JEE Main 2021 (31 Aug, Shift 1) Maths Q63: Roots of Unity

Question

If $\mathrm{a_r}=\cos\frac{2\mathrm{r}\pi}{9}+i\sin\frac{2\mathrm{r}\pi}{9}$, $\mathrm{r}=1, 2, 3, ..., i=\sqrt{-1}$, then the determinant
$\begin{vmatrix}\mathrm{a_1} & \mathrm{a_2} & \mathrm{a_3}\ \mathrm{a_4} & \mathrm{a_5} & \mathrm{a_6}\ \mathrm{a_7} & \mathrm{a_8} & \mathrm{a_9}\end{vmatrix}$ is equal to :

Accepted Answer

Option 2: $\mathrm{a_1a_9-a_3a_7}$

Answer

$\mathrm{a_9}$

Answer

$\mathrm{a_1a_9-a_3a_7}$

Answer

$\mathrm{a_2a_6-a_4a_8}$

Answer

$\mathrm{a_5}$

JEE Main 31 August 2021, Shift 1, Maths Q63

Same idea in other shifts