The given system of equations is:

x + y + z = 6

x - y - z = -4

x + 2y - 2z = -1



We can write the system of equations in matrix form as:

| 1 1 1 | | x | | 6 |

| 1 -1 -1 | x | y | = |-4 |

| 1 2 -2 | | z | |-1 |

Let A be the coefficient matrix and X be the matrix of variables. Then, the system of equations can be written as AX = B, where B is the matrix of constants.

| 1 1 1 | | x | | 6 |

| 1 -1 -1 | x | y | = |-4 |

| 1 2 -2 | | z | |-1 |

| A | x | X | = | B |

We can solve for X by multiplying both sides by the inverse of A.

A^-1 (AX) = A^-1 B

X = A^-1 B



To find the inverse of A, we first need to find the determinant of A.

| 1 1 1 |

| 1 -1 -1 |

| 1 2 -2 |

det(A) = 1(-1)(2) + 1(-1)(-1) + 1(1)(-2) = -1

Since the determinant of A is not equal to zero, we can find the inverse of A as follows:

A^-1 = adj(A)/det(A)

where adj(A) is the adjugate (transpose of the cofactor matrix) of A.

| -4 3 -1 |

| -2 -1 1 |

| 3 -2 0 |

adj(A) = | C11 C12 C13 |

C11 = (-1)^(1+1) * |-1 1|

C12 = (-1)^(1+2) * | 1 -1|

C13 = (-1)^(1+3) * |-1 -1|

C21 = (-1)^(2+1) * | 1 -1|

C22 = (-1)^(2+2) * |-1 -2|

C23 = (-1)^(2+3) * | 1 1|

C31 = (-1)^(3+1) * |-1 -1|

C32 = (-1)^(3+2) * |-1 1|

C33 = (-1)^(3+3) * | 1 -1|

C11 = -1, C12 = -2, C13 = 0, C21 = 1, C22 = 1, C23 = -2, C31 = -1, C32 = 2, C33 = 1

adj(A) = |-4 3 -1|

| -2 -1 1|

| 3 -2 0|

A^-1 = adj(A)/det(A) = |-4 3 -1