Problem 1044

Let $(x, y, z)$ be a solution of the system of equations (1). Then it holds that

$$\begin{aligned} & \left(a_{2} c_{1}-a_{1} c_{2}\right) x+\left(b_{2} c_{1}-b_{1} c_{2}\right) y=c_{1}-c_{2} \\ & \left(a_{3} c_{2}-a_{2} c_{3}\right) x+\left(b_{3} c_{2}-b_{2} c_{3}\right) y=c_{2}-c_{3} \\ & \left(a_{1} c_{3}-a_{3} c_{1}\right) x+\left(b_{1} c_{3}-b_{3} c_{1}\right) y=c_{3}-c_{1} \end{aligned}$$

From this, by multiplying with $b_{3}, b_{2}$, and $b_{1}$ respectively and adding, we obtain

$$D \cdot x=\left(c_{1}-c_{2}\right) b_{3}+\left(c_{2}-c_{3}\right) b_{1}+\left(c_{3}-c_{1}\right) b_{2}$$

where

$$D=a_{1} b_{2} c_{3}-a_{1} b_{3} c_{2}+a_{2} b_{3} c_{1}-a_{2} b_{1} c_{3}+a_{3} b_{1} c_{2}-a_{3} b_{2} c_{1}$$

Furthermore, we obtain

$$D \cdot y=\left(a_{1}-a_{2}\right) c_{3}+\left(a_{2}-a_{3}\right) c_{1}+\left(a_{3}-a_{1}\right) c_{2} \quad \text { and }$$

$$D \cdot z=\left(b_{1}-b_{2}\right) a_{3}+\left(b_{2}-b_{3}\right) a_{1}+\left(b_{3}-b_{1}\right) a_{2}$$

Therefore, the system of equations (1) has, due to (2), (4), and (85), exactly one real solution if $D \neq 0$, so that A wins; in the case $D=0$, no or infinitely many real solutions, so that B wins.

Hence, B can enforce the win with the following strategy, i.e., achieve that $D=0$:

1. Without loss of generality, assume that A first assigns the coefficient $a_{1}$ (all other possibilities can be reduced to this by swapping equations or swapping unknowns). Then B assigns the coefficient $c_{2}$ with 0 and achieves

$$D=a_{1} b_{2} c_{3}+a_{2} b_{3} c_{1}-a_{2} b_{1} c_{3}-a_{3} b_{2} c_{1}$$

where the coefficient $a_{1}$ is now fixed.

2. If A now assigns the coefficient $c_{3}$, B assigns the coefficient $b_{2}$ with 0 and achieves

$$D=a_{2} b_{3} c_{1}-a_{2} b_{1} c_{3}$$

If, on the other hand, A assigns another coefficient, B assigns the coefficient $c_{3}$ with 0 and achieves

$$D=a_{2} b_{3} c_{1}-a_{3} b_{2} c_{1}$$

In both cases, in the obtained representation of $D$, at most one coefficient is fixed (in the first case, this is exactly the coefficient $c_{3}$).

3. If A now assigns another coefficient, in at most one of the two products (from the obtained representation of $D$) two fixed coefficients will occur. B can then assign another coefficient with 0 in a product with the maximum number of fixed coefficients and achieve

$$D=a_{2} b_{3} c_{1} \quad \text { or } \quad D=-a_{3} b_{2} c_{1}$$

where in each case at most one coefficient is fixed.

4. Regardless of which coefficient A now assigns, B can achieve that $D=0$ by assigning the still free coefficient with 0.

Regardless of any further assignments made until the end of the game, B has thus enforced the win, as in the case $D=0$, the system of equations (1) has no or infinitely many solutions.

## Adapted from [3]