If $A^2 = I, B^2 = \begin{bmatrix}3 & 2\\-2 & -1 \end{bmatrix}, AB = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$, find $BA$.
What I did was the following: $A^2 = I ; AA^{-1} = I {\implies} A^2 = AA^{-1}$ $(AB)^2 = ABAB {\implies} \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}^2 = I {\implies} ABAB = I {\implies} BA = A^{-1}B^{-1} {\implies} BA = (BA)^{-1} {\implies} BA = \begin{bmatrix}±1 & 0 \\ 0 & ±1\end{bmatrix}$
which I'm not sure whether it's correct or not and if it's correct, which of these ±s to select from.
Thanks for your helps in advance.
Hint: Use that $0={\rm tr}(AB)={\rm tr}(BA)$ and $-1=\det(AB)=\det(BA)$.