MHB Is a Latin Square always invertible?

[Sudharaka]

Hi everyone, :)

An interesting question I thought about recently. Is it true that a Latin Square of integers (or real numbers) treated as a matrix is always invertible? If not can anybody give a counterexample. I think latin squares are invertible but I am unable to prove it. Hope you can help me out. :)

 

[Opalg]

Hi everyone, :)

An interesting question I thought about recently. Is it true that a Latin Square of integers (or real numbers) treated as a matrix is always invertible? If not can anybody give a counterexample. I think latin squares are invertible but I am unable to prove it. Hope you can help me out. :)

The smallest counterexample I can come up with is the $6\times6$ circulant matrix $$\begin{bmatrix} 12&4&8&13&3&9 \\4&8&13&3&9&12 \\ 8&13&3&9&12&4 \\ 13&3&9&12&4&8 \\ 3&9&12&4&8&13 \\ 9&12&4&8&13&3 \end{bmatrix}\ .$$ This has determinant zero and is therefore not invertible.
If negative numbers are allowed, the problem is much easier, and the smallest counterexample would be $\begin{bmatrix}1&-1 \\ -1&1\end{bmatrix}.$ In fact, a Latin square matrix whose elements have sum zero is never invertible. Reason: if you add all the other rows of the matrix to the top row then each element of the top row becomes the sum of the elements. If that is zero then the determinant is zero.

More generally, if $x_1,x_2,\ldots,x_n$ are the elements in a circulant matrix, and $\omega$ is a primitive $n$th root of unity, then $x_1 + \omega x_2 + \omega^2x_3 + \ldots + \omega^{n-1}x_n$ is a factor of the determinant of the matrix. If that sum is zero then the matrix will not be invertible. The smallest example I could find where that happens with distinct positive integers for $x_1,x_2,\ldots,x_n$ was when $n=6$, using the numbers $12,4,8,13,3,9$ as in the above matrix.