Matrices and Determinants 4 Question 33

34. For a real number $\alpha$, if the system

$ \begin{bmatrix} 1 & \alpha & \alpha^{2} \\ \alpha & 1 & \alpha \\ \alpha^{2} & \alpha & 1 \end{bmatrix} $ $ \begin{bmatrix} x\\ y\\ z \end{bmatrix}$ =$ \begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix}$

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

(2017 Adv.)

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Answer:

Correct Answer: 34. (1)

Solution:

  1. $\begin{bmatrix}1 & \alpha & \alpha^{2} \\ \alpha & 1 & \alpha \\ \alpha^{2} & \alpha & 1\end{bmatrix}=0$

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

$\Rightarrow \quad \alpha^{2}=1$

$\Rightarrow \quad \alpha= \pm 1$

But $\quad \alpha=1$ not possible

$\therefore \quad \alpha=-1$

[Not satisfying equation]

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



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