`A=[(1,0,0),(0,1,1),(0,-2,4)],I=[(1,0,0),(0,1,0),(0,0,1)]`
`A^(-1)=1/6(A^(2)+CA+DI)` then C and D equal to

Options

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

Answer

To find the values of C and D in the equation \( A^{-1} = \frac{1}{6}(A^2 + CA + DI) \), we will follow these steps:

### Step 1: Find the characteristic polynomial of matrix A.
The characteristic polynomial of a matrix \( A \) is given by \( \text{det}(A - \lambda I) = 0 \).

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

We need to compute \( A - \lambda I \):
\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 1 \\ 0 & -2 & 4 - \lambda \end{pmatrix} \]

Now, we compute the determinant:
\[ \text{det}(A - \lambda I) = (1 - \lambda) \text{det}\begin{pmatrix} 1 - \lambda & 1 \\ -2 & 4 - \lambda \end{pmatrix} \]

Calculating the 2x2 determinant:
\[ \text{det}\begin{pmatrix} 1 - \lambda & 1 \\ -2 & 4 - \lambda \end{pmatrix} = (1 - \lambda)(4 - \lambda) - (-2)(1) = (1 - \lambda)(4 - \lambda) + 2 \]

Expanding this:
\[ = 4 - 5\lambda + \lambda^2 + 2 = \lambda^2 - 5\lambda + 6 \]

Thus, the characteristic polynomial becomes:
\[ \text{det}(A - \lambda I) = (1 - \lambda)(\lambda^2 - 5\lambda + 6) = 0 \]

### Step 2: Solve for the eigenvalues.
Setting the characteristic polynomial to zero:
\[ (1 - \lambda)(\lambda^2 - 5\lambda + 6) = 0 \]

The roots of \( \lambda^2 - 5\lambda + 6 = 0 \) can be found using the quadratic formula:
\[ \lambda = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} \]
This gives us:
\[ \lambda = 3 \quad \text{and} \quad \lambda = 2 \]

Thus, the eigenvalues are \( \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 3 \).

### Step 3: Write the characteristic equation.
From the eigenvalues, we can write the characteristic equation:
\[ A^3 - 6A^2 + 11A - 6I = 0 \]

### Step 4: Relate the characteristic equation to the given equation.
We have:
\[ A^{-1} = \frac{1}{6}(A^2 + CA + DI) \]
Multiplying both sides by \( A \):
\[ I = \frac{1}{6}(A^3 + CA^2 + DIA) \]
This can be rearranged to:
\[ 6I = A^3 + CA^2 + DIA \]
Now, we can compare this with the characteristic equation:
\[ A^3 - 6A^2 + 11A - 6I = 0 \]

### Step 5: Compare coefficients.
From the comparison:
\[ A^3 + CA^2 + DIA = 6I \]
We can equate coefficients:
- Coefficient of \( A^2 \): \( C = -6 \)
- Coefficient of \( A \): \( D = 11 \)

### Final Result:
Thus, the values of \( C \) and \( D \) are:
\[ C = -6, \quad D = 11 \]