If a _ r = cos(2rπ)/(9) + isin(2rπ)/(9), r = 1, 2, 3, , i = - 1 ,...

Given that

${a}_{r} = cos\dfrac{2r\pi}{9} + isin\dfrac{2r\pi}{9}, r = 1, 2, 3, \ldots , i = \sqrt{- 1}$

$⇒ {a}_{r} = {e}^{i\dfrac{2r\pi}{9}} - - - - ( I )$

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

\left|$\begin{cases}{a}_{1} & {a}_{2} & {a}_{3} \\ {a}_{4} & {a}_{5} & {a}_{6} \\ {a}_{7} & {a}_{8} & {a}_{9}\end{cases}$\right|$</p><p>$= \left|$\begin{cases}{e}^{i\dfrac{2\pi}{9}} & {e}^{i\dfrac{4\pi}{9}} & {e}^{i\dfrac{6\pi}{9}} \\ {e}^{i\dfrac{8\pi}{9}} & {e}^{i\dfrac{i10\pi}{9}} & {e}^{i\dfrac{12\pi}{9}} \\ {e}^{i\dfrac{14\pi}{9}} & {e}^{i\dfrac{i16\pi}{9}} & {e}^{i\dfrac{18\pi}{9}}\end{cases}$\right|$</p><p>Taking${e}^{i\dfrac{2\pi}{9}}, {e}^{i\dfrac{8\pi}{9}}, {e}^{i\dfrac{14\pi}{9}}$common form each row</p><p>$= {e}^{i\left(\dfrac{2\pi}{9} + \dfrac{8\pi}{9} + \dfrac{14\pi}{9}\right)}\left|$\begin{cases}1 & {e}^{i\dfrac{2\pi}{9}} & {e}^{i\dfrac{4\pi}{9}} \\ 1 & {e}^{i\dfrac{2\pi}{9}} & {e}^{i\dfrac{4\pi}{9}} \\ 1 & {e}^{i\dfrac{2\pi}{9}} & {e}^{i\dfrac{4\pi}{9}}\end{cases}$\right|$</p><p>We know that If two rows are identical the value of determinant will be equal to zero.</p><p>$= 0 - - - ( II )$</p><p>Now From equation$( I )$</p><p>${a}_{1} = {e}^{i\dfrac{2\pi}{9}}, {a}_{9} = {e}^{i ( \dfrac{2 \times 9\pi}{9} )}, {a}_{3} = {e}^{i ( \dfrac{2 \times 3\pi}{9} )}, {a}_{7} = {e}^{i ( \dfrac{2 \times 7\pi}{9} )}$</p><p>${a}_{1}{a}_{9} - {a}_{3}{a}_{7} = {e}^{i\left(\dfrac{20\pi}{9}\right)} - {e}^{i\left(\dfrac{20\pi}{9}\right)} = 0 - - - - ( III )$</p><p>From equation$( II ) \text{ and } ( III )</p><p>\left|\begin{cases}{a}_{1} & {a}_{2} & {a}_{3} \\ {a}_{4} & {a}_{5} & {a}_{6} \\ {a}_{7} & {a}_{8} & {a}_{9}\end{cases}\right| = {a}_{1}{a}_{9} - {a}_{3}{a}_{7}