If the system of equations x + ay = 0, az + y = 0, ax + z = 0 has

The given system of homogeneous equations x + ay = 0, az + y = 0, ax + z = 0 has infinitely many solutions.

$\therefore\left|\begin{array}{lll}1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1\end{array}\right|=0$

$\Rightarrow 1(1-0)-a\left(0-a^{2}\right)+0(0-a)=0$

$\Rightarrow 1+a^{3}=0$

$\Rightarrow(1+a)\left(1-a+a^{2}\right)=0$

$\Rightarrow a+1=0 \quad\left(a^{2}-a+1>0 \forall a \in \mathrm{R}\right)$

$\Rightarrow a=-1$

Thus, the value of $a$ is $-1$.

If the system of equations x + ay = 0, az + y = 0, ax + z = 0 has infinitely many solutions then a = __−1__.