I've been using the method of getting $R$ by getting $A$ to reduced row echelon form. Then multiplying $R$ by $U$, where: $$U=\begin{bmatrix}a&&b&&c\\d&&e&&f\\g&&h&&i\end{bmatrix}$$ so that the product of $U$ and $R$ equals $A$: $$A=U*R$$ So far this method has produced answers for most entries of $U$ but not all so I will show the steps I used for one example.
If $$A=\begin{bmatrix}3&-3&6&9&3\\ 2&-2&9&6&-3\\ -1&1&-5&-3&3\end{bmatrix}$$ therefore, $$R=\begin{bmatrix}3&-3&6&9&3\\ 0&0&5&0&-5\\ 0&0&-3&0&4\end{bmatrix}$$ and $$\begin{bmatrix}3&-3&6&9&3\\ 2&-2&9&6&-3\\ -1&1&-5&-3&3\end{bmatrix}=\begin{bmatrix}a&&b&&c\\d&&e&&f\\g&&h&&i\end{bmatrix}*\begin{bmatrix}3&-3&6&9&3\\ 0&0&5&0&-5\\ 0&0&-3&0&4\end{bmatrix}$$ $$\begin{bmatrix}3&-3&6&9&3\\ 2&-2&9&6&-3\\ -1&1&-5&-3&3\end{bmatrix}=\begin{bmatrix}3a&-3a&6a+5b-3c&9a&3a-5b+4c\\ 3d&-3d&6d+5e-3f&9d&3d-5e+4f\\ 3g&-3g&6g+5h-3i&9g&3g-5h+4i\end{bmatrix}$$ I'm not sure where to go on from here or whether this is the right method.
Performing an elementary row operation on a matrix is equivalent to left-multiplying it by an invertible matrix. The product of invertible matrices is also invertible, so the entire process of reducing a matrix to its RREF is equivalent to multiplying it by some invertible matrix. That is, $R=EA$ for invertible matrix $E$. Clearly, then, $U=E^{-1}$.
Now, because of the way matrix multiplication works, if you augment $A$ with the identity matrix and row-reduce, you’ll transform $[A\mid I]$ into $[EA\mid EI] = [R\mid E]$. So, one way to find $U$ is to augment $A$ with the identity matrix, row-reduce, then extract the right-hand matrix and invert it.