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Let a,b and c be three real numbers sa...

Let a,b and c be three real numbers satisfying
[a b c ] `|(1,9,7),(8,2,7),(7,3,7)|`=[ 0 0 0 ] ….(i)
Let `omega` be a solution of `x^3-1=0` with `lim(omega) > 0` . If a=2 with b and c satisfying Eq.(i) then the value of `3/omega^4+ 1/omega^b + 1/omega^c` is :

A

`-2`

B

2

C

3

D

`-3`

Text Solution

Verified by Experts

The correct Answer is:
A

We have,
`[abc][{:(1,9,7),(2,8,7),(7,3,7):}]=[0,0,0]`
`rArr [a+2b+7c, 9a+8b+3c, 7a+7b+7c]=[0 0 0]`
`rArr a+2b+7c=0, 9a+8b+3c=0`and `7a+7b+7c=0`
`rArr a=5,b=-6,c=1`
Clearly, `omega=-1/sqrt(2)+isqrt(3)/2` and it satisfies `omega^(3)=1, 1/omega=omega^(2)`
`therefore 3/omega^(a)+1/omega^(b)+3/omega^(c)=3/omega^(5)+1/omega^(-6)+3/omega^(1)=3omega+1+3omega^(2)`
`=3(omega^(2)+omega)+1=-3+1=-2`
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