`A= [{:( 1,0,0) ,( 0,1,1) , ( 0,-2,4) :}] ,I= [{:( 1,0,0) ,( 0,1,0),( 0,0,1) :}]and A^(-1) =[(1)/(6) (A^(2)+cA +dt)]` then , the value of c and d are

Options

A. `-6-11`
B. ` 6,11`
C. `-6,11`
D. `6,-11`

Answer

To solve the problem, we need to find the values of \( c \) and \( d \) in the expression for the inverse of matrix \( A \). We start by using the characteristic polynomial of matrix \( A \).

### Step-by-Step Solution:

1. **Write down the matrix \( A \)**:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{pmatrix} \]

2. **Set up the characteristic polynomial**:
The characteristic polynomial is given by \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix.
\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 1 \\ 0 & -2 & 4 - \lambda \end{pmatrix} \]

3. **Calculate the determinant**:
We can calculate the determinant using the first row:
\[ \text{det}(A - \lambda I) = (1 - \lambda) \cdot \text{det}\begin{pmatrix} 1 - \lambda & 1 \\ -2 & 4 - \lambda \end{pmatrix} \]
Now, calculate the determinant of the 2x2 matrix:
\[ = (1 - \lambda)((1 - \lambda)(4 - \lambda) - (-2)(1)) = (1 - \lambda)((1 - \lambda)(4 - \lambda) + 2) \]

4. **Expand the determinant**:
Expanding the expression:
\[ = (1 - \lambda)(4 - 5\lambda + \lambda^2 + 2) = (1 - \lambda)(\lambda^2 - 5\lambda + 6) \]

5. **Factor the polynomial**:
The quadratic \( \lambda^2 - 5\lambda + 6 \) can be factored as:
\[ = (1 - \lambda)(\lambda - 2)(\lambda - 3) \]
Thus, the characteristic polynomial is:
\[ (1 - \lambda)(\lambda - 2)(\lambda - 3) = 0 \]

6. **Formulate the Cayley-Hamilton theorem**:
According to the Cayley-Hamilton theorem, the matrix \( A \) satisfies its own characteristic equation:
\[ A^3 - 6A^2 + 11A - 6I = 0 \]

7. **Express \( A^{-1} \)**:
Rearranging gives:
\[ A^3 = 6A^2 - 11A + 6I \]
Multiplying by \( A^{-1} \):
\[ A^2 = 6A - 11I + 6A^{-1} \]
Thus:
\[ A^{-1} = \frac{1}{6}(A^2 - 6A + 11I) \]

8. **Identify \( c \) and \( d \)**:
Comparing with the given form \( A^{-1} = \frac{1}{6}(A^2 + cA + dI) \):
We see that:
\[ c = -6 \quad \text{and} \quad d = 11 \]

### Final Answer:
The values of \( c \) and \( d \) are:
\[ c = -6, \quad d = 11 \]