We have $A,B$ $(2×2)$ matrices with complex entries. We know $AB≠BA$. Show that $A+B=AB+BA$ if and only if $\text{Tr}(A)=\text{Tr}(B)=\text{Tr}(AB)=1$.
I tried writing $A=X+Y$ and $B=X-Y$ so we can get $X=X^{2}-Y^{2}$ but from here I know what to do.
Let $S=A+B$. Define $\tau_s=\operatorname{tr}(S),\,\delta_s=\det(S)$ and define $\tau_a,\tau_b,\delta_a,\delta_b$ analogously. Then $$ \begin{align} &AB+BA-(A+B)\\ &=S^2-(A^2+B^2)-S\\ &=(S^2-S)-A^2-B^2\\ &=(\tau_s-1)S-\delta_sI-(\tau_aA-\delta_aI)-(\tau_b B-\delta_bI)\\ &=(\tau_b-1)A+(\tau_a-1)B+(\delta_a+\delta_b-\delta_s)I.\tag{1}\\ \end{align} $$ Now suppose $AB+BA=A+B$, so that the sum on line $(1)$ is zero. If one of $\tau_b-1$ and $\tau_a-1$ is nonzero, then one of $A$ and $B$ is a linear combination of the other and the identity matrix, but this is impossible because $A$ and $B$ by assumption do not commute. Hence we must have $\tau_a=\tau_b=1$ and $\operatorname{tr}(AB)=\frac{1}{2}\operatorname{tr}(AB+BA)=\frac{1}{2}\operatorname{tr}(A+B)=1$.
Conversely, if $\tau_a=\tau_b=\operatorname{tr}(AB)=1$, then $AB+BA-(A+B)=(\delta_a+\delta_b-\delta_s)I$, by $(1)$. Since the LHS is traceless, the RHS, which is a scalar multiple of $I$, must be zero. Therefore $AB+BA=A+B$.