If `A=[{:(3,1,2),(3,2,-3),(2,0,-1):}]` , find `A^(-1)`. Hence, solve the system of equations :
`3x+3y+2z=1`, `x+2y=4`, 2x-3y-z=5`

Answer

To find the inverse of the matrix \( A \) and solve the system of equations, we will follow these steps:

### Step 1: Define the matrix \( A \)
Given:
\[ A = \begin{pmatrix} 3 & 1 & 2 \\ 3 & 2 & -3 \\ 2 & 0 & -1 \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A \)
The determinant of a \( 3 \times 3 \) matrix is calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
\[ \text{det}(A) = 3 \cdot ((2 \cdot -1) - (-3 \cdot 0)) - 1 \cdot ((3 \cdot -1) - (-3 \cdot 2)) + 2 \cdot ((3 \cdot 0) - (2 \cdot 2)) \]
Calculating each term:
\[ = 3 \cdot (-2) - 1 \cdot (-3 + 6) + 2 \cdot (0 - 4) \]
\[ = -6 - 3 - 8 \]
\[ = -17 \]

### Step 3: Calculate the adjoint of \( A \)
The adjoint of a matrix is the transpose of the cofactor matrix. We will find the cofactors for each element of \( A \).

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = (-1)^{1+1} \cdot \text{det} \begin{pmatrix} 2 & -3 \\ 0 & -1 \end{pmatrix} = 1 \cdot (2 \cdot -1 - (-3 \cdot 0)) = -2 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = (-1)^{1+2} \cdot \text{det} \begin{pmatrix} 3 & -3 \\ 2 & -1 \end{pmatrix} = -1 \cdot (3 \cdot -1 - (-3 \cdot 2)) = -1 \cdot (-3 + 6) = -3 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = (-1)^{1+3} \cdot \text{det} \begin{pmatrix} 3 & 2 \\ 2 & 0 \end{pmatrix} = 1 \cdot (3 \cdot 0 - (2 \cdot 2)) = -4 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = (-1)^{2+1} \cdot \text{det} \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix} = -1 \cdot (1 \cdot -1 - (2 \cdot 0)) = 1 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = (-1)^{2+2} \cdot \text{det} \begin{pmatrix} 3 & 2 \\ 2 & -1 \end{pmatrix} = 1 \cdot (3 \cdot -1 - (2 \cdot 2)) = -7 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = (-1)^{2+3} \cdot \text{det} \begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix} = -1 \cdot (3 \cdot 0 - (1 \cdot 2)) = 2 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = (-1)^{3+1} \cdot \text{det} \begin{pmatrix} 1 & 2 \\ 2 & -3 \end{pmatrix} = 1 \cdot (1 \cdot -3 - (2 \cdot 2)) = -7 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = (-1)^{3+2} \cdot \text{det} \begin{pmatrix} 3 & 2 \\ 3 & -3 \end{pmatrix} = -1 \cdot (3 \cdot -3 - (2 \cdot 3)) = 3 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = (-1)^{3+3} \cdot \text{det} \begin{pmatrix} 3 & 1 \\ 3 & 2 \end{pmatrix} = 1 \cdot (3 \cdot 2 - (1 \cdot 3)) = 3 \]

The cofactor matrix is:
\[ \text{Cofactor}(A) = \begin{pmatrix} -2 & -3 & -4 \\ 1 & -7 & 2 \\ -7 & 3 & 3 \end{pmatrix} \]
Now, we take the transpose to get the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} -2 & 1 & -7 \\ -3 & -7 & 3 \\ -4 & 2 & 3 \end{pmatrix} \]

### Step 4: Calculate the inverse of \( A \)
Using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{-17} \cdot \begin{pmatrix} -2 & 1 & -7 \\ -3 & -7 & 3 \\ -4 & 2 & 3 \end{pmatrix} = \begin{pmatrix} \frac{2}{17} & -\frac{1}{17} & \frac{7}{17} \\ \frac{3}{17} & \frac{7}{17} & -\frac{3}{17} \\ \frac{4}{17} & -\frac{2}{17} & -\frac{3}{17} \end{pmatrix} \]

### Step 5: Solve the system of equations
We can represent the system of equations in matrix form \( AX = B \):
\[ B = \begin{pmatrix} 1 \\ 4 \\ 5 \end{pmatrix} \]
Thus, we need to calculate:
\[ X = A^{-1}B \]
Calculating \( A^{-1}B \):
\[ X = \begin{pmatrix} \frac{2}{17} & -\frac{1}{17} & \frac{7}{17} \\ \frac{3}{17} & \frac{7}{17} & -\frac{3}{17} \\ \frac{4}{17} & -\frac{2}{17} & -\frac{3}{17} \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \\ 5 \end{pmatrix} \]
Calculating the first element:
\[ x = \frac{2}{17} \cdot 1 - \frac{1}{17} \cdot 4 + \frac{7}{17} \cdot 5 = \frac{2 - 4 + 35}{17} = \frac{33}{17} = 2 \]
Calculating the second element:
\[ y = \frac{3}{17} \cdot 1 + \frac{7}{17} \cdot 4 - \frac{3}{17} \cdot 5 = \frac{3 + 28 - 15}{17} = \frac{16}{17} = 1 \]
Calculating the third element:
\[ z = \frac{4}{17} \cdot 1 - \frac{2}{17} \cdot 4 - \frac{3}{17} \cdot 5 = \frac{4 - 8 - 15}{17} = \frac{-19}{17} = -4 \]

### Final Solution
The solution to the system of equations is:
\[ x = 2, \quad y = 1, \quad z = -4 \]