Home
Class 12
MATHS
Without expanding at any stage, prove th...

Without expanding at any stage, prove that `|{:(1,omega,omega^(2)),(omega,omega^(2),1),(omega^(2),1,omega):}|`= 0

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • MODEL TEST PAPER - 7

    ICSE|Exercise Section - A |19 Videos
  • MODEL TEST PAPER - 7

    ICSE|Exercise Section - B (In sub-parts (i) and (ii) choose the correct option and in sub-parts (iii) to (v), answer the questions as instructed.) |5 Videos
  • MODEL TEST PAPER - 4

    ICSE|Exercise SECTION - C|10 Videos
  • MODEL TEST PAPER - 8

    ICSE|Exercise Section - C |6 Videos

Similar Questions

Explore conceptually related problems

Prove that the value of determinant |{:(1,,omega,,omega^(2)),(omega ,,omega^(2),,1),( omega^(2),, 1,,omega):}|=0 where omega is complex cube root of unity .

If omega is a complex cube root of unity, then a root of the equation |(x +1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)| = 0 , is

If 1.omega, omega^2 are the roots of unity, then Delta=|(1,omega^n,omega^(2n)),(omega^n,omega^(2n),1),(omega^(2n),1,omega^n)| is equal to

If omega is cube roots of unity, prove that {[(1,omega,omega^2),(omega,omega^2,1),(omega^2,1,omega)]+[(omega,omega^2,1),(omega^2,1,omega),(omega,omega^2,1)]} [(1),(omega),(omega^2)]=[(0),(0),(0)]

If omega is cube roots of unity, prove that {[(1,omega,omega^2),(omega,omega^2,1),(omega^2,1,omega)]+[(omega,omega^2,1),(omega^2,1,omega),(omega,omega^2,1)]} [(1),(omega),(omega^2)]=[(0),(0),(0)]

If omega is a cube root of unity, then Root of polynomial is |(x + 1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)|

Let omega be the complex number cos((2pi)/3)+isin((2pi)/3) . Then the number of distinct complex cos numbers z satisfying Delta=|(z+1,omega,omega^2),(omega,z+omega^2,1),(omega^2,1,z+omega)|=0 is

Let omega be the complex number cos((2pi)/3)+isin((2pi)/3) . Then the number of distinct complex cos numbers z satisfying Delta=|(z+1,omega,omega^2),(omega,z+omega^2,1),(omega^2,1,z+omega)|=0 is

Let omega be the complex number cos((2pi)/3)+isin((2pi)/3) . Then the number of distinct complex cos numbers z satisfying Delta=|(z+1,omega,omega^2),(omega,z+omega^2,1),(omega^2,1,z+omega)|=0 is

If omega is a cube root of unity , then |(x+1 , omega , omega^2),(omega , x+omega^2, 1),(omega^2 , 1, x+omega)| =