If AB = E, then BA = E.

Let $A$ and $B$ be $n\times n$ matrices. Also let $f$ and $g$ be lineat operators such that $f(x) = Ax$, $g(x) = Bx$. Because $AB = E$, $A$ and $B$ are invertible, and $f$ and $g$ are bijective. (Proof is left as an exercise.) Now we have,

\begin{align}
f \circ g = id \to (f \circ g)(x) = x
\end{align}

then for all $x\in \mathbb{F}^n$,

\begin{align}
g(x) = g ((f\circ g)(x)) = (g \circ f)(g(x))
\end{align}

is satisfied. Because $g$ is bijective, this applies to all the elements in $\mathbb{F}^n$. Therefore, $g \circ f$ is also identity map and $BA = E$.