A commutator that dictates the dimension

Set $M=A+\omega B$, where $\omega=-\frac{1}{2}+i \frac{\sqrt{3}}{2}$. We have
$$\begin{aligned} M \bar{M} & =(A+\omega B)(A+\bar{\omega} B)=A^2+\omega B A+\bar{\omega} A B+B^2 \\ & =A B+\omega B A+\bar{\omega} A B=\omega(B A-A B), \end{aligned}$$because $\bar{\omega}+1=-\omega$. Since $\det(M \bar{M})=\det M . \det \bar{M}$ is a real number and $\operatorname{det} \omega(B A-A B)=\omega^n \operatorname{det}(B A-A B)$ and $\operatorname{det}(B A-A B) \neq 0$, then $\omega^n$ is a real number. This is possible only when $n$ is divisible by 3 .