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?
I'm not sure what kind of terminology is familiar to you, but one way to see this is by talking about the pivot positions (wikipedia link).  If $A$ is an invertible square matrix, then $A$ has a pivot position in every row.  This is because there is a unique solution to $Ax = b$ for every $b$.  But if $A$ is a square matrix with a pivot in every row, that means $A$ is row-equivalent to the identity matrix $I$.  That means $[A \;0]$ can be rewritten as $[I\; 0]$ using row operations.
Gauss-Jordan eliminition works if and only if an inverse exists. It doesn't work for the null element matrix (matrix of zeros).

They said that [A 0] can be rewritten as [I 0] using elementary row operations so there are operations so that works.