# Solve this

Question:

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) $\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$

Solution:

(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,

$A^{4 n}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$