The number of solutions of the system of equations $\alpha^2x+\alpha y = - 1$ $\alpha x+ \alpha^2y = 1$ is infinite. Then $\alpha $ is :

0 1 -1 2

-1

The correct answer is Option (3) → -1 for infinite no. of solution to exist $\frac{α^2}{α}=\frac{α}{α^2}=\frac{-1}{1}$ so $α=-α^2⇒α^2+α=0$ so $α=0,1$ for $α=0$ not exists so $α=-1$