Question

If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$, find $A^{-1}$ and prove that $A^2-4A-5I=O$

Answer 1

$$\begin{gathered} A=\left[122 \\ 212 \\ 221\right] \\ \Rightarrow \left|A\right|=\left|122 \\ 212 \\ 221\right|=1\left(1-4\right)-2\left(2-4\right)+2\left(4-2\right)=-3+4+4=5 \\ \mathrm{Since},\left|A\right|\ne 0 \\ \mathrm{Hence},A\mathrm{is}\mathrm{invertible}. \\ \mathrm{Now}, \\ {A}^2=\left[122 \\ 212 \\ 221\right]\left[122 \\ 212 \\ 221\right]=\left[1+4+42+2+42+4+2 \\ 2+2+44+1+44+2+2 \\ 2+4+24+2+21+4+4\right]=\left[988 \\ 898 \\ 889\right] \\ \mathrm{Now},A^2-4A-5I=\left[988 \\ 898 \\ 889\right]-4\left[122 \\ 212 \\ 221\right]-5\left[100 \\ 010 \\ 001\right]=\left[9-4-58-8-08-8-0 \\ 8-8-09-4-58-8-0 \\ 8-8-08-8-09-4-5\right]=\left[000 \\ 000 \\ 000\right]=O \\ \Rightarrow A^2-4A-5I=O\left[\mathrm{Proved}\right] \\ \mathrm{Again},A^2-4A-5I=O \\ \Rightarrow A^{-1}\left(A^2-4A-5I\right)=A^{-1}O[\mathrm{Pre}-\mathrm{multiplying}\mathrm{with}{A}^{-1}] \\ \Rightarrow A^{-1}{A}^2-4A^{-1}A-5A^{-1}=O \\ \Rightarrow A-4I=5A^{-1} \\ \Rightarrow 5A^{-1}=\left[122 \\ 212 \\ 221\right]-4\left[100 \\ 010 \\ 001\right]=\left[1-42-02-0 \\ 2-01-42-0 \\ 2-02-01-4\right]=\left[-322 \\ 2-32 \\ 22-3\right] \\ \Rightarrow A^{-1}=\frac{1}{5}\left[-322 \\ 2-32 \\ 22-3\right] \end{gathered}$$