For a ult (unit low. tri.) matrix A, can we transform $[A|I]$ into $[I|A^{-1}]$ only using row operations that correspond to ult elementary matrices?

A unit lower triangular (ult) matrix $A$ is a matrix whose entries are $a_{ij}=0$ for $j>i$, and $a_{ii}=1$.

I am trying to prove that the inverse of a ult matrix is another ult matrix. (There are different proofs here). I have already proved that the product of ult matrices is also a ult matrix. So I want to use this property by defining $A^{-1}=E_n...E_2E_1I$, where $E_k$ are elementary matrices that correspond to the row operations of the transformation $[A|I] \rightarrow [I|A^{-1}]$. If we can show that $E_k$ are all ult elementary matrices, then the proof is complete. Therefore, the question is:

Considering a ult matrix A, can we transform $[A|I]$ into $[I|A^{-1}]$ only using row operations that correspond to ult elementary matrices?

Hint. Row operations corresponding to unit lower triangular matrices means adding a multiple of a row to a row that is below it. If you think about how you go from $A$ to $I$ in the simplest way, where $A$ is ult, you will see that you are only using operations of this kind.
Let me suggest another proof however. Let $A$ be a ult matrix. Let $V$ be the space of lower triangular matrices. The mapping $X \mapsto AX$ is an endomorphism of the finite-dimensional space $V$. Moreover this mapping is injective, because $A$ is invertible (its columns are linearly independent). Since this mapping is injective, it must also be surjective. Thus $I$ is in its image, meaning that there is a lower triangular matrix $X$ such that $AX = I$, which can only be the inverse of $A$. It is not difficult to see that $X$ must be ult.
• @i.ozturk: Follow the hint, and ignore the alternative proof (which is not an answer to the questions). Show that an appropriate composition of row operations (namely, subtractions of multiples of the $1$-st, $2$-nd, etc., $n-1$-st rows from the $n$-th row) turns the $n$-th row into $\left(0,0,\ldots,0,1\right)$. Then, argue by induction. – darij grinberg Jul 8 '15 at 14:22