设
\[
A=\begin{pmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}\\
a_{31} & a_{32}\\
\end{pmatrix},\quad
B=\begin{pmatrix}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23}\\
\end{pmatrix}
\]
则
\[
A\cdot B=\begin{pmatrix}
a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} & a_{11}b_{13}+a_{12}b_{23}\\
a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} & a_{21}b_{13}+a_{22}b_{23}\\
a_{31}b_{11}+a_{32}b_{21} & a_{31}b_{12}+a_{32}b_{22} & a_{31}b_{13}+a_{32}b_{23}\\
\end{pmatrix}=
\begin{pmatrix}
8 & 2 & -2\\
2 & 5 & 4\\
-2 & 4 & 5
\end{pmatrix}
\]
并设 $BA=\begin{pmatrix}x & y\\ u & v\end{pmatrix}$, 即
\[
B\cdot A=\begin{pmatrix}
b_{11}a_{11}+b_{12}a_{21}+b_{13}a_{31} & b_{11}a_{12}+b_{12}a_{22}+b_{13}a_{32}\\
b_{21}a_{11}+b_{22}a_{21}+b_{23}a_{31} & b_{21}a_{12}+b_{22}a_{22}+b_{23}a_{32}\\
\end{pmatrix}=
\begin{pmatrix}
x & y\\
u & v
\end{pmatrix}
\]
易见 $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$ (这实际上是一般性的结果, 见问题2747), 即有
\[
x+v=8+5+5=18
\]
利用问题1798 中的结论, $|I_3-AB|=|I_2-BA|$, 可推出 $|BA|=81$.
具体的, 经过计算,
\[
|I_3-AB|=\begin{vmatrix}
-7 & -2 & 2\\
-2 & -4 & -4\\
2 & -4 & -4
\end{vmatrix}=64
\]
因此,
\[
\begin{vmatrix}
1-x & -y\\
-u & 1-v
\end{vmatrix}=64.
\]
这推出
\[
\begin{split}
&(1-x)(1-v)-uy=64\\
\Rightarrow\ &1-(x+v)+xv-uy=64\\
\Rightarrow\ &1-18+|BA|=64\\
\Rightarrow\ &|BA|=81.
\end{split}
\]