Let
$\mathrm{S}=\left\{\mathrm{n} \in \mathbf{N} \mid\left(\begin{array}{ll}0 & \mathrm{i} \\ 1 & 0\end{array}\right)^{\mathrm{n}}\left(\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right)=\left(\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right) \forall \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathbf{R}\right\}$
where $\mathrm{i}=\sqrt{-1}$. Then the number of 2 -digit numbers in the set $\mathrm{S}$ is_________.
Let $X=\left(\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right) \& A=\left(\begin{array}{ll}\mathrm{o} & \mathrm{i} \\ 1 & 0\end{array}\right)^{\mathrm{n}}$
$\Rightarrow \mathrm{AX}=\mathrm{IX}$
$\Rightarrow \mathrm{A}=\mathrm{I}$
$\Rightarrow\left(\begin{array}{ll}0 & i \\ 1 & 0\end{array}\right)^{n}=I$
$\Rightarrow A^{8}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow \mathrm{n}$ is multiple of 8
So number of 2 digit numbers in the set
$\mathrm{S}=11(16,24,32, \ldots \ldots, 96)$
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