Given any real numbers α1, α2, ... , αm, β, show that for m, n > 1 we can find m real n x n matrices A1, ... , Am such that det Ai = αi, and det(∑ Ai) = β.
Start by setting Ai to be the matrix with 1, 1, ..., 1, αi down the main diagonal and zeros elsewhere. Modify A1 by changing the n, n-1 element to 1. Modify Am by changing the n-1, n element to m(α1 + ... + αm) - β/mn-2. It is clear that this gives det Ai = αi.
A1 + ... + Am has m, m, ... , m, (α1 + ... + αm) down the main diagonal. The only other non-zero elements are n, n-1, which is 1 and n-1, n, which is m(α1 + ... + αm) - β/mn-2. Hence its determinant evaluates to β.
20th Putnam 1959
© John Scholes
15 Feb 2002