Question

If $a_r=\cos \frac{2r\pi }{9}+i\sin \frac{2r\pi }{9},r=1,2,3,...,i=\sqrt{-1},$ then the determinant $\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|$ is equal to

• A. $a_1{a}_9-a_3{a}_7$
• B. $a_2{a}_6-a_4{a}_8$
• C. $a_9$
• D. $a_5$

Answer 1

The correct option is A $a_1{a}_9-a_3{a}_7$

$a_r=\cos \frac{2r\pi }{9}+i\sin \frac{2r\pi }{9}\Rightarrow a_r=e^{i\frac{2r\pi }{9}}\text{where}r=1,2,3,...$
Now,
$\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|=\left|\begin{array}{ccc}{e}^{i\frac{2\pi }{9}}& e^{i\frac{4\pi }{9}}& e^{i\frac{6\pi }{9}}\\ e^{i\frac{8\pi }{9}}& e^{i\frac{10\pi }{9}}& e^{i\frac{12\pi }{9}}\\ e^{i\frac{14\pi }{9}}& e^{i\frac{16\pi }{9}}& e^{i\frac{18\pi }{9}}\end{array}\right|$
$=e^{i(\frac{2\pi }{9}+\frac{8\pi }{9}+\frac{14\pi }{9})}\left|\begin{array}{ccc}1& e^{i\frac{2\pi }{9}}& e^{i\frac{4\pi }{9}}\\ 1& e^{i\frac{2\pi }{9}}& e^{i\frac{4\pi }{9}}\\ 1& e^{i\frac{2\pi }{9}}& e^{i\frac{18\pi }{9}}\end{array}\right|=0$
Now, checking from the options
$a_1\cdot a_9-a_3\cdot a_7=e^{i\frac{2\pi }{9}}\cdot e^{i\frac{18\pi }{9}}-e^{i\frac{6\pi }{9}}\cdot e^{i\frac{14\pi }{9}}=e^{i\frac{20\pi }{9}}-e^{i\frac{20\pi }{9}}=0$