MCQ
For a real number $\alpha$, if the system

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

of linear equations, has infinitely many solutions, then $1+\alpha+\alpha^2=$

  • A
    $5$
  • B
    $8$
  • C
    $2$
  • $1$

Answer

Correct option: D.
$1$
d
Here, $D=\left|\begin{array}{lll}1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1\end{array}\right|=0$

which gives $\alpha=-1$ or +1

For $\alpha=1$, the equations become

$ x+y+z=1$

$x+y+z=-1$

$and x+y+z=1$

which give no solution

For $\alpha=-1$, the equations become

$x-y+z=1$

$-x+y-z=-1$

$x-y+z=1$

which are all same and hence infinitely many solutions

Hence, $\alpha=-1$

$\Rightarrow 1+\alpha+\alpha^2=1$

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