Question

If the system of equations $x+\alpha y+{\alpha }^{2}z=1,\alpha x+y+\alpha z=-1,{\alpha }^{2}x+\alpha y+z=1$ has infinitely many solutions then $1+\alpha +{\alpha }^2=$

• A. 0
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is D

1

For infinitely many solutions, $\left|\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right|=0$

$\Rightarrow 1(1-{\alpha }^2)-\alpha (\alpha -{\alpha }^3)+{\alpha }^2({\alpha }^2-{\alpha }^2)=0$

$\therefore 1-{\alpha }^2-{\alpha }^2+{\alpha }^4\Rightarrow ({\alpha }^2-1{)}^2=0\Rightarrow {\alpha }^2=1\Rightarrow \alpha =\pm 1$

If $\alpha =1,$ then no solution

If $\alpha =-1,$ then all equations reduce to $x-y+z=1\Rightarrow$ infinitely many solutions.

$\Rightarrow 1+\alpha +{\alpha }^2=1-1+1=1$