Matrix problem

  • Thread starter skymariner
  • Start date
  • #1
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.
 

Answers and Replies

  • #2
disregardthat
Science Advisor
1,866
34
"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?
 
  • #3
Yes, the entries consist of all the numbers 1 to n^2 and each number occurs only once.
 
  • #4
2,981
5
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:
[tex]
\left|\begin{array}{cc}
1 & 2 \\

3 & 4
\end{array}\right| = 4 - 6 = -2
[/tex]

[tex]
\left|\begin{array}{cc}
1 & 2 \\

4 & 3
\end{array}\right| = 3 - 8 = -5
[/tex]

[tex]
\left|\begin{array}{cc}
1 & 3 \\

2 & 4
\end{array}\right| = 4 - 6 = -2
[/tex]

[tex]
\left|\begin{array}{cc}
1 & 3 \\

4 & 2
\end{array}\right| = 2 - 12 = -10
[/tex]

[tex]
\left|\begin{array}{cc}
1 & 4 \\

2 & 3
\end{array}\right| = 3 - 8 = -5
[/tex]

[tex]
\left|\begin{array}{cc}
1 & 4 \\

3 & 2
\end{array}\right| = 2 - 12 = -10
[/tex]

I don't see how to generalize it at this point. I would think we need to add the biggest numbers [itex]n^{2}, (n - 1)^{2}, \ldots, n^{2} - n +1[/itex] along the main diagonal, then the next along the third to the main diagonal and start inserting the smallest elements along the odd diagonals.
 
  • #5
AlephZero
Science Advisor
Homework Helper
7,002
293
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.
 

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