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Let omega=-1/2+i(sqrt(3))/2, then value ...

Let `omega=-1/2+i(sqrt(3))/2,` then value of the determinant `[[1, 1, 1],[ 1,-1,-omega^2],[omega^2, omega^2,omega]]` is (a) `3omega` (b) `3omega(omega-1)` (c) `3omega^2` (d) `3omega(1-omega)`

A

`3 omega`

B

`3 omega ( omega -1)`

C

`3 omega^2`

D

`3 omega (1- omega)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `Delta=|{:(1" "1" "1),(1" "-1-omega^2" " omega^2),(1" " omega^(2) " "omega):}|`
Applying `R_2 rarr R_2-R_1: R_3 rarr R_3-R_3-R_1`
`=|{:(1" "1" "1),(0" "-2-omega^2" "omega^2-1),(0 " "omega^2-1" "omega-1):}|`
`=(-2 -omega^2)(omega-1)-(omega^2-1)^2`
`=-2omega+2-omega^2+omega^2-(omega^4-2omega^2+1)`
`3omega^2-3 omega=3 omega(omega-1)" "([because omega^4=omega)`
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