# Matrices: left inverse is also right inverse? [duplicate]

If $$AA$$ and $$BB$$ are square matrices, and $$AB=IAB=I$$, then I think it is also true that $$BA=IBA=I$$. In fact, this Wikipedia page says that this “follows from the associativity of matrix multiplication”. I assume there’s a nice simple one-line proof, but can’t seem to find it.

Nothing exotic, here — assume that the matrices have finite size and their elements are real numbers.

This isn’t homework (if that matters to you). My last homework assignment was about 50 years ago.

Since $AB=I$ then $B=B(AB)=(BA)B$. Note from $AB=I$ that $1=\det(AB)=\det(A)\det(B)$ so $\det(B)\neq0$.
So by $(BA)B=B$ we have:
$(BA-I)B=0$. Since $\det(B)\neq0$ then $B$ is not a $0$ divisor. So $BA=I$