Prove or Disprove :
$$\operatorname{adj}(AB-BA) = \operatorname{adj}(AB) - \operatorname{adj}(BA)$$
where $A$ and $B$ are arbitrary $n \times n$ matrices, and $\operatorname{adj}$ is the adjugate(classical adjoint) operator.
The statement is false, of course, since $\operatorname{adj}(A) = I = \begin{bmatrix} 1 \end{bmatrix}$ for any $1 \times 1$ matrix $A$.
I am now trying to show that the identity holds only for $n = 2$, as the title says. I have checked it for $2 \times 2$ matrices by brute force calculation($A = [a_{ij}]$, $B = [b_{ij}]$) and it does hold. Also, I found a counterexample for $n = 3$, which is just a random $3 \times 3$ matrix because I was pretty sure that it would not hold for $n > 2$.
Counterexample for $M_{33}$ :
$$A = \begin{bmatrix} 1 & 2 & 3 \\\ 4 & 5 & 6 \\\ 7 & 8 & 9 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 3 & 6 & 1 \\\ 2 & 5 & -1 \\\ -3 & 4 & -7 \end{bmatrix}$$
How can I show that the identity holds only for $n = 2$? The following is my approach.
We know that
$$\begin{align} (AB-BA) \operatorname{adj}(AB-BA) &= \det(AB-BA) I \\\ AB \operatorname{adj}(AB) &= \det(AB) I \\\ BA \operatorname{adj}(BA) &= \det(BA) I \end{align}$$
for all square matrices $A, B$. If the given identity is satisfied, then
$$\begin{align} \det(AB-BA)I &= (AB-BA) \operatorname{adj}(AB-BA) \\\ &= (AB-BA) (\operatorname{adj}(AB) - \operatorname{adj}(BA)) \end{align}$$
Hence, we have
$$\left( 2 \det(AB) - \det(AB-BA) \right) I = BA \operatorname{adj}(B) \operatorname{adj}(A) + AB \operatorname{adj}(A) \operatorname{adj}(B)$$
I thought about evaluating the trace of both sides since the trace is a linear operator - so you can separate two terms using the additivity property. However, I couldn't proceed further after the trace.