I'm trying to understand the best way to approach this problem. Short of writing every combination of matrices, I'm wondering if anyone can help me learn how to solve this problem.

How many $3\times 2$ matrices with entries in $\mathbb{Z}/3\mathbb{Z}$ are in reduced row echelon form?

  • $\begingroup$ So it is clear, $\mathbb{Z}/3\mathbb{Z}$ is another notation for $\mathbb{Z}_3$ (in the context of the OP's question) $\endgroup$ – apnorton Oct 6 '14 at 19:36
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    $\begingroup$ As a first step, consider where the leading ones might be located (and what these imply about the rest of the matrix entries if in reduced row-echelon form). $\endgroup$ – hardmath Oct 6 '14 at 19:36
  • $\begingroup$ @anorton: I'm not trying to be picky, just reflecting that it should be a field. The notation police got after me for using $\mathbb{Z}_p$ recently, saying it was wrong because that meant $p$-adic arithmetic! $\endgroup$ – hardmath Oct 6 '14 at 19:38
  • $\begingroup$ @hardmath If I understand you correctly, you're referring to the leading 1's in the first column, since it's reduced there can only be one row with a 1 in the first column? $\endgroup$ – Russ Oct 6 '14 at 19:40
  • $\begingroup$ @hardmath I know. I wrote that comment primarily for the OP's benefit, so they know that you didn't (really) change the content of their question, but rather just changed the formatting. :) $\endgroup$ – apnorton Oct 6 '14 at 19:40

In a matrix of size $m\times n$ (over any field), the rank is at most the minimum of $m,n$. In the present case then, a $3\times 2$ matrix has rank at most two, and thus a reduced row-echelon form can have at most two leading ones.

Some of the cases then turn out to be quite simple. If there are no leading ones, it must be because all the entries are zero, so there is only one matrix like that.

At the other extreme of having two leading ones, they must appear staggered in different columns, and because all the remaining entries in the two columns are zeros (because of reduced row-echelon form), there is only one matrix like that.

This leaves the slightly more interesting cases with one leading one. Which column is it in? If it is in the second column, that can only be because the entries in the first column were already zeros. And since the leading one in the second column clears out all the other entries there, this again is a case of only one matrix like that.

Finally suppose the first column has a leading one and the second column does not. A picture here is worth a thousand words:

$$ \begin{bmatrix} 1 & x \\ 0 & 0 \\ 0 & 0 \end{bmatrix} $$

The other entry in the first row, following its leading one, that can be anything! Well, anything in the field $\mathbb{Z}/3\mathbb{Z}$ is a short list of things: three possibilities.

Altogether there are, as Russ pointed out in a now removed Comment, six possibilities.


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