linear algebra - If $A$ and $B$ are real matrices and $X,Y$ are are non-singular square matrices such that $XA=BY$

If $A$ and $B$ are real matrices and $X,Y$ are are non-singular square matrices with real entries such that $XA=BY$ then which of the following is true?

$1. \dim(X)=\dim(Y)$

$2. \dim(A)=\dim(B)$

$3.$If $X$ and $A$ commute then $B$ is square.

$4.$If $Y$ and $A$ commute then $B$ is square.

$5.$If $A$ is non-singular then $B$ is also non-singular.

My try: Here, dimension means order of matrix. So I suppose that $$\dim(X)=m\times m, \quad\dim(Y)=n\times n, \quad\dim(A)=m\times a, \quad\dim(B)=n\times b$$

Then I try to think about the options but it becomes difficult for me. Last option is obviously true and I think first two options are also true. What you think? What should be the answer?

Answer

Let $A$ be $m\times n$. Then $X$, being non singular, must be square so it has to be $m\times m$. Since $XA=BY$ and $Y$ is non singular (hence square), $B$ must be $m\times n$ too and $Y$ is $n\times n$.

If $X$ and $A$ commute, then $A$ must be $m\times m$ or $AX$ is undefined. Thus $m=n$ and $B$ is square. The same if $Y$ and $B$ commute.

If $A$ is non singular, then $B=XAY^{-1}$, so…