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Question #41287

Difficulty: 8/10 | 🟢 Free

The Question

Let $\omega = e^{i \frac{2\pi}{7}}$ . Let $A$ be a $6 \times 6$ lower triangular matrix with diagonal entries $A_{kk} = 1 - \omega^k + \frac{1}{1 - \omega^k}$ for $k = 1, 2, \dots, 6$ . Let $B$ be a $6 \times 6$ upper triangular matrix with diagonal entries $B_{kk} = 1 - \omega^{6k} + \frac{1}{1 - \omega^{6k}}$ for $k = 1, 2, \dots, 6$ . Find the value of $49 \det(AB)$ .

Options

A.
12769
B.
16384
C.
11449
D.
14161

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

Triangular/Diagonal Matrix Shortcut|Determinant Product Rule|De Moivre’s Theorem Applications

Why This Question Is Tricky

Common Mistakes

This question tests the ability to decouple the determinant of a product of triangular matrices and evaluate the resulting product of complex factors using cyclotomic polynomials.

The primary trap lies in attempting to multiply $A_{kk}$ and $B_{kk}$ directly, which leads to an intractable trigonometric expression.

A secondary trap is mishandling the roots of the quadratic $x^2-2x+2$ when applying the identity $\prod_{k=1}^{n-1} (x - \omega^k) = \frac{x^n-1}{x-1}$ , particularly confusing the signs or modulus properties of the complex numbers $(1 \pm i)^n$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Triangular/Diagonal Matrix Shortcut
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free