B
因初等变换不改变矩阵的秩,
$r_{1}=r\left[\begin{array}{cc}0 & A \\ B C & E\end{array}\right]=r\left[\begin{array}{cc}-A B C & 0 \\ B C & E\end{array}\right]=r\left[\begin{array}{cc}0 & 0 \\ B C & E\end{array}\right]=n$,
$r_{2}=r\left[\begin{array}{cc}A B & C \\ 0 & E\end{array}\right]=r\left[\begin{array}{cc}A B & 0 \\ 0 & E\end{array}\right]=r(A B)+n$,
$r_{3}=r\left[\begin{array}{cc}E & A B \\ A B & 0\end{array}\right]=r\left[\begin{array}{cc}E & 0 \\ A B & -A B A B\end{array}\right]=r\left[\begin{array}{cc}E & 0 \\ 0 & -A B A B\end{array}\right]=r(A B A B)+n$,
故选(B).