Compute `A^(-1)`, if `A=[{:(,3,-2,3),(,2,1,-1),(,4,-3,2):}]`. Hence sove thje matri equations,
`[{:(,3,0,3),(,2,1,0),(,4,0,2):}]{:[(,x),(,y),(,z):}]={:[(,8),(,1),(,4):}]+{:[(,2y),(,z),(,3y):}]`

Answer

To compute the inverse of the matrix \( A \) and solve the matrix equation, we will follow these steps:

### Step 1: Write down the matrix \( A \)
The given matrix \( A \) is:
\[ A = \begin{pmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A \)
The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \text{det}(A) = 3 \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} - (-2) \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} = (1)(2) - (-1)(-3) = 2 - 3 = -1 \)
2. \( \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = (2)(2) - (-1)(4) = 4 + 4 = 8 \)
3. \( \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} = (2)(-3) - (1)(4) = -6 - 4 = -10 \)

Now substituting back:
\[ \text{det}(A) = 3(-1) + 2(8) + 3(-10) = -3 + 16 - 30 = -17 \]

### Step 3: Check if the determinant is non-zero
Since \( \text{det}(A) = -17 \neq 0 \), the inverse of \( A \) exists.

### Step 4: Find the adjoint of \( A \)
The adjoint of a matrix is the transpose of the cofactor matrix. We will calculate the cofactors for each element of \( A \).

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} = -1 \quad \Rightarrow \quad A_{11} = -1 \]
2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = -8 \quad \Rightarrow \quad A_{12} = -8 \]
3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} = -10 \quad \Rightarrow \quad A_{13} = -10 \]
4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\begin{vmatrix} -2 & 3 \\ -3 & 2 \end{vmatrix} = 6 - 9 = -3 \quad \Rightarrow \quad A_{21} = -3 \]
5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \begin{vmatrix} 3 & 3 \\ 4 & 2 \end{vmatrix} = 6 - 12 = -6 \quad \Rightarrow \quad A_{22} = -6 \]
6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\begin{vmatrix} 3 & -2 \\ 4 & -3 \end{vmatrix} = 9 - 8 = 1 \quad \Rightarrow \quad A_{23} = 1 \]
7. **Cofactor \( C_{31} \)**:
\[ C_{31} = \begin{vmatrix} -2 & 3 \\ 1 & -1 \end{vmatrix} = 2 + 3 = 5 \quad \Rightarrow \quad A_{31} = 5 \]
8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\begin{vmatrix} 3 & 3 \\ 2 & -1 \end{vmatrix} = -(-3 - 6) = 9 \quad \Rightarrow \quad A_{32} = 9 \]
9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = 3 - 4 = -1 \quad \Rightarrow \quad A_{33} = -1 \]

Now, the cofactor matrix is:
\[ \text{Cofactor}(A) = \begin{pmatrix} -1 & -8 & -10 \\ -3 & -6 & 1 \\ 5 & 9 & -1 \end{pmatrix} \]
The adjoint of \( A \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{pmatrix} -1 & -3 & 5 \\ -8 & -6 & 9 \\ -10 & 1 & -1 \end{pmatrix} \]

### Step 5: Calculate the inverse of \( A \)
The inverse of \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{-17} \cdot \begin{pmatrix} -1 & -3 & 5 \\ -8 & -6 & 9 \\ -10 & 1 & -1 \end{pmatrix} \]
Thus,
\[ A^{-1} = \begin{pmatrix} \frac{1}{17} & \frac{3}{17} & -\frac{5}{17} \\ \frac{8}{17} & \frac{6}{17} & -\frac{9}{17} \\ \frac{10}{17} & -\frac{1}{17} & \frac{1}{17} \end{pmatrix} \]

### Step 6: Solve the matrix equation
The matrix equation given is:
\[ \begin{pmatrix} 3 & 0 & 3 \\ 2 & 1 & 0 \\ 4 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} + \begin{pmatrix} 2y \\ z \\ 3y \end{pmatrix} \]

Rearranging gives:
\[ \begin{pmatrix} 3 & 0 & 3 \\ 2 & 1 & 0 \\ 4 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 2y \\ z \\ 3y \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \]

This leads to the system of equations:
1. \( 3x + 3z - 2y = 8 \)
2. \( 2x + y - z = 1 \)
3. \( 4x + 2z - 3y = 4 \)

### Step 7: Write the system in matrix form
Let \( B = \begin{pmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{pmatrix} \) and \( D = \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \).

Thus, the equation can be expressed as:
\[ B \begin{pmatrix} x \\ y \\ z \end{pmatrix} = D \]

### Step 8: Use the inverse to find \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} \)
We can find \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) as follows:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{-1} D \]

Substituting the values:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{1}{17} & \frac{3}{17} & -\frac{5}{17} \\ \frac{8}{17} & \frac{6}{17} & -\frac{9}{17} \\ \frac{10}{17} & -\frac{1}{17} & \frac{1}{17} \end{pmatrix} \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \]

Calculating the right-hand side:
1. \( x = \frac{1}{17}(8 - 3 + 20) = \frac{25}{17} \)
2. \( y = \frac{8}{17}(8) + \frac{6}{17}(1) - \frac{9}{17}(4) = \frac{64 + 6 - 36}{17} = \frac{34}{17} = 2 \)
3. \( z = \frac{10}{17}(8) - \frac{1}{17}(1) + \frac{4}{17} = \frac{80 - 1 + 4}{17} = \frac{83}{17} = 3 \)

### Final Result
Thus, the solution to the matrix equation is:
\[ x = 1, \quad y = 2, \quad z = 3 \]