Question

Let $a,x,y,z$ be real numbers satisfying the equations

$ax+ay=z$

$x+ay=z$

$x+ay=az,$ where

$x,y,z$ are not all zero, then the number of the possible values of $a$ is

Answer 1

As $x,y,z$ are not all zero, thus the system has a non-trivial solution.
Thus, $D=0$
$\Rightarrow \begin{vmatrix} a & a & -1 \\ 1 & a & -1 \\ 1 & a & -a \end{vmatrix}=0$
$\Rightarrow a\left(a^{2} - a\right)-a\left(a - 1\right)+1\left(a - a\right)=0$ or $a^{3}-a^{2}-a^{2}+a=0$
$\Rightarrow a\left(a^{2} - 2 a + 1\right)=0$
$\Rightarrow a=0,1$