Find all $2\times 2$ matrices that commute with $AX = XA$? Find all $2\times 2$ matrices that commute with  $$ A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$$ where $AX = XA$.
Solution and ask:
$$ X = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$
$$AX = \left( \begin{array}{cc} 1\cdot a + 1\cdot c  & 1\cdot b + 1\cdot d  \\ 0\cdot a + 0\cdot c  & 0\cdot b + 0\cdot d  \end{array} \right) = \left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right)$$
$$XA = \left( \begin{array}{cc} a\cdot 1 + b\cdot 0  & a\cdot 1 + b\cdot 0  \\ c\cdot 1+d\cdot 0  & c\cdot 1+d\cdot 0  \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$
Since $AX = XA$, we obtain
$$\left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$
so that $a+c = a$, $b+d = a$, $0 = c$ and $0 = c$.
Is the calculation correct?
What is the value of $a,b,c,d$ ? 
$c = 0, a = ?, b = ?, d = ?$

Your calculations are correct. To summarize, you have found that $c=0$ and that $a=b+d$. This does not determine $a$, $b$ and $d$ uniquely; it only shows that such a matrix must be of the form
$$X = \left( \begin{array}{cc} b+d & b \\ 0 & d \end{array} \right).$$
Now the question that remains is; does every matrix of this form commute with $A$?
