Putnam 51/A1 solution

11th Putnam 1951

Problem A1

A is a skew-symmetric real 4 x 4 matrix. Show that det A ≥ 0.

Solution

Straightforward.

Let A =

 0  a  b  c


-a  0  d  e


-b -d  0  f


-c -e -f  0

Then det A = a²f² + 2acdf - 2abef + b²e² - 2bcde + c²d².

At this point it is helpful to notice that a only appears with f, b with e, and c with d. So putting X = af, Y = cd, Z = be, we have that det A = X² + Y² + Z² + 2XY - 2XZ - 2YZ. This easily factorizes as (X + Y - Z)².

11th Putnam 1951