Question

If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$, then prove that A² − 4A − 5I = O.

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right] \\ Now, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{ccc}1+4+4& 2+2+4& 2+4+2\\ 2+2+4& 4+1+4& 4+2+2\\ 2+4+2& 4+2+2& 4+4+1\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right] \\ {A}^2-4A-5I \\ \Rightarrow A^2-4A-5I=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-4\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]-5\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ \Rightarrow A^2-4A-5I=\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]-\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right] \\ \Rightarrow A^2-4A-5I=\left[\begin{array}{ccc}9-4-5& 8-8-0& 8-8-0\\ 8-8-0& 9-4-5& 8-8-0\\ 8-8-0& 8-8-0& 9-4-5\end{array}\right] \\ \Rightarrow A^2-4A-5I=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=0 \end{gathered}$$

Hence proved.