Question

If $A=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]$, n ∈ N, then A⁴ⁿ equals

(a) $\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right]$

(b) $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

(c) $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

(d) $\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right]$

Answer 1

(c) $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$


$$\begin{gathered} \mathrm{Here}, \\ A=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]=\left[\begin{array}{cc}{i}^2& 0\\ 0& i^2\end{array}\right] \\ \Rightarrow A^3=A^2.A=\left[\begin{array}{cc}{i}^2& 0\\ 0& i^2\end{array}\right]\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]=\left[\begin{array}{cc}-i& 0\\ 0& -i\end{array}\right] \\ \Rightarrow A^4=A^3.A=\left[\begin{array}{cc}-i& 0\\ 0& -i\end{array}\right]\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

So, $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ is repeated on multiple of 4 and 4n is a multiple of 4.

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