Let A = begin{bmatrix} 2 & -1 & -1 1 & 0 & - | vedclass

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(A) Given $A = \begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix}$. Calculating $A^2$: $A^2 = \begin{bmatrix} 2 & -1 & -1 \ 1 & 0

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$17$

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(A) Given $A = \begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix}$. Calculating $A^2$: $A^2 = \begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & -1 & -1 \ 1 & 0 & -1 \ 1 & -1 & 0 \end{bmatrix} = A$. Thus, $A^n = A$ for all $n \geq 1$. Now, $B = A - I = \begin{bmatrix} 1 & -1 & -1 \ 1 & -1 & -1 \ 1 & -1 & -1 \end{bmatrix}$. Calculating $B^2$: $B^2 = \begin{bmatrix} 1 & -1 & -1 \ 1 & -1 & -1 \ 1 & -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & -1 \ 1 & -1 & -1 \ 1 & -1 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \ -1 & 1 & 1 \ -1 & 1 & 1 \end{bmatrix} = -B$. Then $B^3 = -B^2 = B$, $B^4 = -B$, $B^5 = B$, and in general $B^n = B$ for odd $n$ and $B^n = -B$ for even $n$. The equation is $A + \omega^n B^n = A + B$. This implies $\omega^n B^n = B$. Case $1$: $n$ is odd. Then $B^n = B$, so $\omega^n B = B \Rightarrow \omega^n = 1$. Since $\omega = e^{i2\pi/3}$, $\omega^n = 1$ implies $n$ is a multiple of $3$. So $n \in \{3, 9, 15, \ldots, 99\}$. This is an arithmetic progression with $a=3, d=6, l=99$. $99 = 3 + (k-1)6 \Rightarrow 96 = (k-1)6 \Rightarrow 16 = k-1 \Rightarrow k = 17$. Case $2$: $n$ is even. Then $B^n = -B$, so $\omega^n (-B) = B \Rightarrow \omega^n = -1$. $\omega^n = -1$ implies $n$ is an odd multiple of $3$ (e.g., $3, 9, \ldots$), but $n$ must be even, so no solution here. Thus, there are $17$ values.

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 $\{n \in \{1, 2, \ldots, 100\} : A^n + (\omega B)^n = A + B\}$ is equal to $..........$

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