Using matrix method, solve the following system of equation
`x-2y=10,2x+y+3z=8` and `-2y+z=7`

Answer

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

### Step 1: Write the system of equations in matrix form

The given equations are:
1. \( x - 2y = 10 \)
2. \( 2x + y + 3z = 8 \)
3. \( -2y + z = 7 \)

We can express this in the matrix form \( AX = B \), where:
- \( A \) is the coefficient matrix,
- \( X \) is the variable matrix,
- \( B \) is the constant matrix.

Thus, we have:
\[ A = \begin{pmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \]

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

To find the determinant of \( A \):
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 2 & 3 \\ 0 & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 2 & 1 \\ 0 & -2 \end{vmatrix} \]

Calculating the 2x2 determinants:
\[ \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = (1)(1) - (3)(-2) = 1 + 6 = 7 \]
\[ \begin{vmatrix} 2 & 3 \\ 0 & 1 \end{vmatrix} = (2)(1) - (3)(0) = 2 \]

Thus,
\[ \text{det}(A) = 1 \cdot 7 + 2 \cdot 2 = 7 + 4 = 11 \]

### Step 3: Find the inverse of matrix \( A \)

Since \( \text{det}(A) \neq 0 \), the inverse of \( A \) exists. The inverse can be found using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

To find \( \text{adj}(A) \), we need to find the cofactor matrix and then transpose it.

#### Step 3.1: Find the cofactor matrix

The cofactor matrix is calculated by finding the determinants of the minors:
- \( C_{11} = \begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} = 7 \)
- \( C_{12} = -\begin{vmatrix} 2 & 3 \\ 0 & 1 \end{vmatrix} = -2 \)
- \( C_{13} = \begin{vmatrix} 2 & 1 \\ 0 & -2 \end{vmatrix} = 4 \)

Continuing this process for all elements, we find:
\[ \text{cofactor matrix} = \begin{pmatrix} 7 & -2 & 4 \\ 2 & 1 & 2 \\ -6 & -3 & 5 \end{pmatrix} \]

#### Step 3.2: Transpose the cofactor matrix

Now, we transpose the cofactor matrix to find the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ 4 & 2 & 5 \end{pmatrix} \]

#### Step 3.3: Calculate \( A^{-1} \)

Now we can find \( A^{-1} \):
\[ A^{-1} = \frac{1}{11} \cdot \begin{pmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ 4 & 2 & 5 \end{pmatrix} \]

### Step 4: Solve for \( X \)

Using the equation \( AX = B \), we can solve for \( X \):
\[ X = A^{-1}B \]

Calculating \( A^{-1}B \):
\[ X = \frac{1}{11} \begin{pmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ 4 & 2 & 5 \end{pmatrix} \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \]

Calculating the product:
\[ = \frac{1}{11} \begin{pmatrix} (7 \cdot 10) + (2 \cdot 8) + (-6 \cdot 7) \\ (-2 \cdot 10) + (1 \cdot 8) + (-3 \cdot 7) \\ (4 \cdot 10) + (2 \cdot 8) + (5 \cdot 7) \end{pmatrix} \]
\[ = \frac{1}{11} \begin{pmatrix} 70 + 16 - 42 \\ -20 + 8 - 21 \\ 40 + 16 + 35 \end{pmatrix} \]
\[ = \frac{1}{11} \begin{pmatrix} 44 \\ -33 \\ 91 \end{pmatrix} \]

Thus,
\[ X = \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix} \]

### Final Solution

The solution to the system of equations is:
\[ x = 4, \quad y = -3, \quad z = 1 \]