# Thread: Basis of a 3x3 magic square

1. ## 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?

2. 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.

3. 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?

4. 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.

5. Look at this

6. Thanks, that helps.
How do you know that the middle term is three times the other two terms though?

7. Solve this general matrix and you can prove that result.

$\begin{bmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{bmatrix}$

8. It took awhile, but I understand. Thank you very much for your help.

### basis of 3x3 magic square

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