Q51 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
If $\omega \neq 1$ is a cube root of unity, then
$$
\left|\begin{array}{ccc}
\omega+\omega^2 & \omega^2+\omega^9 & \omega^9+\omega \\
\omega^{27}+\omega^{31} & \omega^{31}+\omega^{17} & \omega^{17}+\omega^{27} \\
\omega^{30}+\omega^{41} & \omega^{41}+\omega^{19} & \omega^{19}+\omega^{30}
\end{array}\right|=
$$
- A$3$
- B$2$
- C$1$
- Correct answerD$0$
Solution
$\begin{aligned} & \left|\begin{array}{ccc}\omega+\omega^2 & \omega^2+\omega^9 & \omega^9+\omega \\ \omega^{27}+\omega^{31} & \omega^{31}+\omega^{17} & \omega^{17}+\omega^{27} \\ \omega^{30}+\omega^{41} & \omega^{41}+\omega^{19} & \omega^{19}+\omega^{30}\end{array}\right| \\ & =\left|\begin{array}{ccc}\omega+\omega^2 & \omega^2+1 & 1+\omega \\ 1+\omega & \omega+\omega^2 & \omega^2+1 \\ 1+\omega^2 & \omega^2+\omega & \omega+1\end{array}\right| \\ & R_1 \rightarrow R_1+R_2+R_3 \\ & =\left|\begin{array}{lll}2\left(1+\omega+\omega^2\right) & \omega^2+1 & 1+\omega \\ 2\left(1+\omega+\omega^2\right) & \omega+\omega^2 & \omega^2+1 \\ 2\left(1+\omega+\omega^2\right) & \omega^2+\omega & \omega+1\end{array}\right| \\ & =0 \quad\left(\because 1+\omega+\omega^2=0\right) \\ & \end{aligned}$
Hence, the correct answer is option: 'D'.