Using matrix method, solve the system of equations:
` {:( x+2y+z=1 ),( x-y -z=2 ),( 2x+3y+z=1):}`

Answer

To solve the system of equations using the matrix method, we will follow these steps:

Given equations:
1. \( x + 2y + z = 1 \) (Equation 1)
2. \( x - y - z = 2 \) (Equation 2)
3. \( 2x + 3y + z = 1 \) (Equation 3)

### Step 1: Write the system in matrix form
We can express the system of equations in the form \( AX = B \), where:
- \( A \) is the coefficient matrix,
- \( X \) is the variable matrix,
- \( B \) is the constant matrix.

The coefficient matrix \( A \) and the constant matrix \( B \) are given by:
\[ A = \begin{pmatrix} 1 & 2 & 1 \\ 1 & -1 & -1 \\ 2 & 3 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \]

### Step 2: Find the determinant of matrix \( A \)
To find the inverse of matrix \( A \), we first need to calculate its determinant \( |A| \):
\[ |A| = 1 \cdot \begin{vmatrix} -1 & -1 \\ 3 & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & -1 \\ 3 & 1 \end{vmatrix} = (-1)(1) - (-1)(3) = -1 + 3 = 2 \)
2. \( \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-1)(2) = 1 + 2 = 3 \)
3. \( \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (-1)(2) = 3 + 2 = 5 \)

Now substituting back:
\[ |A| = 1 \cdot 2 - 2 \cdot 3 + 1 \cdot 5 = 2 - 6 + 5 = 1 \]

### Step 3: Find the adjoint of matrix \( A \)
Next, we need to find the adjoint of \( A \) by calculating the cofactors and then transposing the cofactor matrix.

The cofactor matrix \( C \) is calculated as follows:
\[ C_{11} = 2, \quad C_{12} = -3, \quad C_{13} = 5 \\ C_{21} = -3, \quad C_{22} = 1, \quad C_{23} = 1 \\ C_{31} = -1, \quad C_{32} = 2, \quad C_{33} = -3 \]
Thus, the cofactor matrix is:
\[ C = \begin{pmatrix} 2 & -3 & 5 \\ -3 & 1 & 1 \\ -1 & 2 & -3 \end{pmatrix} \]

Now, taking the transpose of the cofactor matrix gives us the adjoint \( \text{adj}(A) \):
\[ \text{adj}(A) = \begin{pmatrix} 2 & -3 & -1 \\ -3 & 1 & 2 \\ 5 & 1 & -3 \end{pmatrix} \]

### Step 4: Find the inverse of matrix \( A \)
The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) = \text{adj}(A) \quad \text{(since } |A| = 1\text{)} \]
Thus,
\[ A^{-1} = \begin{pmatrix} 2 & -3 & -1 \\ -3 & 1 & 2 \\ 5 & 1 & -3 \end{pmatrix} \]

### Step 5: Multiply \( A^{-1} \) by \( B \) to find \( X \)
Now we compute \( X = A^{-1}B \):
\[ X = \begin{pmatrix} 2 & -3 & -1 \\ -3 & 1 & 2 \\ 5 & 1 & -3 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \]

Calculating the product:
1. First row: \( 2 \cdot 1 + (-3) \cdot 2 + (-1) \cdot 1 = 2 - 6 - 1 = -5 \)
2. Second row: \( -3 \cdot 1 + 1 \cdot 2 + 2 \cdot 1 = -3 + 2 + 2 = 1 \)
3. Third row: \( 5 \cdot 1 + 1 \cdot 2 + (-3) \cdot 1 = 5 + 2 - 3 = 4 \)

Thus, we have:
\[ X = \begin{pmatrix} -5 \\ 1 \\ 4 \end{pmatrix} \]

### Final Solution
The values of \( x, y, z \) are:
\[ x = -5, \quad y = 1, \quad z = 4 \]