Compute `A^(-1)`, if `A=[{:(,3,-2,3),(,2,1,-1),(,4,-3,2):}]`. Hence solve the matrix equations,
`[{:(,3,-2,3),(,2,1,-1),(,4,-3,2):}]{:[(,x),(,y),(,z):}]={:[(,8),(,1),(,4):}]`

Answer

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

### Step 1: Define the Matrix \( A \)
Given the matrix:
\[ A = \begin{pmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{pmatrix} \]

### Step 2: Compute the Determinant of \( A \)
The determinant of a \( 3 \times 3 \) matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
- \( a = 3, b = -2, c = 3 \)
- \( d = 2, e = 1, f = -1 \)
- \( g = 4, h = -3, i = 2 \)

Calculating the determinant:
\[ \text{det}(A) = 3(1 \cdot 2 - (-1) \cdot (-3)) - (-2)(2 \cdot 2 - (-1) \cdot 4) + 3(2 \cdot (-3) - 1 \cdot 4) \]
\[ = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) \]
\[ = 3(-1) + 2(8) + 3(-10) \]
\[ = -3 + 16 - 30 = -17 \]

### Step 3: Compute 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. For \( a_{11} = 3 \):
\[ C_{11} = \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} = (1 \cdot 2 - (-1) \cdot (-3)) = 2 - 3 = -1 \]
2. For \( a_{12} = -2 \):
\[ C_{12} = -\begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = - (2 \cdot 2 - (-1) \cdot 4) = - (4 + 4) = -8 \]
3. For \( a_{13} = 3 \):
\[ C_{13} = \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix} = (2 \cdot -3 - 1 \cdot 4) = -6 - 4 = -10 \]
4. For \( a_{21} = 2 \):
\[ C_{21} = -\begin{vmatrix} -2 & 3 \\ -3 & 2 \end{vmatrix} = -((-2) \cdot 2 - 3 \cdot (-3)) = -(-4 + 9) = -5 \]
5. For \( a_{22} = 1 \):
\[ C_{22} = \begin{vmatrix} 3 & 3 \\ 4 & 2 \end{vmatrix} = (3 \cdot 2 - 3 \cdot 4) = 6 - 12 = -6 \]
6. For \( a_{23} = -1 \):
\[ C_{23} = -\begin{vmatrix} 3 & -2 \\ 4 & -3 \end{vmatrix} = -((3)(-3) - (-2)(4)) = -(-9 + 8) = 1 \]
7. For \( a_{31} = 4 \):
\[ C_{31} = \begin{vmatrix} -2 & 3 \\ 1 & -1 \end{vmatrix} = (-2)(-1) - (3)(1) = 2 - 3 = -1 \]
8. For \( a_{32} = -3 \):
\[ C_{32} = -\begin{vmatrix} 3 & 3 \\ 2 & -1 \end{vmatrix} = -((3)(-1) - (3)(2)) = -(-3 - 6) = 9 \]
9. For \( a_{33} = 2 \):
\[ C_{33} = \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = (3)(1) - (-2)(2) = 3 + 4 = 7 \]

Thus, the cofactor matrix is:
\[ \text{Cof}(A) = \begin{pmatrix} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7 \end{pmatrix} \]

Now, the adjoint of \( A \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{pmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{pmatrix} \]

### Step 4: Compute the Inverse of \( A \)
The inverse of \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{-17} \cdot \begin{pmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{pmatrix} \]
\[ = \begin{pmatrix} \frac{1}{17} & \frac{5}{17} & \frac{1}{17} \\ \frac{8}{17} & \frac{6}{17} & -\frac{9}{17} \\ \frac{10}{17} & -\frac{1}{17} & -\frac{7}{17} \end{pmatrix} \]

### Step 5: Solve the Matrix Equation
We need to solve the equation:
\[ A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \]
To find \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} \), we multiply both sides by \( A^{-1} \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{-1} \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \]

Calculating the multiplication:
\[ \begin{pmatrix} \frac{1}{17} & \frac{5}{17} & \frac{1}{17} \\ \frac{8}{17} & \frac{6}{17} & -\frac{9}{17} \\ \frac{10}{17} & -\frac{1}{17} & -\frac{7}{17} \end{pmatrix} \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \]
Calculating each component:
1. \( x = \frac{1}{17}(8) + \frac{5}{17}(1) + \frac{1}{17}(4) = \frac{8 + 5 + 4}{17} = \frac{17}{17} = 1 \)
2. \( y = \frac{8}{17}(8) + \frac{6}{17}(1) - \frac{9}{17}(4) = \frac{64 + 6 - 36}{17} = \frac{34}{17} = 2 \)
3. \( z = \frac{10}{17}(8) - \frac{1}{17}(1) - \frac{7}{17}(4) = \frac{80 - 1 - 28}{17} = \frac{51}{17} = 3 \)

Thus, the solution is:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \]

### Summary of the Solution
1. The inverse of \( A \) is:
\[ A^{-1} = \begin{pmatrix} \frac{1}{17} & \frac{5}{17} & \frac{1}{17} \\ \frac{8}{17} & \frac{6}{17} & -\frac{9}{17} \\ \frac{10}{17} & -\frac{1}{17} & -\frac{7}{17} \end{pmatrix} \]
2. The solution to the matrix equation is:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \]