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Math Help - Basis of a 3x3 magic square

  1. #1
    Junior Member krtica's Avatar
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    Basis of a 3x3 magic square

    I have 8 equations, therefore 8 rows, but I do not know what a general matrix would look like for a 3x3 magic square. What I have so far is a matrix with letters as entries. Then I set the sum of each row, column, and diagonal equal to x.

    I was told by others that the matrix is 8x11. I am pretty sure I have to augment the zero vector to the matrix and use Gauss-Jordan elimination. Can you please give me any suggestions?
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  2. #2
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    Please look here
    http://en.wikipedia.org/wiki/Magic_square.
    For 3x3 min magic constant is 15.
    If you add const to each cell then magic number will be 15+3*const.


    It is proved here
    Magic constant - Wikipedia, the free encyclopedia.
    Last edited by zzzoak; April 18th 2010 at 04:18 PM.
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  3. #3
    Junior Member krtica's Avatar
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    So, it means that x would equal 15.

    But, what would the matrix be? If I were to use zero's and one's for my matrix, would it be something like this?

    (1 0 0 0 0 0 0 0 0 1 | 15)
    (0 1 0 0 0 0 0 0 1 0 | 15)
    (0 0 1 0 0 0 0 1 0 0 | 15)
    etc.

    Maybe?
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  4. #4
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    As I understood in magic square are numbers from 1 to n^2 if n=3 we get 1, 2, 3 ... to 9,
    and matrix 3x3. Or I am not right?
    I think in your matrix there must be 3 one's in a row, with 9 elements in a row.
    Last edited by zzzoak; April 18th 2010 at 05:10 PM.
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  5. #5
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    Look at this
    Attached Files Attached Files
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  6. #6
    Junior Member krtica's Avatar
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    Thanks, that helps.
    How do you know that the middle term is three times the other two terms though?
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  7. #7
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    Solve this general matrix and you can prove that result.

    \begin{bmatrix}<br />
a & b & c\\ <br />
d & e & f\\ <br />
g & h & i<br />
\end{bmatrix}
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  8. #8
    Junior Member krtica's Avatar
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    It took awhile, but I understand. Thank you very much for your help.
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