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If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, show that $A^2-5 A+7 I=0$
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$\mathrm{A}^2=\left[\begin{array}{ll}3 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$ L.H.S. $=\mathrm{A}^2-5 \mathrm{~A}+7 \mathrm{I}$ $=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ $=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=$ R.H.S. Hence $\mathrm{A}^2-5 \mathrm{~A}+7 \mathrm{I}=0$
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