# If a, = cos2rπ/9 + isin2rπ/9, r = 1, 2, 3, ......, i

Question:

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 :

1. $a_{2} a_{6}-a_{4} a_{8}$

2. $\mathrm{a}_{9}$

3. $a_{1} a_{9}-a_{3} a_{7}$

4. $\mathrm{a}_{5}$

Correct Option: , 3

Solution:

$a_{r}=e^{\frac{i 2 \pi r}{9}}, r=1,2,3, \ldots a_{1}, a_{2}, a_{3}, \ldots$ 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 $a_{1} a_{9}-a_{3} a_{7}=a_{1}{ }^{10}-a_{1}{ }^{10}=0$