Gauss Jordan elimination reduces to row-echelon form always? I am reading this text:

and I'm wondering if gauss-jordan elimination always leads to an identity matrix on the left? If so, that helps me understand this passage:

I'm trying to figure out why [A 0] can be rewritten as [I 0]. Why is this?

Gauss-Jordan elimination leads to identity matrix if the equation system has a unique solution. Book gives an example when this is the case so here is an example when this is not the case:
Example: Take the equation system
$$x+y+z = 2$$
$$x+y = 1$$
$$z = 2$$
which has no solution and can be represented as
$$ \begin{pmatrix}
    1 & 1 & 1 \\
    1 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{pmatrix}\begin{pmatrix}
    x \\
    y \\
    z \\
    \end{pmatrix} = \begin{pmatrix}
    2 \\
    1 \\
    2 \\
    \end{pmatrix}, \text{where }A = \begin{pmatrix}
    1 & 1 & 1 \\
    1 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{pmatrix}
$$
Here, notice that in order to find solution of this equation system, we need to multiply both sides of the equation by $A^{-1}$ from the left. However, applying Gauss-Jordan elimination,
$$\begin{pmatrix}
    1 & 1 & 1 \\
    1 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{pmatrix} \rightarrow \begin{pmatrix}
    1 & 1 & 1 \\
    0 & 0 & -1 \\
    0 & 0 & 1 \\
    \end{pmatrix} \rightarrow \begin{pmatrix}
    1 & 1 & 0 \\
    0 & 0 & -1 \\
    0 & 0 & 0 \\
    \end{pmatrix} \rightarrow \begin{pmatrix}
    1 & 1 & 0 \\
    0 & 0 & 1 \\
    0 & 0 & 0 \\
    \end{pmatrix}$$
so as you can see, we cannot have identity matrix by using row elementary operations. When it can be done, i.e. $A$ is invertible, we can say $[A\ \ 0]$ can be written as $[I\ \ 0]$ since they are row equivalent.
