If `a_(r)="cos"(2rpi)/9+i "sin"(2rpi)/9` then value of the determinant
`Delta=|(1,a_(8),a_(7)),(a_(3),a_(2),a_(1)),(a_(6),a_(5),a_(4))|` is

Options

A. `-1`
B. `1`
C. 0
D. `-2`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & a_8 & a_7 \\ a_3 & a_2 & a_1 \\ a_6 & a_5 & a_4 \end{vmatrix} \) where \( a_r = \cos\left(\frac{2r\pi}{9}\right) + i \sin\left(\frac{2r\pi}{9}\right) \), we can recognize that \( a_r \) can be expressed as \( a_r = e^{\frac{2\pi ir}{9}} \).

### Step 1: Express the elements of the determinant
The elements of the determinant can be rewritten using the exponential form:
- \( a_1 = e^{\frac{2\pi i \cdot 1}{9}} \)
- \( a_2 = e^{\frac{2\pi i \cdot 2}{9}} \)
- \( a_3 = e^{\frac{2\pi i \cdot 3}{9}} \)
- \( a_4 = e^{\frac{2\pi i \cdot 4}{9}} \)
- \( a_5 = e^{\frac{2\pi i \cdot 5}{9}} \)
- \( a_6 = e^{\frac{2\pi i \cdot 6}{9}} \)
- \( a_7 = e^{\frac{2\pi i \cdot 7}{9}} \)
- \( a_8 = e^{\frac{2\pi i \cdot 8}{9}} \)

### Step 2: Substitute into the determinant
Now substitute these values into the determinant:
\[ \Delta = \begin{vmatrix} 1 & e^{\frac{2\pi i \cdot 8}{9}} & e^{\frac{2\pi i \cdot 7}{9}} \\ e^{\frac{2\pi i \cdot 3}{9}} & e^{\frac{2\pi i \cdot 2}{9}} & e^{\frac{2\pi i \cdot 1}{9}} \\ e^{\frac{2\pi i \cdot 6}{9}} & e^{\frac{2\pi i \cdot 5}{9}} & e^{\frac{2\pi i \cdot 4}{9}} \end{vmatrix} \]

### Step 3: Factor out common terms
Notice that each column can be factored by \( e^{\frac{2\pi i}{9}} \):
\[ \Delta = e^{\frac{2\pi i}{9}} \begin{vmatrix} 1 & e^{\frac{2\pi i \cdot 8}{9}} & e^{\frac{2\pi i \cdot 7}{9}} \\ e^{\frac{2\pi i \cdot 3}{9}} & e^{\frac{2\pi i \cdot 2}{9}} & e^{\frac{2\pi i \cdot 1}{9}} \\ e^{\frac{2\pi i \cdot 6}{9}} & e^{\frac{2\pi i \cdot 5}{9}} & e^{\frac{2\pi i \cdot 4}{9}} \end{vmatrix} \]

### Step 4: Simplify the determinant
The determinant can be simplified using properties of determinants. The determinant of a matrix with complex exponentials can be computed using the roots of unity. The values \( e^{\frac{2\pi i k}{9}} \) for \( k = 0, 1, \ldots, 8 \) are the 9th roots of unity.

### Step 5: Calculate the determinant
The determinant evaluates to zero because the rows are linearly dependent (as they are derived from the same set of roots of unity). Thus, we find:
\[ \Delta = 0 \]

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