Consider a matrix `A=[(0,1,2),(0,-3,0),(1,1,1)].` If `6A^(-1)=aA^(2)+bA+cI`, where `a, b, c in and I` is an identity matrix, then `a+2b+3c` is equal to

Options

A. 10
B. `-10`
C. 8
D. 0

Answer

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) such that the equation \( 6A^{-1} = aA^2 + bA + cI \) holds true, where \( A \) is the given matrix and \( I \) is the identity matrix.

### Step 1: Define the matrix \( A \) and the identity matrix \( I \)

Given:
\[ A = \begin{pmatrix} 0 & 1 & 2 \\ 0 & -3 & 0 \\ 1 & 1 & 1 \end{pmatrix}, \quad I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Find the characteristic polynomial of \( A \)

To find the characteristic polynomial, we compute the determinant of \( A - \lambda I \):
\[ A - \lambda I = \begin{pmatrix} -\lambda & 1 & 2 \\ 0 & -3 - \lambda & 0 \\ 1 & 1 & 1 - \lambda \end{pmatrix} \]

The determinant is calculated as follows:
\[ \text{det}(A - \lambda I) = -\lambda \cdot \text{det} \begin{pmatrix} -3 - \lambda & 0 \\ 1 & 1 - \lambda \end{pmatrix} = -\lambda \left((-3 - \lambda)(1 - \lambda) - 0\right) \]
\[ = -\lambda \left(-3 - \lambda + 3\lambda + \lambda^2\right) = -\lambda \left(\lambda^2 + 2\lambda - 3\right) \]

Setting the determinant to zero gives us the characteristic equation:
\[ -\lambda(\lambda^2 + 2\lambda - 3) = 0 \]

### Step 3: Solve the characteristic equation

The roots of the characteristic polynomial are found by solving:
\[ \lambda^2 + 2\lambda - 3 = 0 \]
Factoring gives:
\[ (\lambda + 3)(\lambda - 1) = 0 \]
Thus, the eigenvalues are \( \lambda = -3 \) and \( \lambda = 1 \).

### Step 4: Use the Cayley-Hamilton theorem

According to the Cayley-Hamilton theorem, \( A \) satisfies its own characteristic equation:
\[ A^3 + 2A^2 - 3I = 0 \]
Rearranging gives:
\[ A^3 + 2A^2 = 3I \]

### Step 5: Multiply by \( A^{-1} \)

Multiplying the entire equation by \( A^{-1} \):
\[ A^2 + 2A = 3A^{-1} \]
Thus, we can express \( 6A^{-1} \) as:
\[ 6A^{-1} = 2A^2 + 4A \]

### Step 6: Identify coefficients \( a \), \( b \), and \( c \)

From the equation \( 6A^{-1} = aA^2 + bA + cI \), we can see that:
- \( a = 2 \)
- \( b = 4 \)
- \( c = 0 \)

### Step 7: Calculate \( a + 2b + 3c \)

Now, we compute:
\[ a + 2b + 3c = 2 + 2(4) + 3(0) = 2 + 8 + 0 = 10 \]

### Final Answer

Thus, the value of \( a + 2b + 3c \) is:
\[ \boxed{10} \]