TITLE: Determinant of a $3\times3$ magic square
QUESTION [10 upvotes]: This is my first question with mathOverflow so I hope my etiquette is up to par here.
My question is regarding a $3\times3$ magic square constructed using the la Loubère method (see la Loubère method)
Using the method, I have constructed a magic square and several semimagic squares (where one or both of the diagonals do not add up to a magic sum) with a program on written on my graphing calculator. After playing around with the program, I was shocked that the determinants of these $3\times3$ magic squares are all the same (specifically -360). Why is this so? (I am still an undergraduate so please go easy on the math :] )

REPLY [11 votes]: Kevin has done the magic squares; I'll do the semi-magics. So the rules are, we're using the 9 digits, once each, and all the row sums and all the column sums are equal. Note that this common sum must be 15. Now the only ways to use 1 and get 15 are $1+6+8$ and $1+5+9$. Since exchanging rows doesn't affect the (absolute value of) the determinant, and exchanging columns doesn't, either, and taking the transpose doesn't, and since further these operations don't affect semi-magicity, we may assume the matrix looks like this: $$\pmatrix{1&6&8\cr5&&\cr9&&\cr}$$ Now the only ways to use 9 are $1+5+9$, which we've already used, and $2+4+9$, so the matrix must look like $$\pmatrix{1&6&8\cr5&&\cr9&2&4\cr}\qquad{\rm or}\qquad\pmatrix{1&6&8\cr5&&\cr9&4&2\cr}$$ The first matrix fills out uniquely to $$\pmatrix{1&6&8\cr5&7&3\cr9&2&4\cr}$$ while the second one can't be completed. So, up to transpose, row swaps, and column swaps, there is only one order 3 semi-magic square, hence, only one determinant (up to sign).