Question

If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, show that $A^2-5A+7I=O$. Hence, find A⁻¹.

Answer 1

$$\begin{gathered} A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right] \\ \mathrm{and} \\ {A}^2-5A+7I=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right] \\ \Rightarrow A^2-5A+7I=\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=O \\ \mathrm{Now}, \\ {A}^2-5A+7I=0 \\ \Rightarrow A^2-5A=-7I \\ \Rightarrow A^{-1}{A}^2-5A^{-1}A=-7A^{-1}I\left[\mathrm{Pre}-\mathrm{multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}{A}^{-1}\right] \\ \Rightarrow A-5I=-7A^{-1} \\ \Rightarrow A^{-1}=-\frac{1}{7}\left(A-5I\right) \\ \Rightarrow A^{-1}=-\frac{1}{7}\left\{\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]-\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]\right\}=\frac{1}{7}\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right] \end{gathered}$$