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.

Since inverse/transpose are not allowed we start by writing
$$A = \begin{bmatrix}
    a_{11} & a_{12} & a_{13} & \dots  & a_{1n} \\
    a_{21} & a_{22} & a_{23} & \dots  & a_{2n} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    a_{n1} & a_{n2} & a_{n3} & \dots  & a_{nn}
\end{bmatrix}$$
and similarly
$$B = \begin{bmatrix}
    b_{11} & b_{12} & b_{13} & \dots  & b_{1n} \\
    b_{21} & b_{22} & b_{23} & \dots  & b_{2n} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    b_{n1} & b_{n2} & b_{n3} & \dots  & b_{nn}
\end{bmatrix}$$
Since $AB = I$, using matrix multiplication definition we can write the elements of AB as: $$ \sum_{i,i'} \sum_{j,j'} a_{i,i'}b_{j,j'} = I_{i,j'}=    \begin{cases}
      1, & \text{if}\ i=j'\space and \space j=i' \\
      0, & \text{otherwise}
    \end{cases} $$
such that when $i = j'$ and $j = i'$ we get the diagonal elements and off diagonal otherwise.
But note that since $a_{i,i'}b_{j,j'}$ are scalar and commute we can write.
$$ \sum_{i,i'} \sum_{j,j'} b_{j,j'}a_{i,i'} =   \sum_{i,i'} \sum_{j,j'} a_{i,i'}b_{j,j'} =   \begin{cases}
      1, & \text{if}\ i=j'\space and \space j=i' \\
      0, & \text{otherwise}
    \end{cases} $$ 
Now we observe that the elements of $BA$ can be written as...
$$ \sum_{i,i'} \sum_{j,j'} b_{j,j'}a_{i,i'} $$ 
such that when $i = j'$ and $j = i'$ we get the diagonal elements and off diagonal otherwise. But we showed that $$ \sum_{i,i'} \sum_{j,j'} b_{j,j'}a_{i,i'} =     \begin{cases}
      1, & \text{if}\ i=j'\space and \space j=i' \\
      0, & \text{otherwise}
    \end{cases} $$ 
thus $BA = I$
