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Question

Let a,b and c be three real numbers satisfying
[abc]197827737=[000] . . . (i)

Let ω be a solution of x3 - 1 = 0 with Im ω > 0. If a = 2 with b and c satisfying Eq. (i) then the value of 3ωa+1ωb+3ωc is

A
-2
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B
2
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C
3
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D
-3
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Solution

The correct option is A -2
On multiplying matrices, we have
[a+8b+7c9a+2b+3c7a+7b+7c]=[000]
On solving the 3 equations, we get b=6a and c=7a

If a = 2, we get , b = 12, c = -14
3ωa+1ωb+3ωc
3ω2+1ω12+3ω14=3ω2+1+3ω2=3ω+1+3ω2
= 1 + 3 (ω+ω2)= 1 - 3 = -2

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