Answer
See below
Work Step by Step
Let:
$A_1=\begin{bmatrix}
2 & 1\\
3 & 4
\end{bmatrix}$
$A_2=\begin{bmatrix}
-1 & 2\\
1 & 3
\end{bmatrix}$
Let $c_1,c_2$ be scalars such that
$c_1\begin{bmatrix}
2 & 1\\
3 & 4
\end{bmatrix}+c_2\begin{bmatrix}
-1 & 2\\
1 & 3
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}\\
\begin{bmatrix}
2c_1 & -c_1\\
3c_1 & 4c_1
\end{bmatrix}+\begin{bmatrix}
-c_2 & 2c_2\\
c_2 & 3c_2
\end{bmatrix}=\begin{bmatrix}
0 & 0 \\ 0 & 0
\end{bmatrix}\\\begin{bmatrix}
2c_1-c_2 & -c_1+2c_2\\
3c_1+c_2 & 4c_1+3c_2
\end{bmatrix}=\begin{bmatrix}
0 & 0 \\ 0 & 0
\end{bmatrix}$
We have the system: $2c_1-c_2=0\\
2c_2-c_1=0\\
3c_2+c_2=0\\
4c_1+3c_2=0\\
\rightarrow c_2=2c_1\\
2c_2-c_1=2(2c_1)-c_2=3c_2=0\rightarrow c_2=0$
We can see that the only solution to the system above is $(0,0)$
Hence $\{ A_1,A_2 \}$ is a linearly dependent set in $M_2(R)$
