Let `A=[[1,0,0],[0,4,-1],[0,12,-3]]`.Then the sum of the diagonal elements of the matrix `(A+I)^(11)` is equal to :

Options

A. `2050`
B. `4094`
C. `6144`
D. `4097`

Answer

To solve the problem, we need to find the sum of the diagonal elements of the matrix \((A + I)^{11}\), where \(A\) is given as:

\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix} \]

and \(I\) is the identity matrix of the same size:

\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

### Step 1: Calculate \(A + I\)

First, we add the matrix \(A\) to the identity matrix \(I\):

\[ A + I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 + 1 & 0 & 0 \\ 0 & 4 + 1 & -1 \\ 0 & 12 & -3 + 1 \end{bmatrix} \]

This simplifies to:

\[ A + I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 5 & -1 \\ 0 & 12 & -2 \end{bmatrix} \]

### Step 2: Find the eigenvalues of \(A + I\)

The next step is to find the eigenvalues of the matrix \(A + I\). The eigenvalues can be found by calculating the determinant of \(A + I - \lambda I\):

\[ \begin{vmatrix} 2 - \lambda & 0 & 0 \\ 0 & 5 - \lambda & -1 \\ 0 & 12 & -2 - \lambda \end{vmatrix} \]

Calculating this determinant, we have:

\[ (2 - \lambda) \begin{vmatrix} 5 - \lambda & -1 \\ 12 & -2 - \lambda \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ = (5 - \lambda)(-2 - \lambda) - (-1)(12) = (5 - \lambda)(-2 - \lambda) + 12 \]

Expanding this gives:

\[ = -10 - 5\lambda + 2\lambda + \lambda^2 + 12 = \lambda^2 - 3\lambda + 2 \]

Setting the determinant to zero to find the eigenvalues:

\[ (2 - \lambda)(\lambda^2 - 3\lambda + 2) = 0 \]

The eigenvalues are \(\lambda = 2\) and the roots of \(\lambda^2 - 3\lambda + 2 = 0\), which are \(\lambda = 1\) and \(\lambda = 2\).

Thus, the eigenvalues of \(A + I\) are \(2, 1, 2\).

### Step 3: Calculate the sum of the diagonal elements of \((A + I)^{11}\)

The sum of the diagonal elements of a matrix is equal to the sum of its eigenvalues raised to the power of the matrix:

\[ \text{Sum of diagonal elements of } (A + I)^{11} = 2^{11} + 1^{11} + 2^{11} = 2 \cdot 2^{11} + 1 = 2^{12} + 1 \]

Calculating \(2^{12} + 1\):

\[ 2^{12} = 4096 \implies 2^{12} + 1 = 4096 + 1 = 4097 \]

### Final Answer

Thus, the sum of the diagonal elements of the matrix \((A + I)^{11}\) is:

\[ \boxed{4097} \]