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Question #41281

Difficulty: 8/10 | 🟢 Free

The Question

Let $z_r = \cos\left(\frac{2\pi r}{11}\right) + i\sin\left(\frac{2\pi r}{11}\right)$ for $r = 1, 2, \dots, 10$ . Consider the vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3 \in \mathbb{C}^3$ such that the $k$ -th component of $\vec{v}_j$ is given by: $A_{jk} = \sum_{r=1}^{10} i^{r|j-k|} \left( z_r^{j-k} - z_r^{k-j} \right)$ where $i = \sqrt{-1}$ . If $V$ represents the volume of the parallelepiped formed by the vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$ , which is given by the modulus of their scalar triple product, find the value of $V$ .

Options

A.
0
B.
11
C.
$10 \sqrt{2}$
D.
121

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Skew-Symmetric Determinant|Scalar Triple Product Volume Logic|Algebraic Manipulation of i|De Moivre’s Theorem Applications

Why This Question Is Tricky

Common Mistakes

Students often mistakenly believe that the identity $\det(A) = 0$ for odd-ordered skew-symmetric matrices only applies to real matrices, incorrectly assuming that complex entries introduce non-vanishing imaginary parts due to $i^2 = -1$ .

The trap is failing to realize that the algebraic cancellation in the determinant expansion is universal over any field.

By forcing a skew-symmetric structure using powers of roots of unity and absolute value indices, the problem baits students into attempting a brutal De Moivre's expansion instead of recognizing the $A^T = -A$ property that strictly forces the scalar triple product (volume) to zero.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Skew-Symmetric Determinant
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free