If `a=cos((4pi)/3)+isin((4pi)/3)` then `|(1,1,1),(1,a,a^2),(1,a^2,a)|` (a) purely real (b) purely imaginary (c) `0` (d) none of these

Options

A. purely real
B. purely imaginary
C. `0`
D. none of these

Explanation

To solve the given problem, we need to evaluate the determinant \( |(1,1,1),(1,a,a^2),(1,a^2,a)| \) where \( a = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) \).

### Step-by-Step Solution:

1. **Identify \( a \)**:
\[ a = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) \]
The angle \( \frac{4\pi}{3} \) is in the third quadrant, where:
\[ \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}, \quad \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]
Therefore,
\[ a = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \]

2. **Set up the determinant**:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{vmatrix} \]

3. **Apply column operations**:
Replace \( C_2 \) with \( C_2 - C_1 \) and \( C_3 \) with \( C_3 - C_1 \):
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & a - 1 & a^2 - 1 \\ 1 & a^2 - 1 & a - 1 \end{vmatrix} \]

4. **Expand the determinant**:
Using the first row for expansion:
\[ D = 1 \cdot \begin{vmatrix} a - 1 & a^2 - 1 \\ a^2 - 1 & a - 1 \end{vmatrix} \]
The determinant of a \( 2 \times 2 \) matrix is given by \( ad - bc \):
\[ D = (a - 1)(a - 1) - (a^2 - 1)(a^2 - 1) \]
\[ = (a - 1)^2 - (a^2 - 1)^2 \]

5. **Simplify the expression**:
Let \( x = a - 1 \) and \( y = a^2 - 1 \):
\[ D = x^2 - y^2 = (x - y)(x + y) \]
Where:
\[ x = a - 1 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} - 1 = -\frac{3}{2} - i\frac{\sqrt{3}}{2} \]
\[ y = a^2 - 1 = \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)^2 - 1 \]
Calculate \( a^2 \):
\[ a^2 = \left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right)\left(-i\frac{\sqrt{3}}{2}\right) + \left(-i\frac{\sqrt{3}}{2}\right)^2 \]
\[ = \frac{1}{4} + i\frac{\sqrt{3}}{2} - \frac{3}{4} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \]
Thus:
\[ y = -\frac{1}{2} + i\frac{\sqrt{3}}{2} - 1 = -\frac{3}{2} + i\frac{\sqrt{3}}{2} \]

6. **Calculate \( x - y \) and \( x + y \)**:
\[ x - y = (-\frac{3}{2} - i\frac{\sqrt{3}}{2}) - (-\frac{3}{2} + i\frac{\sqrt{3}}{2}) = -i\sqrt{3} \]
\[ x + y = (-\frac{3}{2} - i\frac{\sqrt{3}}{2}) + (-\frac{3}{2} + i\frac{\sqrt{3}}{2}) = -3 \]

7. **Final result**:
\[ D = (-i\sqrt{3})(-3) = 3i\sqrt{3} \]
Since this is purely imaginary, the answer is:
\[ \text{(b) purely imaginary} \]