Mathematics - Determinant Question with Solution | TestHub

Given that

${\mathrm{a}}_{\mathrm{r}}=\cos \frac{2\mathrm{r}\pi }{9}+i\sin \frac{2\mathrm{r}\pi }{9},\mathrm{r}=1,2,3,\dots ,i=\sqrt{-1}$

$\Rightarrow a_r = e^{i\frac{2r\pi }{9}}----\left(I\right)$

From Euler form $e^{i\theta } = \cos \theta + i \sin \theta$

$\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|$

$=\left|\begin{array}{ccc}{e}^{i\frac{2\pi }{9}}& e^{i\frac{4\pi }{9}}& e^{i\frac{6\pi }{9}}\\ e^{i\frac{8\pi }{9}}& e^{i\frac{i10\pi }{9}}& e^{i\frac{12\pi }{9}}\\ e^{i\frac{14\pi }{9}}& e^{i\frac{i16\pi }{9}}& e^{i\frac{18\pi }{9}}\end{array}\right|$

Taking $e^{i\frac{2\pi }{9}}, e^{i\frac{8\pi }{9}}, e^{i\frac{14\pi }{9}}$ common form each row

$=e^{i\left(\frac{2\pi }{9}+\frac{8\pi }{9}+\frac{14\pi }{9}\right)}\left|\begin{array}{ccc}1& e^{i\frac{2\pi }{9}}& e^{i\frac{4\pi }{9}}\\ 1& e^{i\frac{2\pi }{9}}& e^{i\frac{4\pi }{9}}\\ 1& e^{i\frac{2\pi }{9}}& e^{i\frac{4\pi }{9}}\end{array}\right|$

We know that If two rows are identical the value of determinant will be equal to zero.

$= 0---\left(II\right)$

Now From equation $\left(I\right)$

$a_1=e^{i\frac{2\pi }{9}}, a_9=e^{i\left(\frac{2\times 9\pi }{9}\right)}, a_3=e^{i\left(\frac{2\times 3\pi }{9}\right)}, a_7=e^{i\left(\frac{2\times 7\pi }{9}\right)}$

${\mathrm{a}}_1{\mathrm{a}}_9-{\mathrm{a}}_3{\mathrm{a}}_7={\mathrm{e}}^{\mathrm{i}\left(\frac{20\pi }{9}\right)}-{\mathrm{e}}^{\mathrm{i}\left(\frac{20\pi }{9}\right)}=0 ----\left(III\right)$

From equation $\left(II\right) \& \left(III\right)$

$\left|\begin{array}{lll}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right| = a_1{a}_9 - a_3{a}_7$