-2
$\begingroup$

My attempt is quite handwaivy. But I think this has something to do with permutation matrices. I am absolutely new to this topic. can anyone throw any light on this solution? I know there are $2^{n\times m}$ binary matrices of size $n\times m$, and $n!m!$ possible permutations, but somehow I fail to get an intuition on what this implies for the equivalence classes.

$\endgroup$

closed as off-topic by Saad, Shailesh, Xander Henderson, mrtaurho, InterstellarProbe Apr 12 at 16:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Shailesh, Xander Henderson, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ So the entry could be 1/2? $\endgroup$ – Maria Mazur Feb 18 at 10:28
4
$\begingroup$

If the row sums are all odd, then the total number of $1$s is the sum of these $6$ odd numbers, hence even. If the column sums are all odd, then the total number of $1$s is the sum of these $7$ odd numbers, hence odd. A contradiction; hence the number of such matrices is $0$.

$\endgroup$
  • $\begingroup$ @InterstellarProbe If the column sums are odd, then the sum of sums is odd because there are $7$ columns. $\endgroup$ – Servaes Apr 12 at 17:40
  • $\begingroup$ @InterstellarProbe To reach a contradiction. Have you read my answer? $\endgroup$ – Servaes Apr 12 at 19:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.