# show that there exists one and only one matrix $X$ such that $A=BX$ and one and only one matrix $Y$ such that $A=YB$ if $B$ is non-singular

If $$A$$ is a given matrix and $$B$$ is a non-singular matrix(both of order $$n$$),then show that there exists one and only one matrix $$X$$ such that $$A=BX$$ and one and only one matrix $$Y$$ such that $$A=YB$$.

I don't know how to prove the above. I know that rank if a matrix remains same after pre-multiplication or post multiplication by a non-singular matrix. Does this fact help at all in proving the above? Or is there some other way?

If $$B$$ is non-singular, it means that $$B$$ is invertible: there exists $$B^{-1}$$ with $$B^{-1}B=BB^{-1}=I$$. If you multiply $$A=BX$$ by $$B^{-1}$$ on the left, you get $$B^{-1}A=X$$. That's the only possible value of $$X$$. Similar with the other equation.