If a r = cos (2 r π/9)+i sin (2 r π/9), r =1,2,3, ldots, i=√-1 then the determinant |a1 a2 a3 a4 a5 a6 a7 a8 a9| is equal to :

Question

If a r = cos (2 r π/9)+i sin (2 r π/9), r =1,2,3, ldots, i=√-1 then the determinant |a1 a2 a3 a4 a5 a6 a7 a8 a9| is equal to : (A) a 2 a 6- a 4 a 8 (B) a9 (C) a 1 a 9- a 3 a 7 (D) a 5

Tardigrade Admin · Accepted Answer

Q. If $a_{r} = cos \frac{2 r π}{9} + i sin \frac{2 r π}{9}, r = 1, 2, 3, \dots, i = - 1$
then the determinant $∣ ∣ a_{1} a_{4} a_{7} a_{2} a_{5} a_{8} a_{3} a_{6} a_{9} ∣ ∣$ is equal to :

$a_{2} a_{6} - a_{4} a_{8}$

$a_{9}$

$a_{1} a_{9} - a_{3} a_{7}$

$a_{5}$

Q. If ar​=cos92rπ​+isin92rπ​,r=1,2,3,…,i=−1​ then the determinant ∣∣​a1​a4​a7​​a2​a5​a8​​a3​a6​a9​​∣∣​ is equal to :

a2​a6​−a4​a8​

a9​

a1​a9​−a3​a7​

a5​

Q. If $a_{r} = cos \frac{2 r π}{9} + i sin \frac{2 r π}{9}, r = 1, 2, 3, \dots, i = - 1$
then the determinant $∣ ∣ a_{1} a_{4} a_{7} a_{2} a_{5} a_{8} a_{3} a_{6} a_{9} ∣ ∣$ is equal to :

$a_{2} a_{6} - a_{4} a_{8}$

$a_{9}$

$a_{1} a_{9} - a_{3} a_{7}$

$a_{5}$