Question

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

Answer 1

Given system of linear equations has infinitely many solutions

$\therefore \Delta =0$

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

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

$⟹1-{\alpha }^2-{\alpha }^2+{\alpha }^4=0$

$⟹{\alpha }^4-2{\alpha }^2+1=0$

$⟹({\alpha }^2-1{)}^2=0$

$⟹\alpha =\pm 1$

If $\alpha =1$, then the matrix reduces to

$\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$

which is not possible since $x+y+z$ obtains two different values

$\therefore \alpha =1$ is not possible

$⟹\alpha =-1$

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

Hence, $1+\alpha +{\alpha }^2=1$