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

• A. $1$
• B. $3$
• C. $0$
• D. $-1$

Answer 1

The correct option is A $1$

Given system of linear equation has infinitely many solutions
$\therefore \Delta =0$

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

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

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

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

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

$\Rightarrow \alpha =\pm 1$

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

$\Rightarrow \alpha =-1$

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

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