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Question #40314

Difficulty: 8/10 | 🟢 Free

The Question

Let $M$ and $N$ be two matrices of order $3 \times 3$ with real entries such that $\det(M) = 4$ , and $N^2 = N$ where $N$ is not the identity matrix. Define the following sets: $A = \{ k \in \mathbb{R} \mid \det(kM) = k \det(M) \}$ $B = \{ k \in \mathbb{R} \mid kM \text{ is a singular matrix} \}$ $C = \{ k \in \mathbb{R} \mid \det(kN) = k \det(N) \}$ $D = \{ k \in \mathbb{R} \mid kN \text{ is a singular matrix} \}$ Which of the following represents the set $(A \Delta B) \cap (C \Delta D)$ ? (where $\Delta$ denotes the symmetric difference of two sets)

Options

A.
$\emptyset$
B.
$\{-1, 1\}$
C.
$\{0\}$
D.
$\{-1, 0, 1\}$

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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

Difference vs. Symmetric Difference|Determinant Scaling Property|Singular vs Non-Singular Matrices

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to distinguish between the algebraic scaling property of determinants ( $\det(kA) = k^n \det(A)$ ) and the definition of a singular matrix, particularly when the matrix itself is already singular.

A common trap is blindly solving the equation $k^n = k$ for the scaling property without checking if $\det(A) = 0$ .

For matrix $N$ , since $N^2 = N$ and $N \neq I$ , $\det(N) = 0$ .

This makes the equations for sets $C$ and $D$ degenerate into $0 = 0$ , meaning both sets encompass all real numbers ( $\mathbb{R}$ ), and their symmetric difference is the null set.

Students who mechanically solve $k^3 = k$ for $C$ will incorrectly conclude $C = \{-1, 0, 1\}$ and choose the trap option $\{-1, 1\}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Difference vs. Symmetric Difference
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free