If $A,B\in M(2,\mathbb{F})$ and $AB=I$, then $BA=I$ This is Exercise 7, page 21, from Hoffman and Kunze's book.

Let $A$ and $B$ be $2\times 2$ matrices such that $AB=I$. Prove that
  $BA=I.$

I wrote $BA=C$ and I tried to prove that $C=I$, but I got stuck on that. I am supposed to use only elementary matrices to solve this question.
I know that there is this question, but in those answers they use more than I am allowed to use here. 
I would appreciate your help.

$AB = I$ implies that $ABAB = I$ and $AABB = A(I)B = AB = I$, hence $ABAB = AABB$. Since $\det(AB) = \det(A)\det(B) = 1$, the determinant of $A$ and the determinant of $B$ are units, so that $A$ and $B$ have inverses (using the adjoint matrix thing), hence $AABB = ABAB$ implies  $BA=AB = I$. 
Hope that helps,
Note : This proof works when $\mathbb F$ is an arbitrary commutative ring with unity. You didn't specify what $\mathbb F$ was so that I'm stating in which generality this proof holds.
