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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

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    $\begingroup$ 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. $\endgroup$ – ndhanson3 Nov 11 '20 at 6:03
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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 $i'$).

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