`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 \( d \) given the matrix \( A \), the identity matrix \( I \), and the expression for \( A^{-1} \).

### Step 1: Calculate \( A^2 \)

Given:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{pmatrix} \]

To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{pmatrix} \]

Calculating the elements:
- First row:
- \( (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot -2) = 0 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 4) = 0 \)

- Second row:
- \( (0 \cdot 1 + 1 \cdot 0 + 1 \cdot 0) = 0 \)
- \( (0 \cdot 0 + 1 \cdot 1 + 1 \cdot -2) = -1 \)
- \( (0 \cdot 0 + 1 \cdot 1 + 1 \cdot 4) = 5 \)

- Third row:
- \( (0 \cdot 1 + -2 \cdot 0 + 4 \cdot 0) = 0 \)
- \( (0 \cdot 0 + -2 \cdot 1 + 4 \cdot -2) = -10 \)
- \( (0 \cdot 0 + -2 \cdot 1 + 4 \cdot 4) = 14 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14 \end{pmatrix} \]

### Step 2: Calculate the Determinant of \( A \)

The determinant of \( A \) is calculated as follows:
\[ \text{det}(A) = 1 \cdot (1 \cdot 4 - 1 \cdot -2) = 1 \cdot (4 + 2) = 6 \]

### Step 3: Find the Adjoint of \( A \)

To find the adjoint, we need the cofactor matrix. The cofactor matrix \( C \) is calculated as follows:

- For element \( a_{11} \): \( C_{11} = 4 + 2 = 6 \)
- For element \( a_{12} \): \( C_{12} = 0 \)
- For element \( a_{13} \): \( C_{13} = 0 \)
- For element \( a_{21} \): \( C_{21} = 0 \)
- For element \( a_{22} \): \( C_{22} = 0 \)
- For element \( a_{23} \): \( C_{23} = 0 \)
- For element \( a_{31} \): \( C_{31} = 4 \)
- For element \( a_{32} \): \( C_{32} = -2 \)
- For element \( a_{33} \): \( C_{33} = 1 \)

Thus, the cofactor matrix is:
\[ C = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 0 & 0 \\ 4 & -2 & 1 \end{pmatrix} \]

The adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{pmatrix} 6 & 0 & 4 \\ 0 & 0 & -2 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Calculate \( A^{-1} \)

Using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{6} \cdot \begin{pmatrix} 6 & 0 & 4 \\ 0 & 0 & -2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & \frac{2}{3} \\ 0 & 0 & -\frac{1}{3} \\ 0 & 0 & \frac{1}{6} \end{pmatrix} \]

### Step 5: Set up the equation for \( A^{-1} \)

Given:
\[ A^{-1} = \frac{1}{6}(A^2 + cA + dI) \]

Substituting the known values:
\[ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1 \end{pmatrix} = \frac{1}{6} \left( \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14 \end{pmatrix} + c \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{pmatrix} + d \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right) \]

### Step 6: Compare coefficients

From the equation, we can compare coefficients to find \( c \) and \( d \):
1. \( 1 + c + d = 6 \)
2. \( 4 + c = -1 \)
3. \( -10 + 4c + d = 14 \)

From the second equation:
\[ c = -1 - 4 = -5 \]

Substituting \( c \) into the first equation:
\[ 1 - 5 + d = 6 \implies d = 6 + 5 - 1 = 11 \]

### Final Values
Thus, we find:
\[ c = -5, \quad d = 11 \]