For a real number a., if the system $\left[\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$ of linear equations, has infinitely many solutions, then $1+\alpha +{\alpha }^2=$

 

For $AX=B$ to have infinitely many solutions, a necessary (but NOT sufficient) condition is $\det A=0$,for if $\det A\ne 0$ then $A^{-1}$ 

exists and we get  $X=A^{-1}B$ to be a unique solution.

$\det A=0\Rightarrow \left|\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right|=0$

Applying $R_1\to R_1+R_2+R_3,$ we get 

$\left|\begin{array}{ccc}1+\alpha +{\alpha }^2& 2\alpha +1& {\alpha }^2+\alpha +1\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right|=0$

Applying $C_3\to C_3-C_1$ we get

$\begin{array}{l}\left|\begin{array}{ccc}1+\alpha +{\alpha }^2& 2\alpha +1& 0\\ \alpha & 1& 0\\ {\alpha }^2& \alpha & 1-{\alpha }^2\end{array}\right|=0\\ \Rightarrow \left(1-{\alpha }^2\right)\left(1+\alpha +{\alpha }^2-2{\alpha }^2-\alpha \right)=0\\ \Rightarrow \left(1-{\alpha }^2\right)\left(1-{\alpha }^2\right)=0\\ \therefore {\alpha }^2=1\Rightarrow \alpha =1,-1\end{array}$

For $a=1,$the equation becomes 

$\begin{array}{l}x+y+z=1\\ x+y+z=-1\\ x+y+z=1\end{array}$

which gives no solution

For $a = -1 ,$the equation becomes 

$\begin{array}{l}x-y+z=1\\ -x+y-z=-1\\ x-y+z=1\end{array}$

which are all the same and hence infinitely many solutions.

Now, $1+\alpha +{\alpha }^2=1-1+1=1$