Answer
It has been proved that $AI_2= I_2 A=A$.
Work Step by Step
Let us suppose that $A=\begin{bmatrix} a & b \\ c & d\end{bmatrix}$
Here, we have the value of $AI_{2}$ by matrix multiplication is equivalent to:
$AI_2=\begin{bmatrix} a & b \\ c & d\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}=\begin{bmatrix} a& b \\ c &d \end{bmatrix} =A$
Now, the value of $I_{2}A$ by matrix multiplication is to:
$I_2A= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix} a & b \\ c & d\end{bmatrix}=\begin{bmatrix} a& b \\ c &d \end{bmatrix}=A $
This implies that it has been proved that $AI_2= I_2 A=A$.
