Question

Show that $A=\left[\begin{array}{cc}5& 3\\ -1& -2\end{array}\right]$ satisfies the equation $x^2-3x-7=0$. Thus, find A⁻¹.

Answer 1

$$\begin{gathered} A=\left[53 \\ -1-2\right] \\ {A}^2=\left[229 \\ -31\right] \\ \mathrm{If}{I}_2\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{matrix}\mathrm{of}\mathrm{order}2,\mathrm{then} \\ {A}^2-3A-7I_2=\left[229 \\ -31\right]-3\left[53 \\ -1-2\right]-7\left[10 \\ 01\right] \\ \Rightarrow A^2-3A-7I_2=\left[22-15-79-9-0 \\ -3+3+01+6-7\right]=\left[00 \\ 00\right]=0 \\ \Rightarrow A^2-3A-7I_2=0 \\ \mathrm{Thus},A\mathrm{satisfies}{x}^2-3x-7=0. \\ \mathrm{Now}, \\ {A}^2-3A-7I_2=0 \\ \Rightarrow A^2-3A=7I_2 \\ \Rightarrow A^{-1}\left(A^2-3A\right)=A^{-1}\times 7I_{\mathit{2}}\mathit{}\mathit{}\left[\mathrm{Pre}-\mathrm{multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}{A}^{-1}\right] \\ \Rightarrow A-3I_2=7A^{-1} \\ \Rightarrow \left[53 \\ -1-2\right]-3\left[10 \\ 01\right]=7A^{-1} \\ \Rightarrow A^{-1}=\frac{1}{7}\left[5-33-0 \\ -1-0-2-3\right]=\frac{1}{7}\left[23 \\ -1-5\right] \end{gathered}$$