Gy. 1729. Solve the system of equations,

$$
\begin{aligned}
& a_{11} x+a_{12} y+a_{13} z=0 \\
& a_{21} x+a_{22} y+a_{23} z=0 \\
& a_{31} x+a_{32} y+a_{33} z=0
\end{aligned}
$$

if the coefficients satisfy the following conditions:

a) each one is positive

b) the sum of the coefficients in each row and each column is 1,
c) $a_{11}=a_{22}=a_{33}=1 / 2$.
Let's denote the coefficient $a_{12}$ by $t$, and use conditions b) and c). Then our system of equations becomes:

$$
\begin{aligned}
& \frac{1}{2} x + t y + \left(\frac{1}{2} - t\right) z = 0 \\
& \left(\frac{1}{2} - t\right) x + \frac{1}{2} y + t z = 0 \\
& t x + \left(\frac{1}{2} - t\right) y + \frac{1}{2} z = 0
\end{aligned}
$$

By adding the three equations, we get $x + y + z = 0$. Subtract the $t$-multiple of this equation from the second equation of the system (1) and the $\frac{1}{2}$-multiple from the third equation. We obtain:

$$
\left(t - \frac{1}{2}\right) x - t y = 0
$$

Here, we multiply the first equation by $t$, the second by $\left(\frac{1}{2} - t\right)$, and then add the two equations. After rearranging the coefficient of $x$, we get:

$$
\left(-3 t^{2} + \frac{3}{2} t - \frac{1}{4}\right) x = 0
$$

This last equation's coefficient can be written in the form $\left[-3\left(t - \frac{1}{4}\right)^{2} - \frac{1}{16}\right]$, from which it is clear that this coefficient is negative for all values of $t$, so $x = 0$. Substituting this into any equation of the system (2) immediately yields that $y = 0$, and therefore $z = 0$. This is the only solution to the system of equations.

Remark. This problem is a special case of a problem outside the P. 294. competition (see page 166 of this issue). Unlike the general case, here we did not need to use condition a).