If $\mathrm{a}_{\mathrm{r}}=\cos \frac{2 \mathrm{r} \pi}{9}+i \sin \frac{2 \mathrm{r} \pi}{9}, \mathrm{r}=1,2,3, \ldots, i=\sqrt{-1}$
then the determinant $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is equal to :
Correct Option: , 3
$a_{r}=e^{\frac{i 2 \pi r}{9}}, r=1,2,3, \ldots a_{1}, a_{2}, a_{3}, \ldots$ are in G.P.
$\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{n} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|=\left|\begin{array}{lll}a_{1} & a_{2}^{2} & a_{1}^{3} \\ a_{1}^{4} & a_{1}^{5} & a_{1}^{6} \\ a_{1}^{7} & a_{1}^{8} & a_{1}^{9}\end{array}\right|=a_{1} \cdot a_{1}^{4} \cdot a_{1}^{7}\left|\begin{array}{ccc}1 & a_{1} & a_{1}^{2} \\ 1 & a_{1} & a_{1}^{2} \\ 1 & a_{1} & a_{1}^{2}\end{array}\right|=0$
Now $a_{1} a_{9}-a_{3} a_{7}=a_{1}{ }^{10}-a_{1}{ }^{10}=0$
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