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.

Here we explain how to derive entries of $B$ from the entries of $A$. As the book explains in the beginning of page 19 (section 1.5), if we let $B_1$ and $B_2$ denote the first and second columns of the matrix $B$ then one can write $AB = [AB_1,AB_2]$. Hence AB = I if and only if 
$AB_1=\left[\begin{array}{c}
1\\0\\\end{array}\right]$
and 
$AB_2=\left[\begin{array}{c}
0\\1\\\end{array}\right]$. So we know that the two systems of two linear equations in two unknowns $AX=\left[\begin{array}{c}
1\\0\\\end{array}\right]$ and $AY=\left[\begin{array}{c}
0\\1\\\end{array}\right]$ are both having solutions. We would like to prove that the only solutions are $B_1$ and $B_2$, and we would like to find these solutions in terms of entries of $A$. 
If we let 
$A=
 \left[ \begin{array}{cc}
a & b  \\
c & d  \end{array} \right]$ then the two equations can be written as 
$\left[ \begin{array}{ccc}
a & b & 1\\
c & d & 0 \end{array} \right]$
and
$\left[ \begin{array}{ccc}
a & b & 0\\
c & d & 1 \end{array} \right]$. Since both these equations have solutions ($B_1$ and $B_2$), then the equations 
$\left[\begin{array}{ccc}
ad-bc & 0 & d\\
0 & ad-bc & -c \end{array}\right]$
and $\left[ \begin{array}{ccc}
ad-bc & 0 & -b\\
0 & ad-bc & a \end{array} \right]$, which are linear combinations of the original equations, are also having solutions (section 1.2 of the book). Now based on explanation in page 14, which explains the conditions for which non-homogenous systems of equations have solutions, if $ad-bc=0$ then $a=b=c=d=0$. This is a contradiction since if $A=0$, then $AB=0B=0\neq I$. So we can assume that $ad-bc \neq 0$. This immediately proves that the two equations are having unique solutions, therefore $B$ can only be of the following form 
$$\left[ \begin{array}{cc}
\frac{d}{ad-bc} & \frac{-b}{ad-bc}  \\
\frac{-c}{ad-bc} & \frac{a}{ad-bc}  \end{array} \right].$$
Now using this form for $B$, one can verify that $BA = I$.
