Long solution pls

$A = a_{i,j},~B=b_{i,j}\\ AB-BA = \begin{pmatrix} a_{1,2} b_{2,1}-a_{2,1} b_{1,2} \\ \left(a_{1,1}-a_{2,2}\right) b_{1,2}+a_{1,2} \left(b_{2,2}-b_{1,1}\right) \\ \left(a_{2,2}-a_{1,1}\right) b_{2,1}+a_{2,1} \left(b_{1,1}-b_{2,2}\right) \\ a_{2,1} b_{1,2}-a_{1,2} b_{2,1} \\ \end{pmatrix} = 0$

$\text{From the first row we see }a_{1,2}=a_{2,1}=0\\ \text{Then from row 2, }a_{1,1}=a_{2,2}\\ \text{This agrees with row 3, and row 4 agrees with row 1 }\\ A = \lambda I_{2,2},~\lambda \in \mathbb{C}$