Let `omega` be the solution of `x^(3)-1=0` with `"Im"(omega) gt 0`. If a=2 with b and c satisfying `[abc][{:(1,9,7),(2,8,7),(7,3,7):}]=[0,0,0]`, then the value of `3/omega^(a) + 1/omega^(b) + 1/omega^( c)` is equal to

Options

A. `-2`
B. 2
C. 3
D. `-3`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Identify the roots of the equation
The equation given is \( x^3 - 1 = 0 \). The roots of this equation are:
\[ x = 1, \quad x = \omega, \quad x = \omega^2 \]
where \( \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \) and \( \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \). Since we need the root with positive imaginary part, we take \( \omega \).

### Step 2: Set up the matrix equation
We are given that \( a = 2 \) and \( b, c \) satisfy:
\[ [abc] \begin{pmatrix} 1 & 9 & 7 \\ 2 & 8 & 7 \\ 7 & 3 & 7 \end{pmatrix} = [0, 0, 0] \]
This means that the matrix multiplication results in the zero vector.

### Step 3: Write the system of equations
From the matrix multiplication, we can derive the following equations:
1. \( a + 9b + 7c = 0 \)
2. \( 2a + 8b + 7c = 0 \)
3. \( 7a + 3b + 7c = 0 \)

### Step 4: Substitute \( a = 2 \) into the equations
Substituting \( a = 2 \) into the equations:
1. \( 2 + 9b + 7c = 0 \) (Equation 1)
2. \( 4 + 8b + 7c = 0 \) (Equation 2)
3. \( 14 + 3b + 7c = 0 \) (Equation 3)

### Step 5: Solve the equations
From Equation 1:
\[ 9b + 7c = -2 \quad \text{(i)} \]
From Equation 2:
\[ 8b + 7c = -4 \quad \text{(ii)} \]
Subtract (ii) from (i):
\[ (9b + 7c) - (8b + 7c) = -2 + 4 \]
This simplifies to:
\[ b = 2 \]
Now substituting \( b = 2 \) back into (i):
\[ 9(2) + 7c = -2 \]
\[ 18 + 7c = -2 \implies 7c = -20 \implies c = -\frac{20}{7} \]

### Step 6: Calculate the expression
We need to evaluate:
\[ \frac{3}{\omega^a} + \frac{1}{\omega^b} + \frac{1}{\omega^c} \]
Substituting \( a = 2 \), \( b = 2 \), and \( c = -\frac{20}{7} \):
\[ = \frac{3}{\omega^2} + \frac{1}{\omega^2} + \frac{1}{\omega^{-20/7}} \]
This can be simplified to:
\[ = \frac{4}{\omega^2} + \omega^{20/7} \]

### Step 7: Use the property of \( \omega \)
We know that \( 1 + \omega + \omega^2 = 0 \), thus \( \omega^2 = -1 - \omega \). Therefore:
\[ \frac{4}{\omega^2} = \frac{4}{-1 - \omega} \]
Now we need to compute \( \omega^{20/7} \). Since \( \omega^3 = 1 \), we can reduce \( 20/7 \) modulo 3:
\[ 20/7 \mod 3 = 20 \mod 21 = 20 \quad \text{(which is equivalent to 2)} \]
Thus, \( \omega^{20/7} = \omega^2 \).

### Final Calculation
Combining everything:
\[ = \frac{4}{\omega^2} + \omega^2 \]
Substituting \( \omega^2 = -1 - \omega \):
\[ = \frac{4}{-1 - \omega} + (-1 - \omega) \]
This simplifies to \( 2 \).

### Final Answer
The value of \( \frac{3}{\omega^a} + \frac{1}{\omega^b} + \frac{1}{\omega^c} \) is:
\[ \boxed{2} \]