Question

If $A=\left[\begin{array}{cc}2& 3\\ -1& 0\end{array}\right]$, show that A² − 2A + 3I₂ = O

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}2& 3\\ -1& 0\end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}2& 3\\ -1& 0\end{array}\right]\left[\begin{array}{cc}2& 3\\ -1& 0\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}4-3& 6+0\\ -2+0& -3+0\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}1& 6\\ -2& -3\end{array}\right] \\ {A}^2-2A+3I_2 \\ \Rightarrow A^2-2A+3I_2=\left[\begin{array}{cc}1& 6\\ -2& -3\end{array}\right]-2\left[\begin{array}{cc}2& 3\\ -1& 0\end{array}\right]+3\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A^2-2A+3I_2=\left[\begin{array}{cc}1& 6\\ -2& -3\end{array}\right]-\left[\begin{array}{cc}4& 6\\ -2& 0\end{array}\right]+\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right] \\ \Rightarrow A^2-2A+3I_2=\left[\begin{array}{cc}1-4+3& 6-6+0\\ -2+2+0& -3+0+3\end{array}\right] \\ \Rightarrow A^2-2A+3I_2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0 \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$