Question

Show that $A=\left[\begin{array}{cc}6& 5\\ 7& 6\end{array}\right]$ satisfies the equation $x^2-12x+1=O$. Thus, find A⁻¹.

Answer 1

$$\begin{gathered} A=\left[65 \\ 76\right] \\ \therefore A^2=\left[7160 \\ 8471\right] \\ \mathrm{If}{I}_2\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{matrix}\mathrm{of}\mathrm{order}2,\mathrm{then} \\ {A}^2-12A+I_2=\left[7160 \\ 8471\right]-12\left[65 \\ 76\right]+\left[10 \\ 01\right] \\ \Rightarrow A^2-12A+I_2=\left[71-72+160-60+0 \\ 84-84+071-72+1\right] \\ \Rightarrow A^2-12A+I_2=0 \\ \mathrm{Thus},A\mathrm{satisfies}{x}^2-12x+1=0. \\ \mathrm{Now}, \\ {A}^2-12A+I_2=0 \\ \Rightarrow I_2=12A-A^2 \\ \Rightarrow A^{-1}{I}_2=A^{-1}\left(12A-A^2\right)\left[\mathrm{Pre}-\mathrm{multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}{A}^{-1}\right] \\ \Rightarrow A^{-1}=12I_2-A \\ \Rightarrow A^{-1}=12\left[10 \\ 01\right]-\left[65 \\ 76\right] \\ \Rightarrow A^{-1}=\left[12-60-5 \\ 0-712-6\right] \\ \Rightarrow A^{-1}=\left[6-5 \\ -76\right] \end{gathered}$$