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WBJEE · Maths · Determinants

\(\omega\) 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

  1. A 1
  2. B 0
  3. C -1
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(B) 0

Step-by-step Solution

Detailed explanation

\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 &…