If $\displaystyle \mathrm{A}=\left[\begin{array}{rr}2 & 3 \\ -1 & 2\end{array}\right]$, then show that $\displaystyle \mathrm{A}^{2}-4 \mathrm{~A}+7 \mathrm{I}=0$.

\[\begin{aligned} & \mathrm{A}^{2}=\left[\begin{array}{cc} 2 & 3 \\ -1 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & 12 \\ -4 & 1 \end{array}\right] . \\ & \text { L.H.S. }=\mathrm{A}^{2}-4 \mathrm{A}+7 \mathrm{I}=\left[\begin{array}{cc} 1 & 12 \\ -4 & 1 \end{array}\right]-\left[\begin{array}{cc} 8 & 12 \\ -4 & 8 \end{array}\right]+\left[\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=O=\text { R.H.S. } \end{aligned} \]