### 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 = a^{2}f^{2} + 2acdf - 2abef + b^{2}e^{2} - 2bcde + c^{2}d^{2}.
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 = X^{2} + Y^{2} + Z^{2} + 2XY - 2XZ - 2YZ. This easily factorizes as (X + Y - Z)^{2}.

11th Putnam 1951

© John Scholes

jscholes@kalva.demon.co.uk

24 Nov 1999