3rd Putnam 1940

Problem B6

The n x n matrix (mij) is defined as mij = aiaj for i ≠ j, and ai2 + k for i = j. Show that det(mij) is divisible by kn-1 and find its other factor.




Answer: det(mij) = kn-1(k + Sai2).

Induction on n. Clearly true for n = 1.
Expanding by the first row we get k. kn-2(k + Si>1ai2) + det(m'ij), where m'ij is the same as mij except that m'11 = a12. Subtracting appropriate multiples of the first row from the others we zero all the elements outside the first row except those on the diagonal, which become k. Hence det(m'ij) = kn-2a12.



3rd Putnam 1940

© John Scholes
15 Sep 1999