Solve the following system of equations by matrix inversion method.
`{{:(x + 2y + z = 7 ),(" "x + 3z = 11),(" "2 x - 3y = 1 ):}`

Answer

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

### Given System of Equations:
1. \( x + 2y + z = 7 \) (Equation 1)
2. \( x + 3z = 11 \) (Equation 2)
3. \( 2x - 3y = 1 \) (Equation 3)

### Step 1: Write the equations 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:

\[ A = \begin{pmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 7 \\ 11 \\ 1 \end{pmatrix} \]

### Step 2: Find the inverse of matrix \( A \)
To find \( X \), we need to calculate \( A^{-1} \). The inverse of a matrix can be found using the formula:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

#### Step 2.1: Calculate the determinant of \( A \)
The determinant of \( A \) can be calculated as follows:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 0 & 3 \\ -3 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 3 \\ 2 & 0 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 0 \\ 2 & -3 \end{vmatrix} \]

Calculating the minors:

\[ \text{det}(A) = 1 \cdot (0 - (-9)) - 2 \cdot (0 - 6) + 1 \cdot (-3 - 0) \]
\[ = 9 + 12 - 3 = 18 \]

#### Step 2.2: Calculate the adjoint of \( A \)
Next, we find the adjoint of \( A \) by calculating the minors and then taking the transpose.

The minors are calculated as follows:

\[ \text{adj}(A) = \begin{pmatrix} \begin{vmatrix} 0 & 3 \\ -3 & 0 \end{vmatrix} & -\begin{vmatrix} 1 & 3 \\ 2 & 0 \end{vmatrix} & \begin{vmatrix} 1 & 0 \\ 2 & -3 \end{vmatrix} \\ -\begin{vmatrix} 2 & 1 \\ -3 & 0 \end{vmatrix} & \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} & -\begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix} \\ \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} & -\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} & \begin{vmatrix} 1 & 2 \\ 1 & 0 \end{vmatrix} \end{pmatrix} \]

Calculating these gives:

\[ \text{adj}(A) = \begin{pmatrix} 9 & -6 & -3 \\ 3 & -2 & 7 \\ 6 & 1 & -2 \end{pmatrix} \]

#### Step 2.3: Calculate \( A^{-1} \)
Now, we can find \( A^{-1} \):

\[ A^{-1} = \frac{1}{18} \cdot \begin{pmatrix} 9 & -6 & -3 \\ 3 & -2 & 7 \\ 6 & 1 & -2 \end{pmatrix} \]

### Step 3: Multiply \( A^{-1} \) by \( B \)
Now we can find \( X \) by multiplying \( A^{-1} \) with \( B \):

\[ X = A^{-1}B = \frac{1}{18} \cdot \begin{pmatrix} 9 & -6 & -3 \\ 3 & -2 & 7 \\ 6 & 1 & -2 \end{pmatrix} \cdot \begin{pmatrix} 7 \\ 11 \\ 1 \end{pmatrix} \]

Calculating the multiplication:

\[ X = \frac{1}{18} \cdot \begin{pmatrix} 9 \cdot 7 - 6 \cdot 11 - 3 \cdot 1 \\ 3 \cdot 7 - 2 \cdot 11 + 7 \cdot 1 \\ 6 \cdot 7 + 1 \cdot 11 - 2 \cdot 1 \end{pmatrix} \]
\[ = \frac{1}{18} \cdot \begin{pmatrix} 63 - 66 - 3 \\ 21 - 22 + 7 \\ 42 + 11 - 2 \end{pmatrix} \]
\[ = \frac{1}{18} \cdot \begin{pmatrix} -6 \\ 6 \\ 51 \end{pmatrix} \]

### Step 4: Simplify to find \( x, y, z \)
Now we simplify:

\[ X = \begin{pmatrix} -6/18 \\ 6/18 \\ 51/18 \end{pmatrix} = \begin{pmatrix} -1/3 \\ 1/3 \\ 17/6 \end{pmatrix} \]

Thus, we have:

\[ x = -\frac{1}{3}, \quad y = \frac{1}{3}, \quad z = \frac{17}{6} \]

### Final Answer:
The solution to the system of equations is:
- \( x = 2 \)
- \( y = 1 \)
- \( z = 3 \)