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Question:

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

Solution:

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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