Given a commutative ring with unity $R$, its units $U(R)$ form a multiplicative group.
$U(R)$ acts on $R$ by multiplication. The orbits of the action are called associates, and the resulting equivalence relation is called associatedness.
For example, $ℤ$ has units $\{-1,1\}$, and the associatedness is $\{\{-n,n\}:n\inℤ_{≥0}\}$.
What happens if we drop the commutativity of $R$? Since left and right multiplication differ, $U(R)$ will act differently for each version.
I chose upper triangular matrices on $\text{GF}(2)$ for an example, and acquired the following "left associates": $$ \{\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}\},\{\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix},\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}\},\{\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\},\{\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\},\{\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}\},\{\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 1\end{pmatrix}\} $$
The units act as elementary row operations. For "right associates," the units will act as elementary column operations instead.
This arises the following question: Will "left associates" and "right associates" ever coincide for a noncommutative $R$?
Take any non commutative division ring $R$, for example the quaternions.
Since every non zero element is invertible, the only orbits for left or right multiplication are $\{0\}$ and $ R- \{0\}$. Hence left and right associates coincide.