Question

If $A=\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$, find the value of $\lambda$ so that $A^2=\lambda A-2I$. Hence, find A⁻¹.

Answer 1

$$\begin{gathered} A=\left[3-2 \\ 4-2\right] \\ \therefore {\mathrm{A}}^2=\left[1-2 \\ 4-4\right] \\ \mathrm{Given}: \\ {A}^2=\lambda A-2I...\left(1\right) \\ \Rightarrow \left[1-2 \\ 4-4\right]=\lambda \left[3-2 \\ 4-2\right]-2\left[10 \\ 01\right] \\ \Rightarrow \left[1-2 \\ 4-4\right]=\left[3\lambda -2\lambda \\ 4\lambda -2\lambda \right]-\left[20 \\ 02\right] \\ \Rightarrow \left[1-2 \\ 4-4\right]=\left[3\lambda -2-2\lambda \\ 4\lambda -2\lambda -2\right] \\ \mathrm{On}\mathrm{equating}\mathrm{corresponding}\mathrm{terms},\mathrm{we}\mathrm{get} \\ -2\lambda =-2 \\ \Rightarrow \lambda =1 \\ \mathrm{On}\mathrm{substituting}\mathrm{\lambda }=1\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ {A}^2=A-2I \\ \Rightarrow A^2-A=-2I \\ \Rightarrow A-A^2=2I \\ \Rightarrow A^{-1}\left(A-A^2\right)=A^{-1}\times 2I\left(\mathrm{Pre}-\mathrm{multiplying}\mathrm{both}\mathrm{sides}\mathrm{with}{A}^{-1}\right) \\ \Rightarrow I-A=2A^{-1} \\ 2A^{-1}=\left[10 \\ 01\right]-\left[3-2 \\ 4-2\right]=\left[1-30+2 \\ 0-41+2\right] \\ \Rightarrow A^{-1}=\frac{1}{2}\left[-22 \\ -43\right] \end{gathered}$$