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If $A=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]$, such that $A^{2}-4 A+3 I=0$, then $A^{-1}=$
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The correct answer is:
$\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
$A=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right] \quad \Rightarrow|A|=4-1=3 \quad$ and $(\operatorname{adj} A)=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
$\therefore A^{-1} \quad=\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
$\therefore A^{-1} \quad=\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
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