JEE Main 2021 — Mathematics Algebra

Given that

$${a}_{r}=\mathrm{cos}\frac{2r\pi }{9}+i\mathrm{sin}\frac{2r\pi }{9},r=1,2,3,\ldots ,i=\sqrt{-1}$$ar​=cos92rπ​+isin92rπ​,r=1,2,3,…,i=−1​

$$\Rightarrow {a}_{r}={e}^{i\frac{2r\pi }{9}}----(I)$$⇒ar​=ei92rπ​−−−−(I)

From Euler form ${e}^{i\theta }=\mathrm{cos}\theta +i\mathrm{sin}\theta$eiθ=cosθ+isinθ

$$|\begin{matrix}{a}_{1} & {a}_{2} & {a}_{3} \\ {a}_{4} & {a}_{5} & {a}_{6} \\ {a}_{7} & {a}_{8} & {a}_{9}\end{matrix}|$$∣a1​a4​a7​​a2​a5​a8​​a3​a6​a9​​∣

$$=|\begin{matrix}{e}^{i\frac{2\pi }{9}} & {e}^{i\frac{4\pi }{9}} & {e}^{i\frac{6\pi }{9}} \\ {e}^{i\frac{8\pi }{9}} & {e}^{i\frac{i10\pi }{9}} & {e}^{i\frac{12\pi }{9}} \\ {e}^{i\frac{14\pi }{9}} & {e}^{i\frac{i16\pi }{9}} & {e}^{i\frac{18\pi }{9}}\end{matrix}|$$=∣ei92π​ei98π​ei914π​​ei94π​ei9i10π​ei9i16π​​ei96π​ei912π​ei918π​​∣

Taking ${e}^{i\frac{2\pi }{9}},{e}^{i\frac{8\pi }{9}},{e}^{i\frac{14\pi }{9}}$ei92π​,ei98π​,ei914π​ common form each row

$$={e}^{i(\frac{2\pi }{9}+\frac{8\pi }{9}+\frac{14\pi }{9})}|\begin{matrix}1 & {e}^{i\frac{2\pi }{9}} & {e}^{i\frac{4\pi }{9}} \\ 1 & {e}^{i\frac{2\pi }{9}} & {e}^{i\frac{4\pi }{9}} \\ 1 & {e}^{i\frac{2\pi }{9}} & {e}^{i\frac{4\pi }{9}}\end{matrix}|$$=ei(92π​+98π​+914π​)∣111​ei92π​ei92π​ei92π​​ei94π​ei94π​ei94π​​∣

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

$$=0---(II)$$=0−−−(II)

Now From equation $(I)$(I)

$${a}_{1}={e}^{i\frac{2\pi }{9}},{a}_{9}={e}^{i(\frac{2\times 9\pi }{9})},{a}_{3}={e}^{i(\frac{2\times 3\pi }{9})},{a}_{7}={e}^{i(\frac{2\times 7\pi }{9})}$$a1​=ei92π​,a9​=ei(92×9π​),a3​=ei(92×3π​),a7​=ei(92×7π​)

$${a}_{1}{a}_{9}-{a}_{3}{a}_{7}={e}^{i(\frac{20\pi }{9})}-{e}^{i(\frac{20\pi }{9})}=0----(III)$$a1​a9​−a3​a7​=ei(920π​)−ei(920π​)=0−−−−(III)

From equation (II)&(III)

$$|\begin{matrix}{a}_{1} & {a}_{2} & {a}_{3} \\ {a}_{4} & {a}_{5} & {a}_{6} \\ {a}_{7} & {a}_{8} & {a}_{9}\end{matrix}|={a}_{1}{a}_{9}-{a}_{3}{a}_{7}$$∣a1​a4​a7​​a2​a5​a8​​a3​a6​a9​​∣=a1​a9​−a3​a7​