Proof that $(A*B*A^{-1}*B^{-1})$ has determinant 1. (for 2x2 matrices)

Question

I want to prove the following for real invertible $2 \times 2$ matrices $A$ and $B$:

$\det(ABA^{-1}B^{-1})=1$

I tried to write it out for random numbers, and that seems to work out well. But when I tried to write it out in general, it became far too much paperwork. So I think there must be a simpler/shorter method, but I don't know what.

Thank you in advance for your comments!

Answer 1

Hints: $det(XY)=\det(X) \det(Y)$, hence, if $X$ is invertible: $1=\det(X) \det(X^{-1})$.

Answer 2

Hint: $\det (AB) = \det(A) \det(B)$, $\det(I)=1$.