posted
I was playing around with numbers the other say, and I came up with this problem:
If you construct a 3 x 3 Matrix, where entries in the matrix correspond with the numbers of a magic square composed of the numbers 1 through 9, what does each row and column of the inverse of this matrix add to?
Posts: 2489 | Registered: Jan 2002
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posted
It's pretty much just what you think it would be, according to a couple of tests I just ran using an online matrix calculator. But I'm sure as heck not going to prove it.
(It wouldn't be that hard, just a little long.)
Posts: 26071 | Registered: Oct 2003
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and essentially the inverse of a magic square will be a magic square where all the rows columns, and diagonals add up to the inverse of the sum of the original magic square.
I think this is true for all odd magic squares, although I wouldn't want to make a proof.
posted
Very neat MEC! I *had* to delurk for this one
It does hold true for all magic squares. Turns out the proof is not that long.
Proof : Let A be a nxn magic square with sum S. To sum the rows of A, we can post-multiply it by the vector v of all ones i.e transpose([1 1 1 ...1 ]).
Then, since all of A's rows sum to S, Av = trans([S S S ....S]) = Sv i.e Av = Sv (1)
(For example, in the 3x3 case, A is our magic square, S is 15 and v is | 1 | | 1 | | 1 | )
To sum inv(A)'s rows, we need inv(A)*v. So, multiply Eqn (1) by inv(A).
=> inv(A)*A*v = S*inv(A)*v => v = S*inv(A)*v.
Divide both sides by S
=> inv(A)*v = 1/S*v i.e the Sum of the rows of inv(A) is 1/S. So, it turns out that we only need that all the rows of the matrix sum to the same number for the inverse to have all row sums to be 1/S. The columns work out in the same way.
The general case is easier to prove than the specific 3x3 case! I love math Posts: 4 | Registered: Mar 2008
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