Question

If $a _{ r }=\cos \frac{2 r \pi}{9}+i \sin \frac{2 r \pi}{9}, r =1,2,3, \ldots, i=\sqrt{-1}$
then the determinant $\begin{vmatrix}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{vmatrix}$ is equal to :

• A. $a _{2} a _{6}- a _{4} a _{8}$
• B. $a_{9}$
• C. $a _{1} a _{9}- a _{3} a _{7}$
• D. $a _{5}$

Answer 1

Answer: (C)

$a_{r}=e^{\frac{i 2 \pi r}{9}}, r=1,2,3, \ldots a_{1}, a_{2}, a_{3}, \ldots$ are in G.P.
$\begin{vmatrix}a_{1} & a_{2} & a_{3} \\ a_{n} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{vmatrix}=\begin{vmatrix}a_{1} & a_{2}^{2} & a_{1}^{3} \\ a_{1}^{4} & a_{1}^{5} & a_{1}^{6} \\ a_{1}^{7} & a_{1}^{8} & a_{1}^{9}\end{vmatrix}=a_{1} \cdot a_{1}^{4} \cdot a_{1}^{7}\begin{vmatrix}1 & a_{1} & a_{1}^{2} \\ 1 & a_{1} & a_{1}^{2} \\ 1 & a_{1} & a_{1}^{2}\end{vmatrix}=0$
Now $a_{1} \,a_{9}-a_{3} \,a_{7}=a_{1}{ }^{10}-a_{1}{ }^{10}=0$