If `A\|(1,2,0),(-2,-1,-2),(0,-1,1)|`, then find the value of `A^(-1)`
Using `A^(-1)`, solve the system of linear equations `x-2y=10, 2xy-z=8` and `-2y+z=7`

Explanation

To solve the problem step by step, we will first find the inverse of matrix \( A \) and then use it to solve the given system of linear equations.

### Step 1: Define the matrix \( A \)

Given:
\[ A = \begin{pmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A \)

To find the inverse, we first need to check if the determinant of \( A \) is non-zero.

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

Calculating the minors:
1. \(\begin{vmatrix} -1 & -2 \\ -1 & 1 \end{vmatrix} = (-1)(1) - (-2)(-1) = -1 - 2 = -3\)

2. \(\begin{vmatrix} -2 & -2 \\ 0 & 1 \end{vmatrix} = (-2)(1) - (-2)(0) = -2\)

Now substituting back:
\[ \text{det}(A) = 1 \cdot (-3) - 2 \cdot (-2) + 0 = -3 + 4 = 1 \]

Since \(\text{det}(A) \neq 0\), the inverse exists.

### Step 3: Find the cofactors of \( A \)

We will calculate the cofactors for each element of \( A \).

- \( C_{11} = (-1)^{1+1} \begin{vmatrix} -1 & -2 \\ -1 & 1 \end{vmatrix} = 1 \cdot (-3) = -3 \)

- \( C_{12} = (-1)^{1+2} \begin{vmatrix} -2 & -2 \\ 0 & 1 \end{vmatrix} = -1 \cdot (-2) = 2 \)

- \( C_{13} = (-1)^{1+3} \begin{vmatrix} -2 & -1 \\ 0 & -1 \end{vmatrix} = 1 \cdot (2) = 2 \)

- \( C_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 0 \\ -1 & 1 \end{vmatrix} = -1 \cdot (2) = -2 \)

- \( C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 \cdot (1) = 1 \)

- \( C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} = -1 \cdot (-1) = 1 \)

- \( C_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 0 \\ -1 & -2 \end{vmatrix} = 1 \cdot (-4) = -4 \)

- \( C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 0 \\ -2 & -2 \end{vmatrix} = -1 \cdot (-2) = 2 \)

- \( C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} = 1 \cdot (1) = 1 \)

### Step 4: Form the cofactor matrix

The cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 1 \end{pmatrix} \]

### Step 5: Find the adjoint of \( A \)

The adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 1 \end{pmatrix} \]

### Step 6: Calculate the inverse of \( A \)

Using the formula \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \):
\[ A^{-1} = \frac{1}{1} \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 1 \end{pmatrix} = \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 1 \end{pmatrix} \]

### Step 7: Solve the system of equations

We represent the system of equations in matrix form:
\[ C = \begin{pmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \]

The equation can be written as:
\[ C X = B \]

Thus, we can find \( X \) using:
\[ X = A^{-1} B \]

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

Calculating the product:
1. First row: \(-3(10) + -2(8) + -4(7) = -30 - 16 - 28 = -74\)
2. Second row: \(2(10) + 1(8) + 2(7) = 20 + 8 + 14 = 42\)
3. Third row: \(2(10) + 1(8) + 1(7) = 20 + 8 + 7 = 35\)

Thus, we have:
\[ X = \begin{pmatrix} -74 \\ 42 \\ 35 \end{pmatrix} \]

### Final Answer

The solution to the system of equations is:
\[ x = -74, \quad y = 42, \quad z = 35 \]