Question

If \( a_r = \cos\frac{2r\pi}{9} + i \sin\frac{2r\pi}{9}, r = 1, 2, 3, \dots, i = \sqrt{-1} \), then the determinant \( \begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{vmatrix} \) is equal to :

• A. \( a_9 \)
• B. \( a_1a_9 - a_3a_7 \)
• C. \( a_2a_6 - a_4a_8 \)
• D. \( a_5 \)

Hint:

If the rows or columns of a determinant are in a Geometric Progression with the same common ratio, the determinant is always zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The elements are defined as \( a_r = e^{i \frac{2r\pi}{9}} \). This implies that \( a_r = (a_1)^r \). Let \( a_1 = \alpha \). Then the elements are terms of a Geometric Progression.
Step 2: Detailed Explanation:
The determinant is:
\[ \Delta = \begin{vmatrix} \alpha & \alpha^2 & \alpha^3 \\ \alpha^4 & \alpha^5 & \alpha^6 \\ \alpha^7 & \alpha^8 & \alpha^9 \end{vmatrix} \]
Take out common factors from the rows:
- Factor out \( \alpha \) from \( R_1 \).
- Factor out \( \alpha^4 \) from \( R_2 \).
- Factor out \( \alpha^7 \) from \( R_3 \).
\[ \Delta = \alpha \cdot \alpha^4 \cdot \alpha^7 \begin{vmatrix} 1 & \alpha & \alpha^2 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha & \alpha^2 \end{vmatrix} \]
Since all the rows of the resulting determinant are identical, the value of the determinant is \( 0 \).
Now, check the options to see which one evaluates to \( 0 \):
- (B) \( a_1a_9 - a_3a_7 = \alpha^1 \alpha^9 - \alpha^3 \alpha^7 = \alpha^{10} - \alpha^{10} = 0 \).
- (C) \( a_2a_6 - a_4a_8 = \alpha^8 - \alpha^{12} \neq 0 \).
Step 3: Final Answer:
The determinant is \( 0 \), matching option (B).