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Question #40434

Difficulty: 8/10 | 🟢 Free

The Question

Let $\alpha_1, \alpha_2, \alpha_3$ be the roots of the equation $x^3 - 233x^2 + 17x - 5 = 0$ and $\beta_1, \beta_2, \beta_3$ be the roots of the equation $x^3 + 233x^2 - 19x + 7 = 0$ . A $3 \times 3$ matrix $M$ is defined such that its $(i, j)$ -th entry is given by $m_{ij} = \sin^2(\alpha_i + \beta_j) + \cos^2(\alpha_i - \beta_j)$ for $1 \le i, j \le 3$ . Find the value of $\det(M)$ .

Options

A.
0
B.
1
C.
233
D.
-233

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Factor-Product Conversion|Triangular/Diagonal Matrix Shortcut|Linear Dependency of Rows

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize linear dependency masked by trigonometric factor-product conversions.

The trap lies in the intimidating cubic equations with the coefficient '233', which tempts students into a brute-force cofactor expansion followed by complex Vieta's formula substitutions.

However, by applying double-angle and sum-to-product identities, the matrix entries simplify to a form $1 + \sin(2\alpha_i)\sin(2\beta_j)$ .

This reveals that the $3 \times 3$ matrix is the product of a $3 \times 2$ and a $2 \times 3$ matrix, meaning its column space is spanned by only two vectors.

Consequently, the rows are linearly dependent, the rank is at most 2, and the determinant is identically zero, rendering the cubic equations completely superfluous.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Factor-Product Conversion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free