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Matrix problem

  1. Aug 13, 2011 #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.
     
  2. jcsd
  3. Aug 13, 2011 #2

    disregardthat

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    "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?
     
  4. Aug 13, 2011 #3
    Yes, the entries consist of all the numbers 1 to n^2 and each number occurs only once.
     
  5. Aug 13, 2011 #4
    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.
     
  6. Aug 14, 2011 #5

    AlephZero

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    Homework Helper

    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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