Solve the following equations, using inverse of a matrix :
`{:(x+2y=5),(y+2z=8),(z+2x=5):}`

Answer

To solve the system of equations using the inverse of a matrix, we start with the given equations:

1. \( x + 2y = 5 \)
2. \( y + 2z = 8 \)
3. \( z + 2x = 5 \)

We can express this system in the matrix form \( Ax = b \), where:

- \( A \) is the coefficient matrix,
- \( x \) is the variable matrix,
- \( b \) is the constant matrix.

### Step 1: Formulate the matrices

The coefficient matrix \( A \) and the constant matrix \( b \) can be written as follows:

\[ A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1 \end{bmatrix}, \quad x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad b = \begin{bmatrix} 5 \\ 8 \\ 5 \end{bmatrix} \]

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

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

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

Calculating the determinant:

\[ |A| = 1 \cdot (1) - 2 \cdot (-4) + 0 = 1 + 8 = 9 \]

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

Next, we calculate the cofactor matrix \( C \) of \( A \):

\[ C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} \]

Calculating each cofactor:

- \( C_{11} = | \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} | = 1 \)
- \( C_{12} = -| \begin{bmatrix} 0 & 2 \\ 2 & 1 \end{bmatrix} | = -(-4) = 4 \)
- \( C_{13} = | \begin{bmatrix} 0 & 1 \\ 2 & 0 \end{bmatrix} | = -2 \)

- \( C_{21} = -| \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} | = -2 \)
- \( C_{22} = | \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} | = 1 \)
- \( C_{23} = -| \begin{bmatrix} 1 & 2 \\ 2 & 0 \end{bmatrix} | = -(-4) = 4 \)

- \( C_{31} = | \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} | = 4 \)
- \( C_{32} = -| \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} | = -2 \)
- \( C_{33} = | \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} | = 1 \)

Thus, the cofactor matrix \( C \) is:

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

### Step 4: Find the adjoint of \( A \)

The adjoint of \( A \) is the transpose of the cofactor matrix \( C \):

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

### Step 5: Find the inverse of matrix \( A \)

The inverse of matrix \( A \) is given by:

\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) = \frac{1}{9} \begin{bmatrix} 1 & -2 & 4 \\ 4 & 1 & -2 \\ -2 & 4 & 1 \end{bmatrix} \]

### Step 6: Solve for \( x \)

Now, we can find \( x \) using the equation \( x = A^{-1}b \):

\[ x = \frac{1}{9} \begin{bmatrix} 1 & -2 & 4 \\ 4 & 1 & -2 \\ -2 & 4 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \\ 5 \end{bmatrix} \]

Calculating the product:

\[ x = \frac{1}{9} \begin{bmatrix} 1 \cdot 5 + (-2) \cdot 8 + 4 \cdot 5 \\ 4 \cdot 5 + 1 \cdot 8 + (-2) \cdot 5 \\ -2 \cdot 5 + 4 \cdot 8 + 1 \cdot 5 \end{bmatrix} \]

Calculating each component:

1. \( 5 - 16 + 20 = 9 \)
2. \( 20 + 8 - 10 = 18 \)
3. \( -10 + 32 + 5 = 27 \)

Thus, we have:

\[ x = \frac{1}{9} \begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]

### Final Solution

The solution to the system of equations is:

\[ x = 1, \quad y = 2, \quad z = 3 \]