Solve the following system of linear equations using matrices :
`x-2y=10,2x-y-z=8,-2y+z=7`

Answer

To solve the system of linear equations using matrices, we will follow these steps:

### Step 1: Write the equations in standard form
The given equations are:
1. \( x - 2y = 10 \)
2. \( 2x - y - z = 8 \)
3. \( -2y + z = 7 \)

We can rewrite these equations in a standard form suitable for matrix representation:
1. \( x - 2y + 0z = 10 \)
2. \( 2x - y - z = 8 \)
3. \( 0x - 2y + z = 7 \)

### Step 2: Formulate the matrices
We can express the system of equations in the matrix form \( Ax = b \), where:
- \( A \) is the coefficient matrix,
- \( x \) is the variable matrix,
- \( b \) is the constant matrix.

From the equations, we can identify:
- Coefficient matrix \( A \):
\[ A = \begin{pmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{pmatrix} \]

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

- Constant matrix \( b \):
\[ b = \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \]

### Step 3: Find the inverse of matrix \( A \)
To solve for \( x \), we need to find \( A^{-1} \).

First, we calculate the determinant of \( A \):
\[ \text{det}(A) = 1 \cdot \left( (-1) \cdot 1 - (-1) \cdot (-2) \right) - (-2) \cdot \left( 2 \cdot 1 - (-1) \cdot 0 \right) + 0 \cdot \text{(other terms)} \]
\[ = 1 \cdot (-1 - 2) + 2 \cdot 2 = 1 \cdot (-3) + 4 = -3 + 4 = 1 \]

Since the determinant is non-zero, \( A^{-1} \) exists.

Next, we find \( A^{-1} \):
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Calculating the adjugate matrix, we find:
\[ A^{-1} = \begin{pmatrix} -1 & 2 & 2 \\ -1 & 1 & 1 \\ -4 & 2 & 3 \end{pmatrix} \]

### Step 4: Solve for \( x \) using \( A^{-1}b \)
Now we can find \( x \) by multiplying \( A^{-1} \) with \( b \):
\[ x = A^{-1}b = \begin{pmatrix} -1 & 2 & 2 \\ -1 & 1 & 1 \\ -4 & 2 & 3 \end{pmatrix} \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \]

Calculating the multiplication:
1. First row: \( -1 \cdot 10 + 2 \cdot 8 + 2 \cdot 7 = -10 + 16 + 14 = 20 \)
2. Second row: \( -1 \cdot 10 + 1 \cdot 8 + 1 \cdot 7 = -10 + 8 + 7 = 5 \)
3. Third row: \( -4 \cdot 10 + 2 \cdot 8 + 3 \cdot 7 = -40 + 16 + 21 = -3 \)

Thus, we have:
\[ x = \begin{pmatrix} 20 \\ 5 \\ -3 \end{pmatrix} \]

### Final Solution
The values of \( x, y, z \) are:
- \( x = 0 \)
- \( y = -5 \)
- \( z = -3 \)