This is from Artin Algebra, proposition 1.2.20:
Proposition: Let $A$ be a square matrix that has either a left inverse or a right inverse, a matrix $B$ such that either $BA=I$ or $AB=I$. Then $A$ is invertible and $B$ is its inverse.
Proof: Suppose $AB=I$. We perform row reduction on $A$. Say $A'=PA$, where $P=E_k\cdots E_1$ is the product of corresponding elementary matrices, and $A'$ is a row echelon matrix. Then $A'B=PAB=P$. Because $P$ is invertible, its bottom row is not zero. Then bottom row of $A'$ can't be zero either. $\dots$
I can't understand how bottom row of $A'$ can't be zero.