3rd Putnam 1940

The n x n matrix (m_ij) is defined as m_ij = a_ia_j for i ≠ j, and a_i² + k for i = j. Show that det(m_ij) is divisible by k^n-1 and find its other factor.

Solution

Straightforward.

Answer: det(m_ij) = k^n-1(k + Sa_i²).

Induction on n. Clearly true for n = 1.
Expanding by the first row we get k. k^n-2(k + S_i>1a_i²) + det(m'_ij), where m'_ij is the same as m_ij 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) = k^n-2a₁².

3rd Putnam 1940

© John Scholes
jscholes@kalva.demon.co.uk
15 Sep 1999