Regular matrix definition : counter example with AB=I but BA<>I

Question

Definition : $A$ is a regular matrix (has an inverse) if there exists a matrix $B$ such that $A\times B=I$ and $B\times A=I$.
I am looking for an example of a singular matrix $A$ where there is a matrix $B$ such that $A\times B=I$ but $B\times A\neq I$.
Edit : the context of the question : I wanted to show my students that in order to prove a matrix $A$ is regular, we need to show $A\times B=I$ and $B\times A=I$. I wanted to illustrate this point with a counter-example ... but could not find one for the reason stated below.
But then, since inverse is a concept for square matrices (otherwise we have a generalization : the pseudo-inverse concept for non square matrix), and as showed for square matrices when we have $A\times B=I$ then we have $B\times A=I$, why do we have in the definition the need to show $A\times B=I$ and $B\times A=I$ ?

Answer 1

There is no such example. Since $A$ is singular, $\det(A\times B)=\det(A)\det(B)=0$, which is impossible, since $\det(\operatorname{Id})=1$.

Answer 2

If there exists $B$ such that $AB=I_n$ then $BA=I_n$ because $f:M\mapsto AM$ is linear, if $AM=0$, we have $M=I_n M=BAM=0$, the injectivity of $f$ implies its surjectivity because $\dim\mathcal{M}_n(\mathbb{R})$ is finite. Hence there exists $C$ such that $CA=I_n$, but $B=\underbrace{CA}_{=I_n}B=C\underbrace{AB}_{=I_n}=C$ so $BA=I_n$.

Answer 3

If you are looking for square matrices, such $A$ cannot exist, as other answers show.

If you are looking for non-square matrices, such examples exist. Any $m \times n$ matrix such that $rank(A)=m <n$ satisfies this property.

Answer 4

example :
$\left(\begin{array}{cc} 1 & 0\end{array}\right)\left(\begin{array}{c} 1\\ 0 \end{array}\right)=(1)$ but $\left(\begin{array}{c} 1\\ 0 \end{array}\right)\left(\begin{array}{cc} a & b\end{array}\right)=\left(\begin{array}{cc} a & b\\ 0 & 0 \end{array}\right)$ cannot be $I$ $\forall a,b$