2x+y-z=1
x-y+z=2
3x+y-2z=-1

Answer

To solve the system of linear equations given by:

1. \( 2x + y - z = 1 \)
2. \( x - y + z = 2 \)
3. \( 3x + y - 2z = -1 \)

we will use the matrix method.

### Step 1: Formulate the matrices

We start by identifying the coefficient matrix \( A \), the variable matrix \( x \), and the constant matrix \( B \).

- Coefficient matrix \( A \):
\[ A = \begin{bmatrix} 2 & 1 & -1 \\ 1 & -1 & 1 \\ 3 & 1 & -2 \end{bmatrix} \]

- Variable matrix \( x \):
\[ x = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

- Constant matrix \( B \):
\[ B = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \]

### Step 2: Find the inverse of matrix \( A \)

To find \( x \), we need to compute \( A^{-1} \) and then multiply it by \( B \). The formula for the inverse of a matrix is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

#### Step 2.1: Calculate the determinant of \( A \)

We calculate the determinant of \( A \) using the formula for a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \).

For our matrix:
\[ \text{det}(A) = 2((-1)(-2) - (1)(1)) - 1(1(-2) - (1)(3)) - (-1)(1(1) - (-1)(3)) \]
Calculating this gives:
\[ = 2(2 - 1) - 1(-2 - 3) + 1(1 + 3) = 2(1) + 5 + 4 = 11 \]

#### Step 2.2: Calculate the adjugate of \( A \)

Next, we find the adjugate of \( A \), which involves calculating the cofactor matrix and then transposing it.

The cofactor matrix \( C \) is calculated as follows:
- \( C_{11} = \text{det} \begin{bmatrix} -1 & 1 \\ 1 & -2 \end{bmatrix} = (-1)(-2) - (1)(1) = 2 - 1 = 1 \)
- \( C_{12} = -\text{det} \begin{bmatrix} 1 & 1 \\ 3 & -2 \end{bmatrix} = -((1)(-2) - (1)(3)) = -(-2 - 3) = 5 \)
- \( C_{13} = \text{det} \begin{bmatrix} 1 & -1 \\ 3 & 1 \end{bmatrix} = (1)(1) - (-1)(3) = 1 + 3 = 4 \)

Continuing this for all elements:
\[ C = \begin{bmatrix} 1 & 5 & 4 \\ -1 & 1 & 0 \\ -3 & -3 & 1 \end{bmatrix} \]

Now, we take the transpose of the cofactor matrix to get the adjugate:
\[ \text{adj}(A) = \begin{bmatrix} 1 & -1 & -3 \\ 5 & 1 & -3 \\ 4 & 0 & 1 \end{bmatrix} \]

### Step 3: Calculate \( A^{-1} \)

Now we can find \( A^{-1} \):
\[ A^{-1} = \frac{1}{11} \begin{bmatrix} 1 & -1 & -3 \\ 5 & 1 & -3 \\ 4 & 0 & 1 \end{bmatrix} \]

### Step 4: Multiply \( A^{-1} \) by \( B \)

Now we compute \( x = A^{-1}B \):
\[ x = \frac{1}{11} \begin{bmatrix} 1 & -1 & -3 \\ 5 & 1 & -3 \\ 4 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \]

Calculating the multiplication:
\[ = \frac{1}{11} \begin{bmatrix} 1 \cdot 1 + (-1) \cdot 2 + (-3) \cdot (-1) \\ 5 \cdot 1 + 1 \cdot 2 + (-3) \cdot (-1) \\ 4 \cdot 1 + 0 \cdot 2 + 1 \cdot (-1) \end{bmatrix} \]
\[ = \frac{1}{11} \begin{bmatrix} 1 - 2 + 3 \\ 5 + 2 + 3 \\ 4 + 0 - 1 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 2 \\ 10 \\ 3 \end{bmatrix} \]

### Step 5: Final values of \( x, y, z \)

Thus, we have:
\[ x = \frac{2}{11}, \quad y = \frac{10}{11}, \quad z = \frac{3}{11} \]

### Summary of the solution:
The solution to the system of equations is:
\[ x = \frac{2}{11}, \quad y = \frac{10}{11}, \quad z = \frac{3}{11} \]