Suppose we have the following zero trace matrix:
$$M = \begin {pmatrix}-b_1{_2} & s1 & 0\\-b_2{_2} & b_1{_2} & 0\\-b_3{_2} & s2 & 0 \end {pmatrix}$$
Because it is has zero trace, it has a commutator such that:
$$M = AB-BA$$
I've chosen the elements for $M$ carefully in order to pose this question as clearly as I can.
A solution for $A$ and $B$ in this case is:
$$A = \begin {pmatrix}0 & -(b_3{_2}-s_1)/b_2{_2} & 1\\1 & 0 & 0\\1 & -(b_1{_2}+b_3{_2}-s_2)/b_2{_2} & 1 \end {pmatrix}$$
$$B = \begin {pmatrix}0 & b_1{_2} & 0\\0& b_2{_2} & 0\\0 & b_3{_2} & 0 \end {pmatrix}$$
If I choose $b_2{_2}\neq0$, then there is no problem. But, if I choose $b_2{_2}=0$ then there is division by zero in $A$. Yet, in this case $M$ still has a zero trace.
Does the possible division by zero ruin the solution for $A$ and $B$?
Is $M$ still a commutator if $b_2{_2}=0$?
In generating your solution for A and B, there was clearly a division by $b_{22}$ along the way; going back and re-doing the process that got you A and B would "undo" the divide by zero problem and give you a valid commutator relation.
For example, I solve $ax^2 + bx + c = 0$ with $x = \frac{-b +/-\sqrt{b^2-4ac}}{2a}$. However, my solution will be ruined in a sense if I send $a$ to zero, but that doesn't mean the equation can't be solved.