If $AB = I$. Prove that $BA = I.$(use only linear combination/matrix multiplication) Can anyone prove this using only linear combination/matrix multiplication ?

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

$\left[\begin{matrix} a&b\\c&d \end{matrix}\right]\left[\begin{matrix} e&f\\g&h \end{matrix}\right]=\left[\begin{matrix} 1&0\\0&1 \end{matrix}\right]$ is of no use as I end up with so many unknowns
$AB=\left[\begin{matrix} A_{first row}B_{first column}&A_{first row}B_{second column}\\A_{second row}B_{first column}&A_{second row}B_{second column} \end{matrix}\right]$ is also not helping
 A: The trick is to first show that if $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $AB = I$ then necessarily $B = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. This can be done using only the stuff you say. From there it is fairly easy to compute $BA$ and see that it equals $I$.
Now this is a proof, the only thing unsatisfactory about it is that your very smart 'guess' of the right form of $B$ seems to come out of nowhere. I could write a bit about how you could have found it, but perhaps that is not necessary?
A: $AB=I \implies rg(Ab)=n \implies rg(A)=n=rg(B) \implies A,B$ invertible for dimension reasons $\implies ABA=IA \iff ABA=AI \iff A^{-1}ABA=BA=A^{-1}AI=I$ where we used that I commutes with A. 
Alternatively you can do the same thing for $B$. 
Either way, it is important here that $A$ and $B$ are linear maps between the same finite dimensional vector space in order for the rank argument to work (otherwise there might still be a kernel although of full rank) and that we have that by using this actually $AB=I$ is sufficient to show that $A$ or $B$ is invertible, which is the really essential part, since then the rest is just a standard proof from group theory! 
