If $A$ and $B$ are square matrix such that $AB=I$ then $BA=I$ too. If $A$ is a square matrix which coefficients are the scalar of a field $\Bbb K$ then there exist a linear function $f:\Bbb K^n\rightarrow\Bbb K^n$ such that
$$
f(e_j):=A_j
$$
where $A_j$ is the $j$-th column of $A$ for any $j=1,...,n$.
Then if $I$ is the identity square matrix if $A$ and $B$ are two square matix such that $AB=I$ then there exist two function $f,g:\Bbb K^n\rightarrow\Bbb K^n$ such that the identity square matrix is the matrix of the function $(f\circ g)$ and this means that $(f\circ g)$ is the identity on $\Bbb K^n$. So to prove the statement of the question I have to prove that $(g\circ f)$ is the identity too. So how prove the statement? Could someone help me, please?
$f\circ g = \text{id}$ implies $g$ is injective (if a composition is injective then the &quot;inner&quot; function is always injective). Since $g:\Bbb{K}^n\to \Bbb{K}^n$ is a map between vector spaces of the same dimension, the rank-nullity theorem implies $g$ is also bijective. Thus, $f = g^{-1}$.
Another way to phrase the argument is to say that since $f\circ g = \text{id}$, then $f$ is surjective (again if a composition is surjective, then the &quot;outer&quot; function is surjective). Again, rank-nullity implies $f$ is bijective, so that again $g=f^{-1}$.
Theorem:
If $A$ and $B$ are two square matrices such that $A B=I$, then $B A=I$.
Proof:
By applying the properties of determinants of square matrices, we get that
$\det A\cdot\det B=\det(A B)=\det I=1\ne0$.
So it results that $\;\det A\ne0\;$ and $\;\det B\ne0\;$.
Now we consider the matrix
$C=\frac{1}{\det A}\cdot\text{adj}(A)$
where $\;\text{adj}(A)\;$ is the adjugate matrix of $A$ which is the transpose of the cofactor matrix of $A$.
$\text{adj}(A)= 
\begin{pmatrix}
A_{1,1} &amp; A_{2,1} &amp; \cdots &amp; A_{n,1} \\
A_{1,2} &amp; A_{2,2} &amp; \cdots &amp; A_{n,2} \\
\vdots  &amp; \vdots  &amp; \ddots &amp; \vdots  \\
A_{1,n} &amp; A_{2,n} &amp; \cdots &amp; A_{n,n} 
\end{pmatrix}$
where $\;A_{i,j}\;$ is the cofactor of the element $a_{i,j}$ of the matrix $A$.
So $\;A_{i,j}=(-1)^{i+j}\det M_{i,j}$
where $\;M_{i,j}\;$ is the submatrix of $A$ formed by deleting the $i^{th}$ row and the $j^{th}$ column.
We are going to use the Laplace expansions which are the following equalities:
$a_{i,1}A_{j,1}+a_{i,2}A_{j,2}+\ldots+a_{i,n}A_{j,n}=
\begin{cases}
\det A\;,\quad\text{ if } i=j\\
0\;,\quad\quad\;\,\text{ if } i\ne j
\end{cases}$
$A_{1,i}a_{1,j}+A_{2,i}a_{2,j} +\ldots+A_{n,i}a_{n,j}=
\begin{cases}
\det A\;,\quad\text{ if } i=j\\
0\;,\quad\quad\;\,\text{ if } i\ne j
\end{cases}$
By applying Laplace expansions, we get that
$A C = C A = I$.
Since for hypothesis $\;A B =I\;,\;$ then $\;C(A B)= C I\;,\;$ hence $\;(C A)B=C,\;$ therefore $\;I B = C\;$ and so we get that $\;B=C$.
Consequently,
$B A = C A = I\;,$
so we have proved that
$B A = I$.