Question

if w is a cube root of unity then a root of the following equation
|x+1 w w^2 |
|w x+w 1 | = 0 is?
|w^2 1 x+w|​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

If ω is a cube root of unity then a root of the following equation is

$\displaystyle\begin{vmatrix} \: \: x + 1& \omega & {\omega}^{2} \\ \\ \omega & x + {\omega}^{2} & 1 \\ \\ {\omega}^{2} & 1 & x + \omega \: \: \end{vmatrix} = 0$

(a) x = 1 (b) x = ω (c) x = ω² (d) x = 0

EVALUATION

Here the given equation is

$\displaystyle\begin{vmatrix} \: \: x + 1& \omega & {\omega}^{2} \\ \\ \omega & x + {\omega}^{2} & 1 \\ \\ {\omega}^{2} & 1 & x + \omega \: \: \end{vmatrix} = 0$

We simplify it as below

$\displaystyle\begin{vmatrix} \: \: x + 1& \omega & {\omega}^{2} \\ \\ \omega & x + {\omega}^{2} & 1 \\ \\ {\omega}^{2} & 1 & x + \omega \: \: \end{vmatrix} = 0$

$\displaystyle \sf{ \implies \: \begin{vmatrix} \: \: x + 1 +\omega + {\omega}^{2} & \omega + x + 1 + {\omega}^{2} & {\omega}^{2} + x + 1 +\omega \\ \\ \omega & x + {\omega}^{2} & 1 \\ \\ {\omega}^{2} & 1 & x + \omega \: \: \end{vmatrix}} = 0 \: \: ( \: R_1' = R_1 + R_2 + R_3)$

$\displaystyle \sf{\implies \:\begin{vmatrix} \: \: x + 1 +\omega + {\omega}^{2} & x + 1 +\omega + {\omega}^{2} & x + 1 +\omega + {\omega}^{2} \\ \\ \omega & x + {\omega}^{2} & 1 \\ \\ {\omega}^{2} & 1 & x + \omega \: \: \end{vmatrix}} = 0$

$\displaystyle \sf{\implies \:\begin{vmatrix} \: \: x + 0 & x + 0 & x + 0 \\ \\ \omega & x + {\omega}^{2} & 1 \\ \\ {\omega}^{2} & 1 & x + \omega \: \: \end{vmatrix}} = 0 \: \: \: ( \because \: 1 +\omega + {\omega}^{2} = 0)$

$\displaystyle \sf{\implies \:\begin{vmatrix} \: \: x & x & x \\ \\ \omega & x + {\omega}^{2} & 1 \\ \\ {\omega}^{2} & 1 & x + \omega \: \: \end{vmatrix}} = 0$

$\displaystyle \sf{\implies \: x\:\begin{vmatrix} \: \: 1 & 1 & 1 \\ \\ \omega & x + {\omega}^{2} & 1 \\ \\ {\omega}^{2} & 1 & x + \omega \: \: \end{vmatrix}} = 0$

$\displaystyle \sf{\implies \: x = 0 \:}$

FINAL ANSWER

Hence the correct option is (d) x = 0

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