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Question #41025

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $4 \times 4$ matrix defined as $A = \begin{pmatrix} x_1 & x_1 & x_1 & x_1 \\ x_1 & x_2 & x_2 & x_2 \\ x_1 & x_2 & x_3 & x_3 \\ x_1 & x_2 & x_3 & x_4 \end{pmatrix}$ where $x_1, x_2, x_3, x_4 \in \mathbb{Z}$ . Find the number of distinct quadruplets $(x_1, x_2, x_3, x_4)$ such that $|\det(A)| = 293$ .

Options

A.
32
B.
64
C.
128
D.
256

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

Combinatorial Counting|Factor Theorem in Determinants|Triangular Matrix Shortcut|The Modulus Trap

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to simplify determinants using row operations to reveal a triangular matrix, effectively utilizing the Factor Theorem concept for matrices.

The core trap is the 'Modulus Trap': students often simplify $|\det(A)| = 293$ to $\det(A) = 293$ , forgetting the $\det(A) = -293$ case.

Solving only $\det(A) = 293$ requires an even number of negative factors, yielding exactly half the solutions (32).

Furthermore, the use of a prime number (293) forces students to rely on combinatorial counting of factor permutations and sign choices rather than algebraic expansion, rewarding those who recognize the bijective mapping between the differences $d_i$ and the original variables $x_i$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Combinatorial Counting (P&C Integration)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free