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.
I will give a sketch of a proof. Let $A=
 \left( \begin{array}{cc}
a &amp; b  \\
c &amp; d  \end{array} \right)
$ and  $B=
 \left( \begin{array}{cc}
x &amp; x  \\
z &amp; w  \end{array} \right)
$ such that $AB=I.$ Then we get $\left\{\begin{array}{c}
ax + bz = 1 \\
cx + dz = 0 \\
\end{array}\right.$ and $\left\{\begin{array}{c}
ay + bw = 1 \\
cy + dw = 0 \\
\end{array}\right.$ 

I will assume that $a\neq 0$ (since there is no $B$ such that BO=I.) Then we have $x=\frac{1}{a}-\frac{bz}{a}$ and we get $(ad-bc)z=-c$. Let suppose that $ad=bc$. If $b=0$ or $c=0$ then $d=0$  and  we would have $A=
 \left( \begin{array}{cc}
a &amp; b  \\
0 &amp; 0  \end{array} \right)$, or $A=
 \left( \begin{array}{cc}
a &amp; 0  \\
c &amp; 0  \end{array} \right)$, or $A=
 \left( \begin{array}{cc}
a &amp; 0  \\
0 &amp; 0  \end{array} \right)$ but in any case there is no $B$ such that $BA=I$ (It is easy to prove that).  So we have $(a,b,c,d)\neq (0,0,0,0)$. Then we have $a=\frac{bc}{d}$, but in this case the systems above do not have solution. Then $ad-bc\neq 0$ and we get $z=\frac{-c}{ad-bc}$. In the end we will find that $B=
 \frac{1}{ad-bc}\left( \begin{array}{cc}
d &amp; -b  \\
-c &amp; a  \end{array} \right).$ It is easy to check that $BA=I.$ Now if $a=0$ then we have $b\neq0$ and $\dots$

I don't know how to solve the exercise in a different way. This is my best effort.
$AB = I$ 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.