Question

If $a_r = (\cos \ 2r \pi + i \ \sin \ 2r \pi )^{1/9}$, then the value of $\begin{vmatrix}a_{1}&a_{2}&a_{3}\\ a_{4}&a_{5}&a_{6}\\ a_{7}&a_{8}&a_{9}\end{vmatrix} $ is

• A. 1
• B. -1
• C. 0
• D. 2

Answer 1

Answer: (C)

We have,
$a_{r}=(\cos 2 r \pi+i \sin 2 r \pi)^{1 / 9}=e^{\frac{2 r \pi i}{9}}$
Now, $\begin{vmatrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{vmatrix}=\begin{vmatrix}e^{\frac{2 \pi i}{9}} & e^{\frac{4 \pi i}{9}} & e^{\frac{6 \pi i}{9}} \\ e^{\frac{8 \pi i}{9}} & e^{\frac{10 \pi i}{9}} & e^{\frac{12 \pi i}{9}} \\ e^{\frac{14 \pi i}{9}} & e^{\frac{16 \pi i}{9}} & \frac{18 \pi i}{9}\end{vmatrix}$
$=e^{\frac{2 \pi i}{9}} \times e^{\frac{8 \pi i}{9}}\begin{vmatrix}1 & e^{\frac{2 \pi i}{9}} & e^{\frac{4 \pi i}{9}} \\ 1 & e^{\frac{2 \pi i}{9}} & e^{\frac{4 \pi i}{9}} \\ e^{\frac{14 \pi i}{9}} &e^{\frac{16 \pi i}{9}} & e^{\frac{18 \pi i}{9}}\end{vmatrix}$
$ =e^{\frac{2 \pi i}{9}} \times e^{\frac{8 \pi i}{9}} \times 0 (\because R_{1}$ and $ R_{2} $ are identical )
$=0$