If ar = cos(2rπ/9) + i sin(2rπ/9), r = 1, 2, 3, ..., i = √-1, then the determinant |a1 a2 a3; a4 a5

Q: Why did we get zero? Is it always zero?

In this specific case, yes. When rows of a determinant are in a Geometric Progression, they are linearly dependent. Linearly dependent rows mean the determinant is zero. It is a classic JEE property that rewards those who look for patterns rather than brute force!

Q: Can I just expand the determinant directly?

You could, but you would be trapped in a nightmare of sines and cosines. JEE Advanced tests your ability to spot patterns, not your stamina for tedious calculation. Always look for properties first!

Q: How do I know which option is correct if the answer is zero?

Great question! Once you find the determinant is zero, you must evaluate the options. Option (3) $a_1a_9 - a_3a_7$ simplifies to $k^{10} - k^{10} = 0$. It is the only one that matches.

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 :

Options:

(A) $a_2a_6 - a_4a_8$

(B) $a_9$

(C) $a_1a_9 - a_3a_7$

(D) $a_5$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem presents a determinant involving trigonometric terms:
$$\Delta = \begin{vmatrix} a_1 & a_2 & a_3 \ a_4 & a_5 & a_6 \ a_7 & a_8 & a_9 \end{vmatrix}$$

Expanding this directly using the standard definition would lead to an unmanageable sea of sines and cosines. In JEE Advanced mathematics, such structures almost always hide a deeper symmetry.

### The Euler Transformation

The first step is to move beyond the trigonometric representation of $a_r = \cos \frac{2r\pi}{9} + i \sin \frac{2r\pi}{9}$. By applying **Euler's formula**, $e^{i	heta} = \cos 	heta + i \sin 	heta$, we can rewrite the terms as:
$$a_r = e^{i\frac{2r\pi}{9}}$$

Let us define a constant $k = e^{i\frac{2\pi}{9}}$. Consequently, the expression simplifies to $a_r = k^r$.

This transformation reveals that the sequence $a_1, a_2, a_3, \dots$ is a **Geometric Progression** defined by $k, k^2, k^3, \dots$.

### The Determinant Dance

Substituting these values into our determinant, we obtain:
$$\Delta = \begin{vmatrix} k & k^2 & k^3 \ k^4 & k^5 & k^6 \ k^7 & k^8 & k^9 \end{vmatrix}$$

Observe the rows of the matrix. We can factor out $k$ from the first row, $k^4$ from the second row, and $k^7$ from the third row:
$$\Delta = k \cdot k^4 \cdot k^7 \begin{vmatrix} 1 & k & k^2 \ 1 & k & k^2 \ 1 & k & k^2 \end{vmatrix}$$

A fundamental property of determinants states that if any two rows or columns are identical, the value of the determinant is zero. Since all three rows are identical in this case, we conclude:
$$\Delta = 0$$

### Final Verification

To verify this result against the provided options, we examine the expression $a_1a_9 - a_3a_7$. Substituting our exponential forms:
$$(k^1)(k^9) - (k^3)(k^7) = k^{10} - k^{10} = 0$$

The result matches perfectly. This problem demonstrates that when you encounter sequences within a matrix, you should look for a **Geometric Progression**, and when you see complex trigonometric arguments, you should apply **Euler's formula**.

Animated Solution for Mathematics - Complex Numbers: If $a_{r} = cos \frac{2 r π}{9} + i sin \frac{2 r π}{9}, r = 1, 2, 3, \dots, i = - 1$ , then the determinant $a_{1} a_{4} a_{7} a_{2} a_{5} a_{8} a_{3} a_{6} a_{9}$ is equal to :

Select Answer:

Visualized Solution

Understanding $a_{r}$

Euler's Form Conversion

Identifying the Common Ratio

Substituting into the Determinant

Factoring the Rows

The Zero Result

Checking the Options

Final Conclusion

The Sigma Insight: Euler's Form and De Moivre's Theorem

Analyzing the Setup

The Euler Transformation

The Determinant Dance

Final Verification

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