## Task $4 / 90$

Let $a$ be a real number. Solve the following system of equations in real numbers $x, y, z$. Which values of $a$ are excluded?

$$
\begin{array}{r}
a x+y+z=1 \\
x+a y+z=a \\
x+y+a z=a^{2}
\end{array}
$$
Obviously, $a \neq 1$ must hold, because for $a=1$ the three equations would coincide and thus the system would be indeterminate. Adding the three equations, we get $(a+2)(x+y+z)=1+a+a^{2}$. Clearly, $a \neq -2$ must hold, as $a=-2$ leads to the contradiction $0=3$. Given $a+2 \neq 0$, this equation is equivalent to

$$
x+y+z=\frac{1+a+a^{2}}{a+2}
$$

Subtracting this equation from each equation of the given system, we obtain (with $a$ real, $a \neq 1, a \neq -2$)

$$
x=-\frac{a+1}{a+2} ; \quad y=\frac{1}{a+2} ; \quad z=\frac{(a+1)^{2}}{a+2}
$$