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.

$AB= I$,  $Det(AB) = Det (A) . Det(B) = 1$. Hence $Det(B)\neq 0$ Hence $B$ is invertible.
Now let $BA= C$ then we have $BAB= CB$ which gives $B= CB$ that is $B. B^{-1} = C$ this gives $ C= I$
