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Question #41925

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $3 \times 3$ symmetric matrix and $B$ be a $3 \times 3$ skew-symmetric matrix. Define two matrices $P = A^2 B - B A^2$ and $Q = A B^2 - B^2 A$ . Which of the following statements is ALWAYS true regarding the determinants of $P$ and $Q$ ?

Options

A.
\det(Q) = 0 \text{ and } \exists A, B \text{ such that } \det(P) \neq 0
B.
\det(P) = 0 \text{ and } \exists A, B \text{ such that } \det(Q) \neq 0
C.
\det(P) = 0 \text{ and } \det(Q) = 0 \text{ for all } A, B
D.
\exists A, B \text{ such that } \det(P) \neq 0 \text{ and } \det(Q) \neq 0

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Linear Independence vs Dependence|Matrix Multiplication Non-Commutativity|Symmetry vs. Antisymmetry

Why This Question Is Tricky

Common Mistakes

This question tests the properties of matrix multiplication non-commutativity and its effect on symmetry and linear independence.

The trap lies in assuming that the product of symmetric and skew-symmetric matrices follows simple parity rules without explicitly expanding the transpose of the commutator.

Students often incorrectly assume $A^2B - BA^2$ is skew-symmetric because $B$ is skew-symmetric, leading them to conclude both determinants are zero.

By forcing the evaluation of the transpose, it reveals $A^2B - BA^2$ is symmetric (potentially full rank, preserving linear independence) while $AB^2 - B^2A$ is skew-symmetric (singular in odd dimensions, destroying linear independence).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Linear Independence vs Dependence
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free