Let $A=\left[\begin{array}{ccc}2& -1& -1\\ 1& 0& -1\\ 1& -1& 0\end{array}\right]$ and $B=A-I$. If $\omega =\frac{\sqrt{3}\mathrm{i}-1}{2}$, then the number of elements in the set $\left\{n\in \left\{1,2,\dots ,100\right\}:A^n+{\left(\omega B\right)}^n=A+B\right\}$ is equal to _____ .

Given, $A=\left[\begin{array}{ccc}2& -1& -1\\ 1& 0& -1\\ 1& -1& 0\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{ccc}2& -1& -1\\ 1& 0& -1\\ 1& -1& 0\end{array}\right]\left[\begin{array}{ccc}2& -1& -1\\ 1& 0& -1\\ 1& -1& 0\end{array}\right]=\left[\begin{array}{ccc}2& -1& -1\\ 1& 0& -1\\ 1& -1& 0\end{array}\right]=A$

$\Rightarrow A^n=A$ 

Now $\forall n\in \left\{1,2,\dots ,100\right\}$

Now, $B=A-I=\left[\begin{array}{llr}1& -1& -1\\ 1& -1& -1\\ 1& -1& -1\end{array}\right]$

$B^2=\left[\begin{array}{llr}1& -1& -1\\ 1& -1& -1\\ 1& -1& -1\end{array}\right]\left[\begin{array}{llr}1& -1& -1\\ 1& -1& -1\\ 1& -1& -1\end{array}\right]=-\left[\begin{array}{llr}1& -1& -1\\ 1& -1& -1\\ 1& -1& -1\end{array}\right]=-B$ 

$\Rightarrow B^3=-B^2=B$ 

$\Rightarrow B^5=B$ 

$\Rightarrow {\mathrm{B}}^{99}=\mathrm{B}$

Also, ${\omega }^{3\mathit{k}}=1$

So, $n=$ common of $\left\{1,3,5,\dots ,99\right\}$ and  $\left\{3,6,9,\dots ,99\right\}=17$