For $n\times n$ matrices, is it true that $AB=CD\implies AEB=CED$? 
If $A,B,C,D,E$ are $n\times n$ matrices, does $AB=CD$ imply $AEB=CED$?

I only know that $AB=CD \implies ABE=CDE$, but I don't see how you can sandwhich $E$ within it.

Also, if $AB=CD=0$, does $\det(AB)=\det(CD)=0$?

I think this should be true because $AB$ and $CD$ are the same matrices and $\det(0)=0$

I don't think this works by sandwiching.....
$$A=\begin{pmatrix}
        1 & 3  \\
        0 & 1  \\
        \end{pmatrix},
B=\begin{pmatrix}
        1 & 2  \\
        0 & 1  \\
        \end{pmatrix},
C=\begin{pmatrix}
        1 & 1  \\
        0 & 1  \\
        \end{pmatrix},
D=\begin{pmatrix}
        1 & 4  \\
        0 & 1  \\
        \end{pmatrix},$$
Then we have that $AB=CD$ by matrix multiplication, however if
$$E=\begin{pmatrix}
        1 & 1  \\
        1 & 1  \\
        \end{pmatrix}$$
Then $AEB=\begin{pmatrix} 4 & 12\\ 1 & 3 \\ \end{pmatrix}\neq \begin{pmatrix} 2 & 10\\ 1 & 5 \\ \end{pmatrix}=CED$.
