Hatrack River Forum: Math Puzzle

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Posted by MEC (Member # 2968) on :

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?

Posted by Achilles (Member # 7741) on :

Is this going to be on the test?

Posted by Dagonee (Member # 5818) on :

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. [Smile]

(It wouldn't be that hard, just a little long.)

Posted by Corwin (Member # 5705) on :

What's the magic square again? Same sum on lines, columns and diagonals?

Posted by MEC (Member # 2968) on :

yes.

***Answer***

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.

***Answer***

Posted by Dagonee (Member # 5818) on :

The proof is basic algebra.

It's just a LOT of basic algebra. [Smile]

Posted by MEC (Member # 2968) on :

I didn't say that I couldn't, just that I wouldn't want to.
[Razz]

Posted by NewLurker (Member # 11525) on :

Very neat MEC! I *had* to delurk for this one [Big Grin]

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 [Big Grin]

Posted by Mike (Member # 55) on :

Welcome, NewLurker. Nice Proof! [Smile]