Question

Show that the matrix $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$ satisfies the equation A³ − 4A² + A = O

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \therefore A^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}4+3& 6+6\\ 2+2& 3+4\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right] \\ \Rightarrow A^3=A^{2}A \\ \Rightarrow A^3=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^3=\left[\begin{array}{cc}14+12& 21+24\\ 8+7& 12+14\end{array}\right] \\ \Rightarrow A^3=\left[\begin{array}{cc}26& 45\\ 15& 26\end{array}\right] \\ \mathrm{Now},A^3-4A^2+A \\ \Rightarrow A^3-4A^2+A=\left[\begin{array}{cc}26& 45\\ 15& 26\end{array}\right]-4\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]+\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^3-4A^2+A=\left[\begin{array}{cc}26& 45\\ 15& 26\end{array}\right]-\left[\begin{array}{cc}28& 48\\ 16& 28\end{array}\right]+\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^3-4A^2+A=\left[\begin{array}{cc}26-28+2& 45-48+3\\ 15-16+1& 26-28+2\end{array}\right] \\ \Rightarrow A^3-4A^2+A=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=O \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$