Answer
See below
Work Step by Step
Given $A_1=\begin{bmatrix}
1 & 2 \\ -1 & 3
\end{bmatrix},A_2=\begin{bmatrix}
-2 & 1 \\ 1 & -1
\end{bmatrix},A_3=\begin{bmatrix}
3 & 1 \\ -2 & 4
\end{bmatrix}$
We know that $span\{A_1,A_2\}=\{A \in M_2(R):A=c_1A_1+c_2A_2 \forall c_1,c_2 \in R\}\\
=\{A \in M_2(R):A=c_1\begin{bmatrix}
1 & 2 \\ -1 & 3
\end{bmatrix}+c_2\begin{bmatrix}
-2 & 1 \\ 1 & -1
\end{bmatrix} \forall c_1,c_2,c_3 \in R\}\\
=A \in M_2(R):A=\begin{bmatrix}
c_1 & 2c_1 \\ -c_1 & 3c_1
\end{bmatrix}+\begin{bmatrix}
-2c_2 & c_2 \\ c_2 & -c_2
\end{bmatrix}\forall c_1,c_2\in R\}\\
=A \in M_2(R):A=\begin{bmatrix}
c_1-2c_2 & 2c_1+c_2 \\ -c_1+c_2 & 3c_1-c_2
\end{bmatrix} \forall c_1,c_2\in R\}$
Let $c_1=1,c_2=-1$ we have $\begin{bmatrix}
1-2.1 & 2.1+(-1) \\ -1+(-1) & 3.1-(-1)
\end{bmatrix} =\begin{bmatrix}
3 & 1 \\ -2 & 4
\end{bmatrix} =B$
Hence, $B \in span \{A_1,A_2\}$
