# An Example of Matrices $A$, $B$ such that $\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B)$

## Problem 569

For an $m\times n$ matrix $A$, we denote by $\mathrm{rref}(A)$ the matrix in reduced row echelon form that is row equivalent to $A$.
For example, consider the matrix $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 &2 &2 \end{bmatrix}$
Then we have
$A=\begin{bmatrix} 1 & 1 & 1 \\ 0 &2 &2 \end{bmatrix} \xrightarrow{\frac{1}{2}R_2} \begin{bmatrix} 1 & 1 & 1 \\ 0 &1 & 1 \end{bmatrix} \xrightarrow{R_1-R_2} \begin{bmatrix} 1 & 0 & 0 \\ 0 &1 &1 \end{bmatrix}$ and the last matrix is in reduced row echelon form.
Hence $\mathrm{rref}(A)=\begin{bmatrix} 1 & 0 & 0 \\ 0 &1 &1 \end{bmatrix}$.

Find an example of matrices $A$ and $B$ such that
$\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B).$

## Proof.

Let
$A=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix} \text{ and } B=\begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix}.$ Then $A$ is already in reduced row echelon from.
So we have $\mathrm{rref}(A)=A$.

Applying the elementary row operations, we obtain
$B=\begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix} \xrightarrow{R_1\leftrightarrow R_2}\begin{bmatrix} 1 & 0\\ 0& 0 \end{bmatrix}.$ As the last matrix is in reduced row echelon from, we have $\mathrm{rref}(B)=\begin{bmatrix} 1 & 0\\ 0& 0 \end{bmatrix}$.
Therefore, we see that
$\mathrm{rref}(A) \mathrm{rref}(B)=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0& 0 \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}.$

The product of $A$ and $B$ is
$AB=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix} \begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0& 0 \end{bmatrix}.$ It follows that $\mathrm{rref}(AB)=\begin{bmatrix} 1 & 0\\ 0& 0 \end{bmatrix}$.

In summary, we have
$\mathrm{rref}(AB)=\begin{bmatrix} 1 & 0\\ 0& 0 \end{bmatrix} \neq \begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix} =\mathrm{rref}(A) \mathrm{rref}(B),$ as required.

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