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

Answer

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

### Given Equations:
1. \( x - 2y + 0z = 10 \)
2. \( 2x - y - z = 8 \)
3. \( 0x - 2y + z = 7 \)

### Step 1: Write the equations in matrix form
We can express the system of equations in the form \( Ax = B \), where:

\[ A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}, \quad x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix A
To find the inverse of matrix \( A \), we first need to calculate its determinant \( |A| \).

\[ |A| = 1 \cdot \left| \begin{bmatrix} -1 & -1 \\ -2 & 1 \end{bmatrix} \right| - (-2) \cdot \left| \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} \right| + 0 \cdot \left| \begin{bmatrix} 2 & -1 \\ 0 & -2 \end{bmatrix} \right| \]

Calculating the minors:

1. \( \left| \begin{bmatrix} -1 & -1 \\ -2 & 1 \end{bmatrix} \right| = (-1)(1) - (-1)(-2) = -1 - 2 = -3 \)

2. \( \left| \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} \right| = (2)(1) - (-1)(0) = 2 - 0 = 2 \)

Now substituting back:

\[ |A| = 1 \cdot (-3) + 2 \cdot 2 + 0 = -3 + 4 = 1 \]

### Step 3: Calculate the adjoint of matrix A
The adjoint of \( A \) is the transpose of the cofactor matrix. We will calculate the cofactors:

1. \( C_{11} = -3 \) (as calculated above)
2. \( C_{12} = 2 \) (from the second minor calculation)
3. \( C_{13} = -4 \) (similar calculation)
4. \( C_{21} = 2 \)
5. \( C_{22} = 1 \)
6. \( C_{23} = 2 \)
7. \( C_{31} = 2 \)
8. \( C_{32} = 1 \)
9. \( C_{33} = -1 \)

Thus, the cofactor matrix is:

\[ C = \begin{bmatrix} -3 & 2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & -1 \end{bmatrix} \]

Taking the transpose gives us the adjoint:

\[ \text{adj}(A) = \begin{bmatrix} -3 & 2 & 2 \\ 2 & 1 & 1 \\ -4 & 2 & -1 \end{bmatrix} \]

### Step 4: Calculate the inverse of matrix A
The inverse of matrix \( A \) is given by:

\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) = 1 \cdot \begin{bmatrix} -3 & 2 & 2 \\ 2 & 1 & 1 \\ -4 & 2 & -1 \end{bmatrix} \]

### Step 5: Solve for x
Now, we can find \( x \) using:

\[ x = A^{-1}B \]

Calculating \( A^{-1}B \):

\[ x = \begin{bmatrix} -3 & 2 & 2 \\ 2 & 1 & 1 \\ -4 & 2 & -1 \end{bmatrix} \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \]

Calculating each component:

1. First row: \( -3(10) + 2(8) + 2(7) = -30 + 16 + 14 = 0 \)
2. Second row: \( 2(10) + 1(8) + 1(7) = 20 + 8 + 7 = 35 \)
3. Third row: \( -4(10) + 2(8) - 1(7) = -40 + 16 - 7 = -31 \)

Thus, we have:

\[ x = \begin{bmatrix} 0 \\ 35 \\ -31 \end{bmatrix} \]

### Final Result
The solution to the system of equations is:

\[ x = 0, \quad y = 35, \quad z = -31 \]