Given that: $$AB=\begin{pmatrix} 2020 & 1\\ 2021 & 1 \end{pmatrix}$$ where $A$ and $B$ are $2×2$ matrices, find $BA-(BA)^{-1}$.
I was given this problem in a math contest, but I didn't know how to approach it. My guess is that maybe it has something to do with the identity $(AB)^{-1}=B^{-1}A^{-1}$, but I don't know how I can get from $B^{-1}A^{-1}$ to $(BA)^{-1}$, nor how I can find $BA$ given $AB$.
Any help would be appreciated.
The solution is unique. Hint: Note that $\det(AB)=-1$ so we must also have $\det(BA)=-1$. Then write $BA = \pmatrix{a & b \\ c & d}$ and calculate $BA-(BA)^{-1}$. You realize that the resulting expression just involves the trace of $BA$ (thus the same as the trace of $AB$).
EDIT: Just realized that you get a general formula of the above phenomena in the following way: If $A$ and $B$ are invertible and $M=AB$ satisfies $$ M^n + a_{n-1} M^{n-1} + \cdots a_0 {\bf 1} = 0 $$ then $M'=BA$ is a root of the same polynomial. In the present case you have: $$ M^2 - ({\rm trace} M )\ M + \det(M) \ {\bf 1} = M^2 - 2021 M - {\bf 1} = 0$$ or equivalently $$ M - (M)^{-1} = 2021 \ {\bf 1}$$ and then $M'=BA$ verifies the same equation, thus yielding the wanted result.