Here is the equation I'm working on:
\begin{align} \begin{bmatrix}2a+b&a-2b\\5c-d&4c+3d\end{bmatrix} = \begin{bmatrix}4&-3\\11&24\end{bmatrix}\\ \end{align}
My solutions:
Row 1 column 1 \begin{align} 2a + b = 4\\ 2a = 4 - b\\ a = 2 - \frac 12b\\ \end{align}
Row 1 column 2 \begin{align} a - 2b = -3\\ 2 - \frac 12b - 2b = -3\\ 2 - \frac 52b = -3\\ 2\left(-\frac 52b \right) = 2(-5)\\ -5b = -10\\ b = 2 \end{align}
Row 2 column 1 \begin{align} 5c - d = 11\\ 5c = 11 + d\\ c = \frac {11}{5} + \frac {1}{5}d\\ \end{align}
Row 2 column 2 \begin{align} 4c + 3d = 24\\ 4\left(\frac {11}{5} + \frac {1}{5}d \right) + 3d = 24\\ \frac {44}{5} + \frac {4}{5}d + 3d = 24\\ \frac {44}{5} + \frac {19}{5}d = 24\\ \frac {19}{5}d = 24 - \frac {44}{5}\\ \frac {19}{5}d = \frac {76}{5}\\ 19d = 5\left(\frac {76}{5} \right)\\ d = \frac {76}{19}\\ d = 4\\ \end{align}
Based on b value which is 2, I've found the a value:
\begin{align}
2a + b = 4\\
2a + 2 = 4\\
2a = 2\\
a = 1\\
\end{align}
Likewise, found the c value based on d value which is 4:
\begin{align}
5c - 11d = 11\\
5c - 4 = 11\\
5c = 15\\
c = 3\\
\end{align}
Thus, \begin{align} a = 1, b = 2, c = 3, d = 4\\ \end{align}
I'm new to algebra and matrices. My question is: Is this the right approach to solving the problem?
Yes your approach is correct. As an alternative you could consider the $4$ by $5$ augmented matrix for the system in the unknowns $a$,$b$,$c$ and $d$
$$\left[\begin{array}{cccc|c} 2&1&0&0&4 \\ 1&0&0&-2&-3 \\ 0&0&5&-1&11 \\ 0&0&4&3&24 \end{array}\right]$$
and solve by gaussian elimination.