if w is a complex cube root to unity then value of
` Delta =|{:(a_(1)+b_(1)w,,a_(1)w^(2)+b_(1),,c_(1)+b_(1)bar(w)),(a_(2)+b_(2)w,,a_(2)w^(2)+b_(2),,c_(2)+b_(2)bar(w)),(a_(3)+b_(3)w,,a_(3)w^(2)+b_(3),,c_(3)+b_(3)bar(w)):}|` is

Options

A. 0
B. -1
C. 2
D. none of these

Explanation

To solve the given problem, we need to evaluate the determinant:

\[ \Delta = \begin{vmatrix} a_1 + b_1 \omega & a_1 \omega^2 + b_1 & c_1 + b_1 \bar{\omega} \\ a_2 + b_2 \omega & a_2 \omega^2 + b_2 & c_2 + b_2 \bar{\omega} \\ a_3 + b_3 \omega & a_3 \omega^2 + b_3 & c_3 + b_3 \bar{\omega} \end{vmatrix} \]

where \(\omega\) is a complex cube root of unity.

### Step 1: Understanding the properties of cube roots of unity
The complex cube roots of unity are \(1, \omega, \omega^2\), where:
- \(\omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i\)
- \(\omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i\)
- The important property is that \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

### Step 2: Multiply the second column by \(\omega\)
We can multiply the second column of the determinant by \(\omega\). According to the properties of determinants, multiplying a column by a scalar multiplies the determinant by that scalar.

\[ \Delta = \frac{1}{\omega} \begin{vmatrix} a_1 + b_1 \omega & a_1 \omega^3 + b_1 \omega & c_1 + b_1 \bar{\omega} \\ a_2 + b_2 \omega & a_2 \omega^3 + b_2 \omega & c_2 + b_2 \bar{\omega} \\ a_3 + b_3 \omega & a_3 \omega^3 + b_3 \omega & c_3 + b_3 \bar{\omega} \end{vmatrix} \]

Since \(\omega^3 = 1\), we can simplify the determinant:

\[ \Delta = \frac{1}{\omega} \begin{vmatrix} a_1 + b_1 \omega & a_1 + b_1 \omega & c_1 + b_1 \bar{\omega} \\ a_2 + b_2 \omega & a_2 + b_2 \omega & c_2 + b_2 \bar{\omega} \\ a_3 + b_3 \omega & a_3 + b_3 \omega & c_3 + b_3 \bar{\omega} \end{vmatrix} \]

### Step 3: Identifying identical columns
Now we can see that the first and second columns are identical:

\[ \begin{vmatrix} x & x & z \\ y & y & w \\ z & z & v \end{vmatrix} \]

where \(x = a_i + b_i \omega\), \(y = a_i + b_i \omega\), and \(z = c_i + b_i \bar{\omega}\).

### Step 4: Conclusion
The determinant of a matrix with two identical columns is zero. Therefore, we conclude:

\[ \Delta = 0 \]

### Final Answer
The value of the determinant \(\Delta\) is \(0\).

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