How would I solve the following problem involving elementary matrices? 
Let $A=\begin{pmatrix}0&5\\ 7&4\end{pmatrix}$.
1) Write $A$ as a product of $4$ elementary matrices.
2) Write $A^{-1}$ as a product of $4$ elementary matrices.

My work. I have managed to find $A^{-1}$, which came out to be this:
\begin{pmatrix}-\frac{4}{35}&\frac{1}{7}\\ \frac{1}{5}&0\end{pmatrix}
However, I am struggling to figure out how I would split each of these matrices into $4$ elementary matrices. Any help?
 A: For $a\not=0$ we have that
$$\begin{pmatrix} a & b \\ c & 0 \end{pmatrix}=
\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ c & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 0 & -bc/a \end{pmatrix}
\begin{pmatrix} 1 & b/a \\ 0 & 1 \end{pmatrix}$$
then $A^{−1}$ is a product of 4 elementary matrices with $a=-4/35$, $b=1/7$, and $c=1/5$.
Note that elementary matrices generate a group of invertible matrices. Hence if $A^{-1}=E_1E_2E_3E_4$ where $E_i$ for $i=1,2,3,4$ are some elementary matrices then $A=E_4^{-1}E_3^{-1}E_2^{-1}E_1^{-1}$.
A: Hint: I will write $E_1,E_2,\cdots$ to denote some elementary matrices. You are wanting to find $E_1E_2\cdots E_n A =I$ and then you can simply inverse all of these one at a time from the left:
$$E_1^{-1}E_1E_2\cdots E_n A=E_1^{-1}I\iff E_2\cdots E_nA=E_1^{-1}$$
$$\iff A= E_n^{-1}E_{n-1}^{-1}\cdots E_1^{-1}$$
Then to find $A^{-1}$, what do you know about inverse of products?

As a first step though. We can row swap:
$$A=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}7&4\\ 0&5\end{bmatrix}$$
Then you need two more row operations and you are done.
