Dimension of a subspace of $2\times2$ Matrices The question is asking to find the dimension of the subspace $W$, where, $V = M_{2,2}$,
$$
W = \{A \in V: AB= BA\}
$$
where
$$B=\begin{bmatrix}1&2\\3&4\\\end{bmatrix}$$
I defined an arbitrary matrix $A$ which contains the entries $a,b,c,d$. Then I considered the equailty and multiplied the matrices in both sides then got:
$$\begin{bmatrix}a+3b&2a+4b\\c+3d&2c+4d\\\end{bmatrix} = \begin{bmatrix}a+2c&b+2d\\3a+4c&3b+4d\\\end{bmatrix}$$
Then I set:
1)$a+3b = a+2c$
2) $2a+4b = b+2d$
3) $c+3d = 3a+4c$
4) $2c+4d = 3b+4d$
The problem is that things got complicated to find $a,b,c$ and $d$, which allow us to find the dimension! What is the next step?

Your equations give
\begin{align*}
3b&=2c\\
a+c&=d\\
\end{align*}
Meaning that once $a$ and $c$ are known you can deduce $b$ an $d$. 
Consequently the matrix $A$ has the following form
\begin{pmatrix} 
a & 2c/3 \\
c & a+c 
\end{pmatrix}
Can you conclude from there ?
