Question

If $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$, verify that $A^2-4A+I=O,\mathrm{where}I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\mathrm{and}O=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$. Hence, find A⁻¹.

Answer 1

$$\begin{gathered} A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \therefore A^2=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right] \\ \mathrm{and} \\ {A}^2-4A+I=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]-\left[\begin{array}{cc}8& 12\\ 4& 8\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A^2-4A+I=\left[\begin{array}{cc}7-8+1& 12-12+0\\ 4-4+0& 7-8+1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=O \\ \Rightarrow A^2-4A+I=0 \\ \Rightarrow A^2-4A=-I \\ \Rightarrow A^{-1}{A}^2-4A{A}^{-1}=-I{A}^{-1}\left[\mathrm{Pre}-\mathrm{multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}{A}^{-1}\right] \\ \Rightarrow A-4I=-A^{-1} \\ \Rightarrow A^{-1}=4I-A \\ \Rightarrow A^{-1}=\left\{\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]-\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\right\}=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right] \end{gathered}$$