6.5]

Simplifying:
[1 -1 0 | -2/6.5]
[0 2 -2 | -4]
[0 0 1 | -24/13]

Step 7: Subtract -2 times R3 from R2
[1 -1 0 | -2/6.5]
[0 2 0 | -4 + 48/13]
[0 0 1 | -24/13]

Simplifying:
[1 -1 0 | -2/6.5]
[0 2 0 | -20/13]
[0 0 1 | -24/13]

Step 8: Divide R2 by 2
[1 -1 0 | -2/6.5]
[0 1 0 | -10/13]
[0 0 1 | -24/13]

Step 9: Subtract -1 times R2 from R1
[1 0 0 | -2/6.5 + 10/13]
[0 1 0 | -10/13]
[0 0 1 | -24/13]

Simplifying:
[1 0 0 | 14/13]
[0 1 0 | -10/13]
[0 0 1 | -24/13]

Therefore, the solution to the system of equations is x = 14/13, y = -10/13, and z = -24/13.