Matrices - given $AB$, how to find determinant of $BA$ ?? 
Let A$_{3×2}$ and B$_{2×3}$ be matrices such that their product $AB$ is
  $$AB=\begin{pmatrix}
8&2 & -2\\
2&5&4 \\
-2&4&5 \\
\end{pmatrix}$$
  And $BA$ is nonsingular
Find the determinant of $BA$.

I have no idea , how to solve this type of question. All I could notice is that $|AB| = 0$ and it's a symmetric matrix. I tried assuming a general matrix , but I get simply too many unknowns and very few equations.

A direct calculation is also possible in case one  wants to find out the answer easily. If we denote the coefficients of $A$ by $a_i$ and the ones of $B$ by $b_j$, and the given matrix by $C$, then the matrix equation $AB=C$ is equivalent to equations in $a_i,b_j$. We can solve them case by case. 
The first equation is $a_1b_1 + a_2b_4 =8$. 
For $a_1=0$ we obtain $a_2\neq 0$ and
$$
b_4=\frac{8}{a_2}, b_5= \frac{2}{a_2}, b_6=-\frac{2}{a_2},
a_4= \frac{a_2a_3b_3 - 4a_2}{2}, b_3= -\frac{1}{a_3(a_3b_2 - 9}, b_2= \frac{1}{(4a_3)(a_3b_1 + 18)}, a_6= -\frac{a_2a_5b_1 + 2a_2}{8}, a_5=a_3.
$$
Then we obtain
$$
BA=\begin{pmatrix}9 & 0 \cr 0 & 9\end{pmatrix}.
$$
The other case $a_1\neq 0$ is similar. Note that $\det(AB)\neq \det(BA)$ in general, but $tr(AB)=tr(BA)$ is true in general. So the determinant is $81$.
