Question

If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, show that A² − 5A + 7I₂ = O

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right] \\ {A}^2-5A+7I_2 \\ \Rightarrow A^2-5A+7I_2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A^2-5A+7I_2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right] \\ \Rightarrow A^2-5A+7I_2=\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right] \\ \Rightarrow A^2-5A+7I_2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \Rightarrow A^2-5A+7I_2=0 \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$