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અહી $S=\left\{n \in N \mid\left(\begin{array}{ll}0 & i \\ 1 & 0\end{array}\right)^{n}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) \forall a, b, c, d \in R\right\}$ કે જ્યાં $i=\sqrt{-1} $ છે. તો ગણ $\mathrm{S}$ માં $2$ અંકની કેટલી સંખ્યા હશે.
$11$
$15$
$19$
$21$
Solution
Lex $X=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) \;and\; A=\left(\begin{array}{ll}0 & i \\ 1 & 0\end{array}\right)^{n}$
$\Rightarrow \mathrm{AX}=\mathrm{IX}$
$\Rightarrow \mathrm{A}=\mathrm{I}$
$\Rightarrow\left(\begin{array}{ll}0 & \mathrm{i} \\ 1 & 0\end{array}\right)^{n}=\mathrm{I}$
$\Rightarrow \mathrm{A}^{8}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow \mathrm{n}$ is multiple of $8$
So number of $2$ digit numbers in the set
$\mathrm{S}=11(16,24,32, \ldots \ldots, .96)$
