Square matrices and elementary matrices theorem. Linear algebra. I'm reading this text:

and the theorem it's referencing is here:

I don't understand this part:

From Theorem 2.11 you know that the system of linear equations represented by $Ax = O$ has only the trivial solution. But this implies that the augmented matrix [A O] can be rewritten in the form [I O] (using elementary row operations corresponding to $E_1$, $E_2$, . . . , and $E_k$). So, $E_k$. . .$E_3E_2E_1A=I$ and it follows that $A = E_1^{-1}E_2^{-1}E_3^{-}1. . .E_k^{-1}$. A can be written as the product of elementary matrices.

I don't understand any of that. How can the augmented matrix (this just means it's a matrix that includes the constants and the coefficients right?) $[A | 0]$ be rewritten in the form $[I |0]$ using those row operations?
 A: 
augmented matrix (this just means it's a matrix that includes the constants and the coefficients right?) 

A short hand notation for $A x = b$ is the augmented matrix $[A \lvert b]$.
In the above case $A x = b \iff [A\mid 0]$ and $[I \mid 0 ] \iff I x = 0 \iff x = 0$.

How can the augmented matrix (..) $[A|0]$ be rewritten in the form $[I|0]$ using those row operations?

For invertible $A$ you can use Gauss elimination to turn $A$ into $I$. 
Each elimination operation (multiply a row with a scalar non-zero multiple, exchange two rows, add a row to another row) can be expressed as multiplication with an matrix $E_i$ from the left.
So 
$$
E_k \dotsb E_1 A = I
$$
is a way to write down the successful Gauss elimination in $k$ steps from $A$ into $I$.
Further each operation and thus each $E_i$ is invertible. So we have
$$
\begin{align}
E_k E_{k-1} \dotsb E_1 A 
&= I \iff \\
E_k^{-1} E_k E_{k-1} \dotsb E_1 A = E_{k-1} \dotsb E_1 A 
&= E_k^{-1} \iff \\
E_{k-1}^{-1} E_{k-1} \dotsb E_1 A = E_{k-2} \dotsb E_1 A 
&= E_{k-1}^{-1} E_k^{-1} \iff \\
& \,\,\, \vdots \\
A &= E_1^{-1} \dotsb E_k^{-1}
\end{align}
$$
We multiplied both sides of the initial equation with $E_k$ from the left. Then with $E_{k-1}$ from the left and so until we multiply both sides with $E_1$ from the left. So we peeled free $A$. 
