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यदि $S =\left\{ n \in N \mid\left(\begin{array}{ll}0 & i \\ 1 & 0\end{array}\right)^{ n }\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \forall a , b , c , d \in R \right\}$ जहाँ $i=\sqrt{-1}$ है, तो समुच्चय $S$ में दो अंकों वाली संख्याओं की संख्या है
$11$
$15$
$19$
$21$
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
$\mathrm{S}=11(16,24,32, \ldots \ldots, .96)$
