**Problem A2**

The coefficients a_{ij} of the following equations

a_{11}x_{1} + a_{12} x_{2}+ a_{13} x_{3} = 0

a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} = 0

a_{31}x_{1} + a_{32}x_{2} + a_{33}x_{3} = 0

satisfy the following: (a) a_{11}, a_{22}, a_{33} are positive, (b) other a_{ij} are negative, (c) the sum of the coefficients in each equation is positive. Prove that the only solution is x_{1} = x_{2} = x_{3} = 0.

**Solution**

The slog solution is to multiply out the determinant and show it is non-zero. A slicker solution is to take the x_{i} with the largest absolute value. Say |x_{1}| ≥ |x_{2}|, |x_{3}|. Then looking at the first equation we have an immediate contradiction, since the first term has larger absolute value than the sum of the absolute values of the second two terms.

Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

© John Scholes

jscholes@kalva.demon.co.uk

27 Sep 1998

Last corrected/updated 26 Sep 2003