Question

If $A=\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right],\mathrm{then}{A}^n=$

(a) $A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, if n is an even natural number

(b) $A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, if n is an odd natural number

(c) $A=\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right],\mathrm{if}n\in N$

(d) none of these

Answer 1

Disclaimer: In all option, the power of A (i.e. n is missing)

(a) $A^n=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, if n is an even natural number

$$\begin{gathered} A=\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right] \\ {A}^2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A\times A=I \\ \Rightarrow A^{-1}=A \end{gathered}$$

Generally,

$$\begin{gathered} A^n=(A{A}^{-1}{)}^{n/2}\mathrm{when}n\mathrm{is}\mathrm{even}. \\ {A}^n=A(A{A}^{-1}{)}^{n/2}=A\mathrm{when}n\mathrm{is}\mathrm{odd}. \\ \mathrm{Thus},A^n=I\mathrm{when}n\mathrm{is}\mathrm{even}. \end{gathered}$$