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We have $A(3×3)$ matrix (the sum of the elements on each row, column and diagonal are the same) with non-zero natural entries. Show that $det(A)$ is divisible with the sum of all elements in $A$.

I tried doing in the classical way, writing every term of $A$ and try finding a relation between them, but nothing interesting. I tried to split the determinant so I can get a sum to rewrite it in terms of $S$, where $S$ is the sum of the elements on each row, column and diagonal. If $det(A)$ is divisible with the sum of all elements in $A$, then is divisible with $3S$. I saw that this type of matrix is called 'magic square'. Maybe someone knows some interesting proprieties about this type of matrix?

Stefan Solomon
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    You have posted the same question here already. Please do not delete and repost. And have a look at such questions here, e.g., this one. – Dietrich Burde Feb 28 '23 at 20:55
  • Hint: Your approach works, but you need to reduce the number of variables. In particular, all of the entries can be uniquely determined by $A_{1, 1}, A_{2, 2}, A_{3, 2} $ (as opposed to saying that there is nothing interesting in the relation). Using that, calculate the determinant by expanding out all of the terms, and show that $ 9 A_{2, 2}$ is a factor. – Calvin Lin Feb 28 '23 at 21:38
  • Alternative approach: Do elementary row operations to pull out $3 S$ as a factor, then do elementary column operations to pull out $3$ as a factor, hence the determinant is a multiple of $9S$. I'd leave you to work out the details and show the work. – Calvin Lin Feb 28 '23 at 21:43
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    Connected : https://www.jstor.org/stable/pdf/2299242.pdf – Jean Marie Feb 28 '23 at 22:10
  • @CalvinLin Isn't enough to pull out $3S$ as a factor? – Stefan Solomon Feb 28 '23 at 23:19
  • @StefanSolomon Ah sorry, I mixed up what S referred to. I was thinking it was the center element, but you treated it as the Sum. So the ERO allows us to pull out $S$ as a factor, and then the ECO allows us to pull out $3$ as a factor. – Calvin Lin Feb 28 '23 at 23:26
  • @CalvinLin How we can pull out $3$ as a factor with ECO? – Stefan Solomon Mar 01 '23 at 00:07
  • Can you show your work? IE What happens with the ERO to pull out S as a factor? What are you left with? What kinds of ECO could we next do? – Calvin Lin Mar 01 '23 at 00:09
  • @CalvinLin I pulled $S$ as a factor just by adding all the rows to the top one, then I struggle how we can pull $3$ as a factor after I add all the columns to the first one. Maybe $S$ is divisible with 3? – Stefan Solomon Mar 01 '23 at 00:11
  • Can you write out the matrix that we're left with? $\quad$ Yes, $S$ is a multiple of 3. Why? (And that's why I was confused, because I was treating $S$ as the center element instead) – Calvin Lin Mar 01 '23 at 00:37
  • Just to be sure : when you say in the first line diagonal (singular, not plural), do you intend to say the 2 diagonals in fact ? – Jean Marie Mar 01 '23 at 07:29
  • @JeanMarie Yes, both diagonals have the same propriety as every row and column. – Stefan Solomon Mar 01 '23 at 12:19

1 Answers1

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The 9-dimensional vector space of $3 \times 3$ matrices has the set of magic matrices as a subspace. This subspace has dimension 3 with the following general structure (see here)

$$M=\begin{bmatrix} e+h-c & 2e-h & c \\ 2c-h & e & 2e-2c+h \\ 2e-c & h & e-h+c \\ \end{bmatrix}\tag{1}$$

Please note that the magic sum is equal to $3e$, i.e., three times the central entry.

The determinant of $M$ has a nice factorization :

$$\det(M)=9e(e-h)(2c-e-h)\tag{2}$$

on which we can "read" the divisibility property by the magic sum $3e$, and moreover indeed by $3$ times this magic sum.

(of course, this is valid if $e \ne h$ and $e+h \ne 2c$).

N.B. : I just found this similar question with answers proving only that the determinant is divisible by the magic sum, but not by $3$ times the magic sum.

Edit : If one tries to extend the study to the $4 \times 4$ case (see here), with general structure depending upon 8 parameters instead of $3$ :

$$M=\begin{bmatrix} a & b & c & d\\ e&f&g&(a+b+c+d-e-f-g)\\ h & (a-d+e-g+h) & (b+c+2d-e-f-h) & (f+g-h)\\ (b+c+d-e-h) & (c+2d-e-f+g-h) & (a-c-d+e+f-g+h)&(-d+e+h) \end{bmatrix}\tag{1}$$

we find again that the determinant $D$ of matrix $M$ is divisible by the magic sum (here $s=a+b+c+d$), but there is not anymore divisibility property ; in particular, there doesn't exist a fixed integer constant (like $3$ above) such that $\frac{D}{s}$ is always divisible by this constant.

Jean Marie
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  • Can I use the fact that the magic sum is equal to three times the central entry as a theorem? – Stefan Solomon Mar 01 '23 at 16:54
  • Yes, the structure of matrix $M$ I give proves this fact ; but if you are speaking to people who aren't familiar with magic squares, they will not be familiar with this property, therefore, it's better to recall the proof. – Jean Marie Mar 01 '23 at 17:16