I have worked our the solution to a problem, but I want to explain the solution in a mathematical way. I have the following matrix:
$$ \begin{pmatrix} 1 & 2 \\ 1 & 1 \\ \end{pmatrix} $$
And I get the following elementary matrices , call them E$_1$,E$_2$,E$_2$,E$_4$
$$ \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ \end{bmatrix} $$
$$ \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} $$ $$ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} $$$$ \begin{bmatrix} 1 & -1 \\ 0 & 1 \\ \end{bmatrix} $$
I have verified that E$_4$E$_3$E$_2$E$_1$A=I$_n$
So I know that A$^-$=E$_4$E$_3$E$_2$E$_1$
Now if I want to write A as a product of elementary matrices, would I write it as A=(E$_4$E$_3$E$_2$E$_1$)$^-$ or do I have to write it in a form that has no inverse.
For example, on the right hand side, from $I$ to $E_1$, the row operation can be written as $$\begin{align*} \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} &= \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ \end{bmatrix}\\ A_1I &= E_1 \end{align*}$$ Where $A_1$ as a matrix transforms the rows of $I$ to $E_1$. Similarly the transform from $E_{i-1}$ to $E_i$ can be written as a matrix $A_i$.
Then if there were a series of row operations to the left hand side that transforms the left hand side to $I$:
$$A_n\cdots A_2\,A_1 \begin{pmatrix} 1 & 2 \\ 1 & 1 \\ \end{pmatrix} = I$$
Then $A_n\cdots A_2\,A_1$ is the inverse of $\begin{pmatrix} 1 & 2 \\ 1 & 1 \\ \end{pmatrix}$. So on the right hand side:
$$A_n\cdots A_2\,A_1\,I = \begin{pmatrix} 1 & 2 \\ 1 & 1 \\ \end{pmatrix}^{-1}$$