Give a counterexample showing that there exist two different matrices $X$ and $Y$ such that $X\cdot A=I_n=Y\cdot A$

Question

If $X$ and $Y$ two matrix of order $n\times m$ such that $X\cdot A=I_n=Y\cdot A$ for some matrix $A$ of order $m\times n$ then it must be $X=Y$? I am sure not but unfortunately I did not find any counterexample: in particular if $X\cdot A=I_n$ then the linear function $f:\Bbb K^n\rightarrow\Bbb K^m$ generated by $A$ is injective and thus $A$ has rank $n$ so that if $\phi:\Bbb K^n\rightarrow\Bbb K^m$ and $\psi:\Bbb K^n\rightarrow\Bbb K^m$ are the linear functions generated by $X$ and $Y$ then $$ \phi(y)=\psi(y) $$ for any $y\in f[\Bbb K^n]$ but if $f[\Bbb K^n]\neq\Bbb K^m$ then could be exist $y\in\Bbb K^m$ such that $$ \phi(y)\neq\psi(y) $$ but I am not able to make a counterexample. Could be that $X$ and $Y$ are equal? So could someone help me, please?

Answer 1

So let be $A:=\begin{pmatrix}1\\2\end{pmatrix}$ and we suppose that $B:=\begin{pmatrix}u&v\end{pmatrix}$ is such that $B\cdot A=I$ so that $$ 1=u+2vB=u\cdot 1+v\cdot2=B\cdot A $$ and thus taking $u=1$ and $v=0$ for $X$ and $u=3$ and $v=-1$ for $Y$ then $X\neq Y$ and $X\cdot A=I=Y\cdot B$ as it is easy verify.