$AB = I$ implies $BA = I$
Suppose \(AB=I\). Then \(\det B\ne0\), so \(\exists C,BC=I\) (think elementary column operations). Now \(ABC=A\), so \(C=A\).
Suppose \(AB=I\). Then \(\det B\ne0\), so \(\exists C,BC=I\) (think elementary column operations). Now \(ABC=A\), so \(C=A\).
