20th Putnam 1959

Given any real numbers α₁, α₂, ... , α_m, β, show that for m, n > 1 we can find m real n x n matrices A₁, ... , A_m such that det A_i = α_i, and det(∑ A_i) = β.

Solution

Start by setting A_i to be the matrix with 1, 1, ..., 1, α_i down the main diagonal and zeros elsewhere. Modify A₁ by changing the n, n-1 element to 1. Modify A_m by changing the n-1, n element to m(α₁ + ... + α_m) - β/m^n-2. It is clear that this gives det A_i = α_i.

A₁ + ... + A_m has m, m, ... , m, (α₁ + ... + α_m) 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 + ... + α_m) - β/m^n-2. Hence its determinant evaluates to β.

20th Putnam 1959

© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002