Solve for matrices $A$ and $B$:
$2A + B = \begin{bmatrix}
3 & -4 \\
2 & 7
\end{bmatrix}$ and $A - 2B = \begin{bmatrix}
4 & 3 \\
1 & 1
\end{bmatrix}$
Given matrix equations:
$2A + B = \begin{bmatrix}
3 & -4 \\
2 & 7
\end{bmatrix}$ $\;\;\; \cdots \; (1)$; $\;\;$ $A - 2B = \begin{bmatrix}
4 & 3 \\
1 & 1
\end{bmatrix}$ $\;\;\; \cdots \; (2)$
We have from equation $(2)$
$A = \begin{bmatrix}
4 & 3 \\
1 & 1
\end{bmatrix} + 2 B$ $\;\;\; \cdots \; (3)$
$\implies$ $2A = \begin{bmatrix}
8 & 6 \\
2 & 2
\end{bmatrix} + 4 B$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(1)$ becomes
$\begin{bmatrix}
8 & 6 \\
2 & 2
\end{bmatrix} + 4 B + B = \begin{bmatrix}
3 & -4 \\
2 & 7
\end{bmatrix}$
i.e. $\;$ $5 B = \begin{bmatrix}
3 & -4 \\
2 & 7
\end{bmatrix} - \begin{bmatrix}
8 & 6 \\
2 & 2
\end{bmatrix} = \begin{bmatrix}
3 - 8 & -4 - 6 \\
2 - 2 & 7 - 2
\end{bmatrix} = \begin{bmatrix}
-5 & -10 \\
0 & 5
\end{bmatrix}$
$\implies$ $B = \begin{bmatrix}
-1 & -2 \\
0 & 1
\end{bmatrix}$ $\;\;\; \cdots \; (5)$
Substituting the value of $B$ from equation $(5)$ in equation $(3)$ gives
$A = \begin{bmatrix}
4 & 3 \\
1 & 1
\end{bmatrix} + 2 \begin{bmatrix}
-1 & -2 \\
0 & 1
\end{bmatrix}$
i.e. $\;$ $A = \begin{bmatrix}
4 & 3 \\
1 & 1
\end{bmatrix} + \begin{bmatrix}
-2 & -4 \\
0 & 2
\end{bmatrix}$
i.e. $\;$ $A = \begin{bmatrix}
4 - 2 & 3 - 4 \\
1 + 0 & 1 + 2
\end{bmatrix} = \begin{bmatrix}
2 & -1 \\
1 & 3
\end{bmatrix}$
$\therefore \;$ $A = \begin{bmatrix}
2 & -1 \\
1 & 3
\end{bmatrix}$; $\;$ $B = \begin{bmatrix}
-1 & -2 \\
0 & 1
\end{bmatrix}$