If $\mathrm{M,N}$ are $3\times 2, 2 \times 3$ matrices such that $\mathrm{MN}=$ is given. Then $\mathrm{det(NM)}$ is? 
If $\mathrm{M,N}$ are $3\times 2, 2 \times 3$ matrices such that $\mathrm{MN}=\pmatrix{8& 2 & -2\\2& 5& 4\\-2& 4&5}$, then $\mathrm{det(NM)}$ is?
($\mathrm{NM}$ is invertible.)

$\mathrm{det(MN)}$ must be (and is) zero. But how to find $\mathrm{det(NM)}$? Any hint?

One way to proceed, not knowing the relevant theorem( new one for me as well) is to guess that, being symmetric, this is a Gram matrix. Look for integer rectangles...
$$
\left(
\begin{array}{rr}
2 & 2 \\
2 & -1 \\
1 & -2 
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 2 & 1 \\
2 & -1 & -2
\end{array}
\right) =
\left(
\begin{array}{rrr}
8 & 2 & -2 \\
2 & 5 & 4 \\
-2 & 4 & 5
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr}
2 & 2 & 1 \\
2 & -1 & -2
\end{array}
\right)
\left(
\begin{array}{rr}
2 & 2 \\
2 & -1 \\
1 & -2 
\end{array}
\right) =
\left(
\begin{array}{rr}
9 & 0 \\
0 & 9
\end{array}
\right)
$$
Indeed, the first characteristic polynomial is $x^3 - 18 x^2 + 81x$ and the second is $x^2 - 18 x + 81$
