Let S= n N ( ll 0 & i 1 & 0 )^ n ( ll a & b c & d )= ( ll a & b c… - Vidyadip

MCQ

Let $S=\left\{n \in N \mid\left(\begin{array}{ll}0 & i \\ 1 & 0\end{array}\right)^{n}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) \forall a, b, c, d \in R\right\}$, where $i=\sqrt{-1} .$ Then the number of $2 -$ digit numbers in the set $\mathrm{S}$ is $......$

✓
$11$
B
$15$
C
$19$
D
$21$

✓

Answer

Correct option: A.

$11$

a
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

$\mathrm{S}=11(16,24,32, \ldots \ldots, .96)$

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