# Invertible matrices and zero rows

I have a doubt related to proof of a theorem:

Theorem: Suppose that $$A$$ and $$B$$ are matrices such that the product $$AB$$ is an identity matrix. Then the reduced row-echelon form of $$A$$ does not have a row of zeros.

Proof. Let $$R$$ be the reduced row-echelon form of $$A$$. Then $$R = EA$$ for some invertible square matrix $$E$$. By hypothesis $$AB = I$$ where $$I$$ is an identity matrix, so we have a chain of equalities $$R(BE^{-1}) = (EA)(BE^{-1}) = E(AB)E^{-1} = EIE^{-1} = EE^{-1} = I$$. If $$R$$ would have a row of zeros, then so would the product $$R(BE^{-1})$$. But since the identity matrix $$I$$ does not have a row of zeros, neither can $$R$$ have one.

Why the statement "But since the identity matrix $$I$$ does not have a row of zeros, neither can $$R$$ have one." is correct?

I have found a similar question here

This theorem is taken from the text

• The previous sentence "If $R$ would have a row of zeros, then so would the product $R(BE^{−1})$", combined with the string of equalities, tells us that if $R$ had a row of zeros then $I$ would have a row of zeros, which we know is false since we know what $I$ is. So I would make sure you see why $R$ having a row of zeros implies $R(BE^{-1})$ has a row of zeros. Run through the calculation of multiplying a matrix $R$ with a row of zeros by any matrix you want, and see how the row of zeros affects the product. – ndhanson3 Nov 11 '20 at 6:03

## 1 Answer

Let $$BE^{-1}=M$$. The $$(i,j)^{\text{th}}$$ entry of $$RM$$ is given by $$\sum_{k}r_{ik}m_{kj}$$.

If $$R$$ has a zero row, then for some $$i=i',r_{i'k}=0\forall k$$. Thus the $$(i',j)^\text{th}$$ entry of $$RM$$ is $$0$$ for all $$j$$ i.e. $$RM$$ has a zero row (row $$i'$$).