Question

If $A=\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]$, find x and y such that $A^2=xA+yI=O$. Hence, evaluate A⁻¹.

Answer 1

$$\begin{gathered} A=\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right] \\ \therefore A^2=\left[\begin{array}{cc}22& 27\\ 18& 31\end{array}\right] \\ \mathrm{Now}, \\ {A}^2-xA+yI=O \\ \Rightarrow \left[\begin{array}{cc}22& 27\\ 18& 31\end{array}\right]-\left[\begin{array}{cc}4x& 3x\\ 2x& 5x\end{array}\right]+\left[\begin{array}{cc}y& 0\\ 0& y\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}22-4x-y& 27-3x\\ 18-2x& 31-5x-y\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \mathrm{Thus},\mathrm{we}\mathrm{have} \\ 22-4x+y=0,27-3x=0,18-2x=0\mathrm{and}31-5x+y=0 \\ \Rightarrow -3x=-27 \\ \Rightarrow x=9 \\ \mathrm{On}\mathrm{putting}x=9\mathrm{in}22-4x+y=0,\mathrm{we}\mathrm{get} \\ 22-36+y=0 \\ \Rightarrow -14=-y \\ \Rightarrow y=14 \\ \mathrm{Now}, \\ {A}^2-9A+14I=0 \\ \Rightarrow A^2-9A=-14I \\ \Rightarrow A^{-1}{A}^2-9A{A}^{-1}=-14I{A}^{-1}\left[\mathrm{Pre}-\mathrm{multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}{A}^{-1}\right] \\ \Rightarrow A-9I=-14A^{-1} \\ \Rightarrow A^{-1}=-\frac{1}{14}\left(A-9I\right) \\ \Rightarrow A^{-1}=-\frac{1}{14}\left\{\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]-\left[\begin{array}{cc}9& 0\\ 0& 9\end{array}\right]\right\}=\frac{1}{14}\left[\begin{array}{cc}5& -3\\ -2& 4\end{array}\right] \end{gathered}$$