Let \(a, b\) and \(c\) be three real numbers satisfying \[ \left[\begin{array}{lll} a & b & c \end{array}\right]\left[\begin{array}{lll} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right] \] Let \(\omega\) be a solution of \(x^{3}-1=0\) with \(\operatorname{Im}(\omega)>0\). If \(a=2\) with \(b\) and \(c\) satisfying (E), then the value of \[ \frac{3}{\omega^{a}}+\frac{1}{\omega^{b}}+\frac{3}{\omega^{c}} \] is equal to

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Complex Numbers

Let

a, b

and

c

be three real numbers satisfying

[a b c] 187923777 = [000]

Let

ω

be a solution of

x^{3} - 1 = 0

with

Im (ω) > 0

. If

a = 2

with

b

and

c

satisfying (E), then the value of

\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}

is equal to

Answer: (A)

From the previous question, we have

b = 6 a

and

c = - 7 a

.

With

a = 2

b = 12

and

c = - 14

.

Since

ω^{3} = 1

, we have

ω^{12} = (ω^{3})^{4} = 1

and

ω^{- 14} = ω^{- 14 + 15} = ω

.

Also,

ω^{2} = \overline{ω}

(the complex conjugate) and

1 + ω + ω^{2} = 0

.

The expression becomes:

\frac{3}{ω ^{2}} + \frac{1}{ω ^{12}} + \frac{3}{ω ^{- 14}} = 3 ω + 1 + 3 ω^{- 1} = 3 ω + 1 + 3 ω^{2}

.

Since

ω + ω^{2} = - 1

, we have

3 (ω + ω^{2}) + 1 = 3 (- 1) + 1 = - 2

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option