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.

An alternative.
Let $\textbf{A}$ be a $n \times n$ matrix, and let $\lambda_k$ be the eigenvalues, then we can write
$$
\prod_{k=1}^n \Big( \textbf{A} - \lambda_k \textbf{I} \Big) = 0
$$
So we obtain
$$
\textbf{A}^n = a_0 \textbf{I} + \sum_{k=1}^{n-1} a_k \textbf{A}^k
$$
Let us write
$$
\textbf{B} = b_0 \textbf{I} + \sum_{k=1}^{n-1} b_k \textbf{A}^k$$
It is clear that
$$
\textbf{A} \textbf{B} = \textbf{B} \textbf{A}
$$
We also find that
$$
\begin{eqnarray}
\textbf{A} \textbf{B}
&=& b_0 \textbf{A} + \sum_{k=1}^{n-1} b_k \textbf{A}^{k+1}\\
&=& b_0 \textbf{A} + \sum_{k=1}^{n-2} b_k \textbf{A}^{k+1} + b_{n-1} \textbf{A}^{n}\\
&=& b_0 \textbf{A} + \sum_{k=1}^{n-2} b_k \textbf{A}^{k+1} + b_{n-1} a_0 \textbf{I}
+ b_{n-1} \sum_{k=1}^{n-1} a_k \textbf{A}^k\\
&=& b_{n-1} a_0 \textbf{I} + \Big( b_0 + b_{n-1} a_1 \Big) \textbf{A} + \sum_{k=2}^{n-1} \Big( b_{k-1} + b_{n-1} a_k \Big) \textbf{A}^k\\
\end{eqnarray}
$$
When we set (for $a_0 \ne 0$)
$$
\begin{eqnarray}
b_{n-1} a_0 &=& 1\\
b_{k-1} + b_{n-1} a_k &=& 0
\end{eqnarray}
$$
we obtain
$$
\textbf{A} \textbf{B} = \textbf{I}
$$
So we can write
$$
\textbf{B} = -a_0^{-1} a_1 \textbf{I}
- \sum_{k=1}^{n-2} a_0^{-1} a_{k+1} \textbf{A}^k
+ a_0^{-1} \textbf{A}^{n-1}
$$
or
$$
\textbf{A}^{-1} = -a_0^{-1} \left( a_1 \textbf{I}
+ \sum_{k=1}^{n-2} a_{k+1} \textbf{A}^k
- \textbf{A}^{n-1} \right)
$$
the inverse of $\textbf{A}$ can be expressed as a linear sum of $\textbf{A}^k$.
