Prove that the inverse of a 3 x 3 magical square is indeed also a magical square.?

Question

all rows, columns, diagonals add up to the same number in a magical square.

once you find the initial perfect square, show the proof that the inverse is still indeed a magical square. Answers that just sum every row, column, and diagonal will not get the best answer. Best of luck, at the end of the question if its not answered I will post the correct solution.

Answer 1

Well, I know that the sum of the rows, columns, diagonals of the inverse of a magic square is the reciprocal of the sum in the original square. The algebraic manipulation to show this is incredibly long, so there's got to be an easier way to prove it. Hmm..

First point to be made is that a TRUE magic square is one that has different values for each of the n^2 terms. Then proof of the conjecture is far beyond the scope of this forum. For simplicity, I'll just assume that they don't all have to be different.

Okay, let n be the sum in the magic square. Let A represent the magic square in matrix form, and |1| represent the identity vector. We then have:

|n| = A |1|

because by matrix operation on vectors, all the column terms add up to the terms of the resultant vector |n|, where all the components are n because A is a magic square. Now, we do the following:

A^-1 |n| = A^-1 A |1| = |1|

but A^-1 |n| = n A^-1 |1|, so that

A^-1 |1| = |1/n|

so that we see that A^-1 has column terms that all add up to 1/n

We can do a transpose on A and repeat to show that it's true for rows as well. The last things are the diagonals. Hmm...

I give up. Do it the hard way. For any a, b, c, we can construct a 3x3 magic square as follows:

2a............b.........a-b+3c
-b+4c......a+c.....2a+b-2c
a+b-c....2a-b+2c....2c

so that the terms add up to 3a+3c. Very laboriously working out the terms of the inverse of this matrix, and finding the trace, we have:

Tr(A^-1) = 1 / (3a+3c)

and the conjecture is proven

Addendum: This is not true for NxN magic squares for N > 3, so there doesn't exist a general matrix proof where the rank isn't specific.