If $AB = I$ then $BA = I$ 
If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$. 

I do not understand anything more than the following.


*

*Elementary row operations.

*Linear dependence.

*Row reduced forms and their relations with the original matrix.


If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?
P.S.: Please avoid using the transpose and/or inverse of a matrix.

$\newcommand{\mat}[1]{\left(\begin{matrix}#1\end{matrix}\right)}$
Here is a proof using only calculations and induction over the size $n$ of the matrices.
Observe that no commutativity is needed; we can work in any division ring.
If $n=1$ then two scalars $a,b$ with $ab=1$ are given. Then $b=1/a$ and we also have $ba=1$.
Now suppose that the statement is true for all matrices of size $n-1$, $n\geq2$.
Given two $n$ by $n$ matrices with $AB=I$,
we can assume without loss of generality that the upper left element
of $A$ is nonzero. Otherwise, since the first row of $A$ cannot be all zero, we can achieve this
by permuting two columns of $A$ and the correponding rows of $B$. We can also assume that this
upper left element $\alpha$ equals 1,
otherwise we multiply the  first row of $A$ by the inverse of this element from the left and
the first column of $B$ by $\alpha$ from the right.
Now we write $A,B$ in block matrix form:
$$A=\mat{1&a_2\\a_3&a_4}, B=\mat{b_1&b_2\\b_3&b_4},$$
where $b_1$ is a scalar, $a_2,b_2$ are matrices of size 1 by $n-1$,
$a_3,b_3$ have size $n-1$ by 1 and $a_4,b_4$ are $n-1$ by $n-1$.
$AB=I$ means that $$b_1+a_2b_3=1,\ b_2+a_2b_4=0,\ a_3b_1+a_4 b_3=0\mbox{ and }a_3b_2+a_4b_4=I.$$ Here $I$ has size $n-1$ only and "0" abbreviates any matrix of zeros).
First, we calculate $(a_4-a_3a_2)b_4=a_4b_4+a_3b_2=I$. Since both matrices
have size $n-1$ we can conclude that  $b_4(a_4-a_3a_2)=I$.
Next, we calculate
$$(a_4-a_3a_2)(b_3+b_4 a_3)=a_4 b_3-a_3a_2b_3+a_3=
(a_4 b_3+a_3b_1)+a_3(-b_1-a_2b_3+1)=0$$
and, multiplying by $b_4$ from the left, we obtain $b_3+b_4 a_3=0$.
Then we obtain that $a_2b_3=-a_2b_4a_3=b_2a_3$ and therefore also $b_1+b_2a_3=1$.
Finally we have $a_2+b_2(a_4-a_3a_2)=(a_2b_4+b_2)(a_4-a_3a_2)=0$ and thus
$b_2a_4=(b_2a_3-1)a_2=-b_1a_2$.
Altogether we obtain $BA=\mat{b_1&b_2\\b_3&b_4}\mat{1&a_2\\a_3&a_4}=I$
which completes the proof.
