Minimum Absolute Value of a nxn Matrix Determinant

[skymariner]

The following matrix problem occurred to me. I figured out the answer and would like to pose the problem. It's easy but would be best for an undergrad math major. The question: Consider a square n by n matrix with entries 1, 2, ..., n squared. Find a way to arrange these entries so that the absolute value of the determinate of this matrix is a minimum.

 

[disregardthat]

"a square n by n matrix with entries 1, 2, ..., n squared"

could you explain this a little better? Do you mean that the matrix have entries 1,2,3,...,n^2, and do all appear and only once?

 

[skymariner]

Yes, the entries consist of all the numbers 1 to n^2 and each number occurs only once.

 

[Dickfore]

Let's see it for a 2x2 matrix. Since a transposition of rows and columns does not change the value of the matrix, and a transposition of row or columns changes the sign, it is sufficient to consider how many different rows we can form. For a 2x2 matrix, we can form the first row in 3 ways (pairing one fixed element with the other 3 elements) and the second row can be formed in the 2 possible permutations of the remaining elements, so 3x2 = 6 possible determinants. Here they are:
<br /> \left|\begin{array}{cc}<br /> 1 &amp; 2 \\<br /> <br /> 3 &amp; 4<br /> \end{array}\right| = 4 - 6 = -2<br />

<br /> \left|\begin{array}{cc}<br /> 1 &amp; 2 \\<br /> <br /> 4 &amp; 3<br /> \end{array}\right| = 3 - 8 = -5<br />

<br /> \left|\begin{array}{cc}<br /> 1 &amp; 3 \\<br /> <br /> 2 &amp; 4<br /> \end{array}\right| = 4 - 6 = -2<br />

<br /> \left|\begin{array}{cc}<br /> 1 &amp; 3 \\<br /> <br /> 4 &amp; 2<br /> \end{array}\right| = 2 - 12 = -10<br />

<br /> \left|\begin{array}{cc}<br /> 1 &amp; 4 \\<br /> <br /> 2 &amp; 3<br /> \end{array}\right| = 3 - 8 = -5<br />

<br /> \left|\begin{array}{cc}<br /> 1 &amp; 4 \\<br /> <br /> 3 &amp; 2<br /> \end{array}\right| = 2 - 12 = -10<br />

I don't see how to generalize it at this point. I would think we need to add the biggest numbers n^{2}, (n - 1)^{2}, \ldots, n^{2} - n +1 along the main diagonal, then the next along the third to the main diagonal and start inserting the smallest elements along the odd diagonals.

 

[AlephZero]

I don't want to spoil other people's fun with this so I'll give the answer without a proof:
When n >= 3, the minimum absolute value of the determinant is 0.