# 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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