Problem 1981

a) Clearly, if $X = \alpha A + \alpha' I_2$, then $X$ and $A$ commute. Conversely, let $A = \begin{pmatrix} a_1 & a_2 \\ a'_1 & a'_2 \end{pmatrix}$ and $X = \begin{pmatrix} x_1 & x_2 \\ x'_1 & x'_2 \end{pmatrix}$. The equality $AX = XA$ implies
$$a_2 x'_1 = a'_1 x_2, \quad (1)$$
$$(a_1 - a'_2)x_2 + a_2(x'_2 - x_1) = 0, \quad (2)$$
$$(a_1 - a'_2)x'_1 + a'_1(x'_2 - x_1) = 0. \quad (3)$$
Since $A \neq aI_2$, $a \in \mathbb{C}$, either one of $a_2, a'_1$ is non-zero, or $a_2 = a'_1 = 0$ and $a_1 \neq a'_2$.
In the first case, if, for instance, $a_2 \neq 0$, we obtain
$$X = \frac{x_2}{a_2} A + \left( x_1 - \frac{a_1}{a_2} x_2 \right) I_2,$$
while in the second case
$$X = \frac{x_1 - x'_2}{a_1 - a'_2} A + \frac{a_1 x'_2 - a'_2 x_1}{a_1 - a'_2} I_2.$$

b) We prove that $C = \gamma I_2$, for some complex number $\gamma$, hence $C$ commutes with all matrices in $M_2(\mathbb{C})$. Suppose the contrary. Since $A$ and $C$ commute, there exist $\alpha$ and $\alpha'$, such that $A = \alpha C + \alpha' I_2$. Similarly, there exist $\beta$ and $\beta'$, such that $B = \beta C + \beta' I_2$, therefore $AB = BA$, a contradiction.