Q46 / 160 · WBJEE 2010 · Mathematics

$\omega$ is an imaginary cube root of unity and $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$ then one of the values of $x$ is

  • A
    1
  • Correct answer
    B
    0
  • C
    -1
  • D
    2

Solution

$$ \begin{aligned} \text { Hints: } & \stackrel{C_1^{\prime} \rightarrow C_1+C_2+C_3}{\longrightarrow}\left|\begin{array}{ccc} x & \omega & 1 \\ x & \omega^2 & 1+x \\ x & x+\omega & \omega^2 \end{array}\right|=x\left|\begin{array}{ccc} 1 & \omega & 1 \\ 1 & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2 \end{array}\right| \\ \quad & =x\left|\begin{array}{ccc} 1 & \omega & 1 \\ 0 & \omega^2-\omega & x \\ 0 & x & \omega^2-1 \end{array}\right|=x\left\{\left(\omega^2-\omega\right)\left(\omega^2-1\right)-x^2\right\}=0 \quad \Rightarrow x=0 \quad \text { One value of } x=0 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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