If $A=\left[\begin{array}{rr}4 & 2 \\ -1 & 1\end{array}\right]$, prove that $(A-2 D)(A-3 l)=0$
Given: $(A-2 I)(A-3 I)$
$\Rightarrow(A-2 I)(A-3 I)=\left(\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]-2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\right)\left(\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]-3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\right)$
$\Rightarrow(A-2 I)(A-3 I)=\left(\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]-\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]\right)\left(\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\right)$
$\Rightarrow(A-2 I)(A-3 I)=\left[\begin{array}{cc}4-2 & 2-0 \\ -1-0 & 1-2\end{array}\right]\left[\begin{array}{c}4-3 & 2-0 \\ -1-0 & 1-3\end{array}\right]$
$\Rightarrow(A-2 I)(A-3 I)=\left[\begin{array}{cc}2 & 2 \\ -1 & -1\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ -1 & -2\end{array}\right]$
$\Rightarrow(A-2 I)(A-3 I)=\left[\begin{array}{cc}2-2 & 4-4 \\ -1+1 & -2+2\end{array}\right]$
$\Rightarrow(A-2 I)(A-3 I)=\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow(A-2 I)(A-3 I)=0$
Hence proved.