If $A=\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right], n \in N$, then $A^{4 n}$ equals
(a) $\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$
(b) $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
(c) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
(d) $\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$
(c) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Here,
$A=\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right]\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right]=\left[\begin{array}{cc}i^{2} & 0 \\ 0 & i^{2}\end{array}\right]$
$\Rightarrow A^{3}=A^{2} . A=\left[\begin{array}{cc}i^{2} & 0 \\ 0 & i^{2}\end{array}\right]\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right]=\left[\begin{array}{cc}-i & 0 \\ 0 & -i\end{array}\right]$
$\Rightarrow A^{4}=A^{3} \cdot A=\left[\begin{array}{cc}-i & 0 \\ 0 & -i\end{array}\right]\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
So, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ is repeated on multiple of 4 and $4 n$ is a multiple of 4 .
Thus,
$A^{4 n}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
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