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:
- $\begin{bmatrix}1 & \alpha & \alpha^{2} \\ \alpha & 1 & \alpha \\ \alpha^{2} & \alpha & 1\end{bmatrix}=0$
$\Rightarrow \alpha^{4}-2 \alpha^{2}+1=0$
$\Rightarrow \quad \alpha^{2}=1$
$\Rightarrow \quad \alpha= \pm 1$
But $\quad \alpha=1$ not possible
$\therefore \quad \alpha=-1$
[Not satisfying equation]
Hence, $1+\alpha+\alpha^{2}=1$
