Question

If $a_r=(\cos 2r\pi +i\sin 2r\pi {)}^{1/9}$, then the value of $\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|$ is

• A. 1
• B. -1
• C. 0
• D. 2

Answer 1

Answer: (C)

We have,
$a_r=(\cos 2r\pi +i\sin 2r\pi {)}^{1/9}=e^{\frac{2r\pi i}{9}}$
Now, $\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|=\left|\begin{array}{ccc}{e}^{\frac{2\pi i}{9}}& e^{\frac{4\pi i}{9}}& e^{\frac{6\pi i}{9}}\\ e^{\frac{8\pi i}{9}}& e^{\frac{10\pi i}{9}}& e^{\frac{12\pi i}{9}}\\ e^{\frac{14\pi i}{9}}& e^{\frac{16\pi i}{9}}& \frac{18\pi i}{9}\end{array}\right|$
$=e^{\frac{2\pi i}{9}}\times e^{\frac{8\pi i}{9}}\left|\begin{array}{ccc}1& e^{\frac{2\pi i}{9}}& e^{\frac{4\pi i}{9}}\\ 1& e^{\frac{2\pi i}{9}}& e^{\frac{4\pi i}{9}}\\ e^{\frac{14\pi i}{9}}& e^{\frac{16\pi i}{9}}& e^{\frac{18\pi i}{9}}\end{array}\right|$
$=e^{\frac{2\pi i}{9}}\times e^{\frac{8\pi i}{9}}\times 0(\because R_1$ and $R_2$ are identical )
$=0$