If `A=[{:(1,-3),(2,-1):}]andA^(3)-6A^(2)+5A+6l=O`, then the value of `A^(-1)` is :

Options

A. `[{:(1,-3),(2,-1):}]`
B. `[{:(1,4),(-9,-2):}]`
C. `[{:(6,-7),(4,-3):}]`
D. `[{:(-2,3),(-1,4):}]`

Answer

To solve the problem, we need to find the inverse of the matrix \( A \) given the equation \( A^3 - 6A^2 + 5A + 6I = 0 \).

### Step-by-Step Solution:

1. **Given Matrix**:
\[ A = \begin{pmatrix} 1 & 2 \\ -3 & -1 \end{pmatrix} \]

2. **Given Equation**:
\[ A^3 - 6A^2 + 5A + 6I = 0 \]

3. **Multiply the Entire Equation by \( A^{-1} \)**:
\[ A^{-1}(A^3 - 6A^2 + 5A + 6I) = A^{-1}0 \]
This simplifies to:
\[ A^2 - 6A + 5I + 6A^{-1} = 0 \]

4. **Rearranging the Equation**:
\[ 6A^{-1} = A^2 - 6A + 5I \]

5. **Finding \( A^{-1} \)**:
\[ A^{-1} = \frac{1}{6}(A^2 - 6A + 5I) \]

6. **Calculating \( 6A \)**:
\[ 6A = 6 \begin{pmatrix} 1 & 2 \\ -3 & -1 \end{pmatrix} = \begin{pmatrix} 6 & 12 \\ -18 & -6 \end{pmatrix} \]

7. **Calculating \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ -3 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ -3 & -1 \end{pmatrix} \]
Performing the multiplication:
- First row, first column: \( 1 \cdot 1 + 2 \cdot (-3) = 1 - 6 = -5 \)
- First row, second column: \( 1 \cdot 2 + 2 \cdot (-1) = 2 - 2 = 0 \)
- Second row, first column: \( -3 \cdot 1 + (-1) \cdot (-3) = -3 + 3 = 0 \)
- Second row, second column: \( -3 \cdot 2 + (-1) \cdot (-1) = -6 + 1 = -5 \)

Thus,
\[ A^2 = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix} = -5I \]

8. **Substituting Back**:
Now substituting \( A^2 \) and \( 6A \) into the equation for \( A^{-1} \):
\[ A^{-1} = \frac{1}{6}((-5I) - 6A + 5I) \]
Simplifying:
\[ A^{-1} = \frac{1}{6}(-6A) = -A \]

9. **Finding \( A^{-1} \)**:
Since \( A^{-1} = -A \):
\[ A^{-1} = -\begin{pmatrix} 1 & 2 \\ -3 & -1 \end{pmatrix} = \begin{pmatrix} -1 & -2 \\ 3 & 1 \end{pmatrix} \]

### Final Answer:
The value of \( A^{-1} \) is:
\[ A^{-1} = \begin{pmatrix} -1 & -2 \\ 3 & 1 \end{pmatrix} \]