Is $\operatorname{rank}(AQB)=\operatorname{rank}(AB)$ if $Q$ is non-singular? $\newcommand{\rank}{\operatorname{rank}}$We know that $\rank(PA)=\rank(AQ)=\rank(PAQ)=\rank(A)$ where $A\in M_{m\times n}(\mathbb F), P, Q$ are $m\times m, n\times n$ invertible matrices. 
mean to say , from the matrix product we can remove the non-singular matrices; rank will not be effected. 
After studying this, it came to my mind, then what will happen in the case of $\rank(PAQB)$ where $B\in M_{n\times m}(\mathbb F)$ ?
No idea. The only thing I got is $\rank(PAQB)=\rank(AQB)$. Can we remove $Q$ as well and write $\rank(AQB)=\rank(AB)$?
Please help. In case it has been solved earlier, provide me the link.

No. Worse, the much weaker property of the product being zero is not preserved by insertion of a nonsingular matrix: If we take $$A = B = \pmatrix{0&1\\0&0}, \quad Q = \pmatrix{0&1\\1&0},$$ then
$$AQB = A \neq 0$$
but $$AB = 0.$$
