Question

Show that the matrix $A=\left[\begin{array}{cc}5& 3\\ 12& 7\end{array}\right]$ is root of the equation A² − 12A − I = O

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}5& 3\\ 12& 7\end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}5& 3\\ 12& 7\end{array}\right]\left[\begin{array}{cc}5& 3\\ 12& 7\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}25+36& 15+21\\ 60+84& 36+49\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}61& 36\\ 144& 85\end{array}\right] \\ {A}^2-12A-I \\ \Rightarrow A^2-12A-I=\left[\begin{array}{cc}61& 36\\ 144& 85\end{array}\right]-12\left[\begin{array}{cc}5& 3\\ 12& 7\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A^2-12A-I=\left[\begin{array}{cc}61& 36\\ 144& 85\end{array}\right]-\left[\begin{array}{cc}60& 36\\ 144& 84\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A^2-12A-I=\left[\begin{array}{cc}61-60-1& 36-36+0\\ 144-144+0& 85-84-1\end{array}\right] \\ \Rightarrow A^2-12A-I=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \mathrm{Since}A\mathrm{is}\mathrm{satisfying}\mathrm{the}\mathrm{equation}{A}^2-12A-I,\mathit{}A\mathrm{is}\mathrm{the}\mathrm{root}\mathrm{of}\mathrm{the}\mathrm{equation}{A}^2-12A-I\mathit{}\mathit{.} \end{gathered}$$