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

To solve the system of equations using matrices and Gaussian elimination with back-substitution, first write the system of equations in matrix form:

| 1 1 1 | | x | | -5 |
| 1 -1 3 | | y | = | -1 |
| 4 1 1 | | z | | -2 |

Step 1: Row 2 = Row 2 - Row 1
| 1 1 1 | | x | | -5 |
| 0 -2 2 | | y | = | 4 |
| 4 1 1 | | z | | -2 |

Step 2: Row 3 = Row 3 - 4 * Row 1
| 1 1 1 | | x | | -5 |
| 0 -2 2 | | y | = | 4 |
| 0 -3 -3 | | z | | 18 |

Step 3: Row 3 = Row 3 - (3/2) * Row 2
| 1 1 1 | | x | | -5 |
| 0 -2 2 | | y | = | 4 |
| 0 0 -6 | | z | | 18 |

Step 4: Row 3 = (-1/6) * Row 3
| 1 1 1 | | x | | -5 |
| 0 -2 2 | | y | = | 4 |
| 0 0 1 | | z | | -3 |

Step 5: Row 2 = Row 2 + 2 * Row 3
| 1 1 1 | | x | | -5 |
| 0 -2 0 | | y | = | -2 |
| 0 0 1 | | z | | -3 |

Step 6: Row 2 = (-1/2) * Row 2
| 1 1 1 | | x | | -5 |
| 0 1 0 | | y | = | 1 |
| 0 0 1 | | z | | -3 |

Step 7: Row 1 = Row 1 - Row 2 - Row 3
| 1 0 0 | | x | | -6 |
| 0 1 0 | | y | = | 1 |
| 0 0 1 | | z | | -3 |

Therefore, the solution to the system of equations is:
x = -6
y = 1
z = -3

To solve the system of equations using matrices and Gaussian elimination with back-substitution, we will first represent the system of equations in matrix form. Let's call the coefficient matrix A, the unknown vector X, and the constant vector B.

The system of equations can be written as:

A * X = B

where A is the coefficient matrix, X is the unknown vector, and B is the constant vector.

For this system of equations:

A = [[1, 1, 1], [1, -1, 3], [4, 1, 1]]
X = [x, y, z]
B = [-5, -1, -2]

Step 1: Augmented Matrix
To perform Gaussian elimination, we represent the system of equations as an augmented matrix. Append the constant vector B to the right side of the coefficient matrix A:

[A | B] = [[1, 1, 1, -5], [1, -1, 3, -1], [4, 1, 1, -2]]

Step 2: Gaussian Elimination
Perform Gaussian elimination to transform the augmented matrix into row-echelon form or upper triangular form:

1. Row 2 - Row 1:
[R2] = [R2 - R1] = [0, -2, 2, 4]

2. Row 3 - 4 * Row 1:
[R3] = [R3 - 4 * R1] = [0, -3, -3, 18]

The augmented matrix becomes:

[[1, 1, 1, -5], [0, -2, 2, 4], [0, -3, -3, 18]]

Step 3: Back-Substitution
Now, we perform back-substitution to find the values of x, y, and z.

1. Solve for z in the last equation:
-3z = 18 => z = -6

2. Substitute the value of z back into the second equation to solve for y:
-2y + 2(-6) = 4 => y = -1

3. Substitute the values of y and z into the first equation to solve for x:
x + (-1) + (-6) = -5 => x = -4

The solution to the system of equations is x = -4, y = -1, and z = -6.