Let $A=\begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix}$ and $B=A-I$. If $\omega=\frac{\sqrt{3}i-1}{2}$, then the number of elements in the set $\{n \,|\, n \in \{1,2,\ldots,100\}: A^n + (\omega B)^n = A + B\}$ is equal to _____ .

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Given, $A=\begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix}$ $\Rightarrow A^{2}= \begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix} = A$ $\Rightarrow A^{n}= A$ Now $\forall n \in \{1,2,...,100\}$ Now, $B=A-I=\begin{bmatrix} 1 & -1 & -1 \ 1 & -1 & -1 \ 1 & -1 & -1 \end{bmatrix}$ $B^{2}= \begin{bmatrix} 1 & -1 & -1 \ 1 & -1 & -1 \ 1 & -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & -1 \ 1 & -1 & -1 \ 1 & -1 & -1 \end{bmatrix} = -\begin{bmatrix} 1 & -1 & -1 \ 1 & -1 & -1 \ 1 & -1 & -1 \end{bmatrix} = -B$ $\Rightarrow B^{3}= -B^{2}= B$ $\Rightarrow B^{5}= B$ $\Rightarrow B^{99}= B$ Also, $\omega^{3k}=1$ So, $n=$ common of $\{1,3,5,...,99\}$ and $\{3,6,9,...,99\}=17$

Let $A=\begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix}$ and $B=A-I$. If…

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