Let $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix}$. Then the sum of the diagonal elements of the matrix $(A + I)^{11}$ is equal to:

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Aicharya · Accepted Answer

Given, $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix}$ $A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix}$ $\Rightarrow A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{bmatrix} = A$ $\Rightarrow A^2 = A$ $\Rightarrow A^3 = A \cdot A^2 = A^2 = A$ $\Rightarrow A^4 = A^2 \cdot A^2 = A^2 = A$ $\Rightarrow A^3 = A^4 = \ldots = A$ $(A+I)^{11} = C_{11}^{0} A^{11} + C_{11}^{1} A^{10} + \ldots + C_{11}^{10} A + C_{11}^{11} I$ $= (C_{11}^{0} + C_{11}^{1} + \ldots + C_{11}^{10}) A + C_{11}^{11} I$ $= (2^{11} - 1) A + I = 2047 A + I$ $\Rightarrow (A+I)^{11} = 2047 A + I$ $\Rightarrow (A+I)^{11} = \begin{bmatrix} 2047+1 & 0 & 0 \ 0 & 4 \cdot 2047+1 & -1 \cdot 2047 \ 0 & 12 \cdot 2047 & -3 \cdot 2047+1 \end{bmatrix}$ $	herefore$ Sum of diagonal elements $= 2047 (1+4-3) + 3$ $= 4094 + 3 = 4097$

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix}$. Then the sum of the diagonal…

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