$B = \left[ {\begin{array}{*{20}{c}}
  0 \\ 
  { - 1} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  0 
\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  { - 1} 
\end{array}} \right]\,\,\,,\,\,A = \left[ {\begin{array}{*{20}{c}}
  { - 1} \\ 
  1 \\ 
  3 
\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}
  { - 4} \\ 
  { - 2} \\ 
  0 
\end{array}} \right]$

$AB = \left[ {\begin{array}{*{20}{c}}
  { - 1} \\ 
  1 \\ 
  3 
\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}
  { - 4} \\ 
  { - 2} \\ 
  0 
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  0 \\ 
  { - 1} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  0 
\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}
  { - 1} \\ 
  { - 1} 
\end{array}} \right]\, = \left[ {\begin{array}{*{20}{c}}
  4 \\ 
  2 \\ 
  0 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  1 \\ 
  { - 1} \\ 
  { - 3} 
\end{array}\,\,\,\,\begin{array}{*{20}{c}}
  5 \\ 
  1 \\ 
  { - 3} 
\end{array}} \right]$

$ \to \left| {AB} \right| = 4(6) - 1( - 6) + 5( - 6) = 0$