I'm reading Linear Algebra by Hoffman, Kunze where the authors explained that a $n\times n$ matrix $A$ being invertible is equivalent to the fact that $A$ is row-equivalent to $n\times n$ matrix $R$ which is an identity matrix.
In the proof of the theorem, they wrote:
$$R= E_k\ldots E_2E_1 A$$ where $E_1,\ldots,E_k$ are elementary matrices. Each $E_j$ is invertible, and so $$A = E_1^{-1}\ldots E_k^{-1}~R\,. $$ ... Since, $R$ is a (square) row-reduced echelon matrix, $R$ is invertible if and only if $R=I\,.$ [...]
I couldn't get the conclusion, since any row of $R$ can't be zero, it has to be identity matrix $I\,.$ Why is it so?
Isn't there any other row-reduced echelon matrix other than the identity matrix having no zero row and invertible? Why is it so?