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if omega!=1 is cube root of unity and x+...

if `omega!=1` is cube root of unity and x+y+z`!=`0 then `|[x/(1+omega),y/(omega+omega^2),z/(omega^2+1)],[y/(omega+omega^2),z/(omega^2+1),x/(1+omega)],[z/(omega^2+1),x/(1+omega),y/(omega+omega^2)]|`=0 if

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To solve the given determinant problem, we start with the determinant: \[ D = \begin{vmatrix} \frac{x}{1+\omega} & \frac{y}{\omega+\omega^2} & \frac{z}{\omega^2+1} \\ \frac{y}{\omega+\omega^2} & \frac{z}{\omega^2+1} & \frac{x}{1+\omega} \\ \frac{z}{\omega^2+1} & \frac{x}{1+\omega} & \frac{y}{\omega+\omega^2} \end{vmatrix} ...
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If omega!=1 is a cube root of unity and x+y+z!=0, then prove that |[x/(1+omega), y/(omega+omega^2),z/(omega^2+1)],[y/(omega+omega^2),z/(omega^2+1),x/(1+omega)],[(z)/(omega^2+1),x/(1+omega),y/(omega+omega^2)]|=0 if x=y=z

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

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 an imaginary cube root of unity, then the value of the determinant |(1+omega,omega^2,-omega),(1+omega^2,omega,-omega^2),(omega+omega^2,omega,-omega^2)|

If omega is a complex cube root of unity then (1-omega+omega^2)(1-omega^2+omega^4)(1-omega^4+omega^8)(1-omega^8+omega^16)

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If omega is a complex cube root of unity, show 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 and /_\ = |(1, 2 omega), (omega, omega^2)| , then /_\ ^2 is equal to (A) - omega (B) omega (C) 1 (D) omega^2

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CENGAGE ENGLISH-DETERMINANTS-All Questions
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