Question

Show that $A=\left[\begin{array}{cc}-8& 5\\ 2& 4\end{array}\right]$ satisfies the equation $A^2+4A-42I=O$. Hence, find A⁻¹.

Answer 1

$$\begin{gathered} A=\left[\begin{array}{cc}-8& 5\\ 2& 4\end{array}\right] \\ \therefore A^2=\left[\begin{array}{cc}74& -20\\ -8& 26\end{array}\right] \\ \mathrm{and} \\ {A}^2+4A-42I=\left[\begin{array}{cc}74& -20\\ -8& 26\end{array}\right]+\left[\begin{array}{cc}-32& 20\\ 8& 16\end{array}\right]-\left[\begin{array}{cc}42& 0\\ 0& 42\end{array}\right] \\ \Rightarrow A^2+4A-42I=\left[\begin{array}{cc}74-32-42& -20+20-0\\ -8+8-0& 26+16-42\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=O \\ \mathrm{Now}, \\ {A}^2+4A-42I=0 \\ \Rightarrow A^2+4A=42I \\ \Rightarrow A^{-1}{A}^2+4A^{-1}A=42I{A}^{-1}\left[\mathrm{Pre}-\mathrm{multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}{A}^{-1}\right] \\ \Rightarrow A+4I=42A^{-1} \\ \Rightarrow A^{-1}=\frac{1}{42}\left(A+4I\right) \\ \Rightarrow A^{-1}=\frac{1}{42}\left\{\left[\begin{array}{cc}-8& 5\\ 2& 4\end{array}\right]+\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]\right\}=\frac{1}{42}\left[\begin{array}{cc}-4& 5\\ 2& 8\end{array}\right] \end{gathered}$$