Question

If $a_r=(\cos 2r\pi +i\sin 2r\pi {)}^{\frac{1}{9}}$, 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. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

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