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 \( c \) in the equation \( 6A^{-1} = aA^2 + bA + cI \) where \( A = \begin{pmatrix} 0 & 1 & 2 \\ 0 & -3 & 0 \\ 1 & 1 & 1 \end{pmatrix} \) and \( I \) is the identity matrix.

### Step 1: Find the characteristic polynomial of matrix \( A \)

We start by calculating the characteristic polynomial using the determinant of \( A - \lambda I \):

\[ A - \lambda I = \begin{pmatrix} 0 - \lambda & 1 & 2 \\ 0 & -3 - \lambda & 0 \\ 1 & 1 & 1 - \lambda \end{pmatrix} = \begin{pmatrix} -\lambda & 1 & 2 \\ 0 & -3 - \lambda & 0 \\ 1 & 1 & 1 - \lambda \end{pmatrix} \]

Now, we calculate the determinant:

\[ \text{det}(A - \lambda I) = -\lambda \cdot \text{det}\begin{pmatrix} -3 - \lambda & 0 \\ 1 & 1 - \lambda \end{pmatrix} - 1 \cdot \text{det}\begin{pmatrix} 0 & 2 \\ 1 & 1 - \lambda \end{pmatrix} \]

Calculating the determinants of the 2x2 matrices:

1. For \( \begin{pmatrix} -3 - \lambda & 0 \\ 1 & 1 - \lambda \end{pmatrix} \):
\[ \text{det} = (-3 - \lambda)(1 - \lambda) - (0) = (-3 - \lambda)(1 - \lambda) \]

2. For \( \begin{pmatrix} 0 & 2 \\ 1 & 1 - \lambda \end{pmatrix} \):
\[ \text{det} = 0 \cdot (1 - \lambda) - 2 \cdot 1 = -2 \]

Putting it all together:

\[ \text{det}(A - \lambda I) = -\lambda((-3 - \lambda)(1 - \lambda)) + 2 \]

Expanding this gives us:

\[ = -\lambda(3 - 3\lambda - \lambda + \lambda^2) + 2 = -\lambda(-\lambda^2 - 2\lambda + 3) + 2 \]

This simplifies to:

\[ \lambda^3 + 2\lambda^2 - 5\lambda - 6 = 0 \]

### Step 2: Use the Cayley-Hamilton theorem

According to the Cayley-Hamilton theorem, the matrix \( A \) satisfies its own characteristic polynomial:

\[ A^3 + 2A^2 - 5A - 6I = 0 \]

Multiplying the entire equation by \( A^{-1} \):

\[ A^2 + 2A - 5I - 6A^{-1} = 0 \implies 6A^{-1} = A^2 + 2A - 5I \]

### Step 3: Compare coefficients

From the equation \( 6A^{-1} = A^2 + 2A - 5I \), we can identify:

- \( a = 1 \)
- \( b = 2 \)
- \( c = -5 \)

### Step 4: Calculate \( a + 2b + 3c \)

Now we compute:

\[ a + 2b + 3c = 1 + 2(2) + 3(-5) = 1 + 4 - 15 = -10 \]

Thus, the final answer is:

\[ \boxed{-10} \]