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Suppose we have $A \in M_3(\Bbb N\cup\{0\})$ s.t. sum of the elements of each row is $k $ for some fixed $k\in \Bbb N\cup\{0\}$. What are all the possibilities of $A$ s.t. $\det(A)=k$?

We can start from

  • $k=0$ here we have to have the matrix to be zero.
  • For $k=2$, I am getting $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ \end{pmatrix} $$ as one such matrix then what are the other possibilities and can you give me a general question.
  • Then what about $k=1,3$

And so on for any $k$ can we give a general structure?

Then this question can be extended to $A\in M_4(\Bbb N \cup{0})$.

By the way, what I want is if someone can give me some of the partial answers as the general answer might be too strong to expect!

For example first, just give me answer/idea for $k=2$ and then $k=3$ for $A \in M_3(\Bbb N\cup\{0\})$.

You can also give me reference of sage or any other tool that you are using.

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  • $\begingroup$ For k=1 and arbitrary orders, the count is half the number of permutation matrices. This could become a good question for MathOverflow if you provide more motivation and originating context. Since you are asking for a relatively small determinant, my guess is the restriction is tight enough that given k there will be few order 3 matrices satisfying the relation. Gerhard "Up To Permutation, Of Course" Paseman, 2019.11.06. $\endgroup$
    – Gerhard Paseman
    Commented Nov 6, 2019 at 16:52
  • $\begingroup$ I retract a previous remark. Since you only have row and not column constraints there will be superexponentially many lower triangular matrices which are solutions. I would be interested in solutions with column constraints with matrices that are not triangular. Gerhard "Let's Make It Even Harder" Paseman, 2019.11.06. $\endgroup$
    – Gerhard Paseman
    Commented Nov 6, 2019 at 16:57
  • $\begingroup$ Can you please give me solutions for $k=2$ and $k=3$ writing elaborately. I will disclose the motivation of thinking about this problem. It has a strong influence with Cremona groups but I won't disclose that once I go with that a bit more. I can chat and discuss those things once I get this part done and you will realise column part is not so important here. Although in my given examples the column sum is also preserved but this is not the case in general. $\endgroup$
    – Ri-Li
    Commented Nov 6, 2019 at 17:37
  • $\begingroup$ "It has a strong influence with Cremona groups but I won't disclose that"—I understand the impulse to protect one's research directions, but I think that, in general, if you want to motivate other people to do some of your research work for you then you need at least to give them some idea of what it's working towards. $\endgroup$
    – LSpice
    Commented Nov 6, 2019 at 18:44
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    $\begingroup$ Also posted to m.se math.stackexchange.com/questions/3424001/… without notice to either site – that's an abuse. $\endgroup$
    – Gerry Myerson
    Commented Nov 7, 2019 at 4:13

1 Answer 1

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I won't elaborate, but I will suggest other solutions. In general, form a lower triangular matrix, with diagonal elements k,1,1,..., and fill in the lower triangle to meet the row requirements. One can also do an upper triangular form. A lower triangular example for k=3 has nonzero entries 3,2,1,1,1,1. One can do this for order 3 and larger k, producing k many examples. As the order grows, the number of lower triangular examples also grows

Gerhard "Giving You More To Explore" Paseman, 2019.11.06.

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