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    \[\text { If } A=\left[\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right], \text { such that } A^{2}-4 A+3 I=0 \text {, then } A^{-1}=\]

(A) \frac{-1}{3}\left[\begin{array}{cc}2 & 1 \\ 1 & 2\end{array}\right] (B) \frac{-1}{3}\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right] (C) \frac{1}{3}\left[\begin{array}{cc}-2 & -1 \\ 1 & -2\end{array}\right] (D) \frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]
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    \[\text { If } A=\left[\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right], \text { such that } A^{2}-4 A+3 I=0 \text {, then } A^{-1}=\]

(A) \frac{-1}{3}\left[\begin{array}{cc}2 & 1 \\ 1 & 2\end{array}\right] (B) \frac{-1}{3}\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right] (C) \frac{1}{3}\left[\begin{array}{cc}-2 & -1 \\ 1 & -2\end{array}\right] (D) \frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]

The correct option is option (D)

\therefore \mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A}=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right] \cdot\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}4+1 & -2-2 \\ -2-2 & 1+4\end{array}\right]=\left[\begin{array}{cc}5 & -4 \\ -4 & 5\end{array}\right] \cdots \cdot

4 \mathrm{~A}-3 \mathrm{I}=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]-3\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}8 & -4 \\ -4 & 8\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]=\left[\begin{array}{cc}5 & -4 \\ -4 & 5\end{array}\right] …… (ii)

From (i) and (ii), we get \mathrm{A}^{2}=4 \mathrm{~A}-3 \mathrm{I}

=>  \mathrm{A}^{2}=4 \mathrm{~A}-3 \mathrm{I}

Pre-multiplying both sides by \mathrm{A}^{-1}
\begin{array}{l} \mathrm{A}^{-1} \cdot \mathrm{A}^{2}=\mathrm{A}^{-1} \cdot(4 \mathrm{~A}-3 \mathrm{I}) \\ \Rightarrow\left(\mathrm{A}^{-1} \cdot \mathrm{A}\right) \cdot \mathrm{A}=4 \mathrm{~A}^{-1} . \mathrm{A}-3 \mathrm{~A}^{-1} \cdot \mathrm{I} \\ \Rightarrow \mathrm{IA}=4 \mathrm{I}-3 \mathrm{~A}^{-1} \\ \Rightarrow \mathrm{A}=4 \mathrm{I}-3 \mathrm{~A}^{-1} \\ \Rightarrow 3 \mathrm{~A}^{-1}=4 \mathrm{I}-\mathrm{A} \\ \Rightarrow \mathrm{A}^{-1}=\frac{1}{3}\left(4\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right]\right)=\frac{1}{3}\left(\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right]-\left[\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right]\right) \Rightarrow \\ \frac{1}{3}\left[\begin{array}{cc} 2 & +1 \\ +1 & 2 \end{array}\right]\ \end{array}

i

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