If $A=\left[\begin{array}{ll}2 & -1 \\ 3 & -2\end{array}\right]$, then $A^{n}=$
(a) $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, if $n$ is an even natural number
(b) $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, if $n$ is an odd natural number
(c) $A=\left[\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right]$, if $n \in N$
(d) none of these
Disclaimer: In all option, the power of $A$ (i.e. $n$ is missing)
(a) $A^{n}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, if $n$ is an even natural number
$A=\left[\begin{array}{ll}2 & -1 \\ 3 & -2\end{array}\right]$
$A^{2}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow A \times A=I$
$\Rightarrow A^{-1}=A$
Generally,
$A^{n}=\left(A A^{-1}\right)^{n / 2}$ when $n$ is even.
$A^{n}=A\left(A A^{-1}\right)^{n / 2}=A$ when $n$ is odd.
Thus, $A^{n}=I$ when $n$ is even.
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