Find all $2 \times 2$ matrices $A$ such that $AB = BA$ for every $2 \times 2$ matrix $B$ Find all possible $2 \times 2$ matrices A that for any $2 \times 2$ matrix B, 
AB = BA.

Hint: AB = BA must hold for all B. Try matrices B that have lots of zero entries.

I'm clueless as to how to solve this problem. How should I start it? I tried plugging in values for B that "have lots of zero entries" but didn't seem to see anything that could help.
Consider the following four matrices:

$$\left(\begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 0\end{array}\right), \quad \left(\begin{array}{cc} 0 &amp; 1 \\ 0 &amp; 0\end{array}\right), \quad \left(\begin{array}{cc} 0 &amp; 0 \\ 1 &amp; 0\end{array}\right), \quad \left(\begin{array}{cc} 0 &amp; 0 \\ 0 &amp; 1\end{array}\right).$$

See what happens when you solve the equation $AB = BA$ for each of those four (let $B$ be each one of those four). To facilitate it, write $A = \left(\begin{array}{cc} a &amp; b \\ c &amp; d\end{array}\right)$. You will get a set of equations for the entries of $a$ which are easily solved. This trick is quite general.
Given a matrix B, consider any matrix that is a polynomial on $B$ with coefficients {$c_k$} in the field you are working with, i.e., take all $A$ with $$A=c_0I+c_1B+c_2B^2+...+c_nB^n$$. I actually think this gives you all such matrices $A$ , but I am not sure.