# If the system of equations $x+ay=0,az+y=0$ and $ax+z=0$ has infinite solution then the value of a is

$(a)\;-1\qquad(b)\;1\qquad(c)\;0\qquad(d)\;no\;real\;values$

The given equation is $x+ay=0,az+y=0,ax+z=0$
It is system of homogeneous equations therefore it will have infinite many solution if determinant of coefficient matrix is zero.
(i.e) $\begin{vmatrix}1 &a&0\\0 &1&a\\a&0&1\end{vmatrix}=0$
$\Rightarrow 1(1-0)-a(0-a^2)=0$
$\Rightarrow 1+a^3=0$
$\Rightarrow a^3=-1$
$\Rightarrow a=-1$
Hence (a) is the correct answer.