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

Consider the matrix $\left[\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right]$

determinant of $\left|\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right|$

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

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

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

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

$\Rightarrow \alpha =\pm 1$

For $\alpha =1\left[\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{cc}1& \\ -1& \alpha \\ 1& \end{array}\right]$ becomes

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

$x+y+z=1,x+y+z=-1,x+y+z=1$ thus this system has no solution.

For $\alpha =-1\left[\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{cc}1& \\ -1& \alpha \\ 1& \end{array}\right]$ becomes

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

$x-y+z=1,-x+y-z=-1,x-y+z=1$ thus this system has solution.

$\therefore \alpha =-1$

Substituting $\alpha =-1$ in $1+\alpha +{\alpha }^2=1-1+1=1$

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