Before You Begin

Question #41240

Difficulty: 8/10 | 🟢 Free

The Question

Let $A, B,$ and $C$ be $3 \times 3$ real skew-symmetric matrices. It is given that $\text{Tr}(A^2) = -2$ , $\text{Tr}(B^2) = -8$ , $\text{Tr}(C^2) = -18$ , and $\text{Tr}(AB) = \text{Tr}(BC) = \text{Tr}(CA) = 0$ . For any vector $\vec{x} \in \mathbb{R}^3$ , the matrices act as cross products: $A\vec{x} = \vec{a} \times \vec{x}$ , $B\vec{x} = \vec{b} \times \vec{x}$ , and $C\vec{x} = \vec{c} \times \vec{x}$ , where $\vec{a}, \vec{b}, \vec{c}$ form a right-handed orthogonal system (i.e., $[\vec{a} \vec{b} \vec{c}] > 0$ ). If $P = AB - BA$ , $Q = BC - CB$ , and $R = CA - AC$ , find the value of $\text{Tr}(PQR)$ .

Options

A.
-36
B.
36
C.
-216
D.
0

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

Matrix Trace Properties|Matrix Multiplication Non-Commutativity|Vector Product Anti-Commutativity

Why This Question Is Tricky

Common Mistakes

This question tests the deep connection between the Lie algebra of skew-symmetric matrices and vector cross products.

The primary trap lies in the evaluation of $\text{Tr}(PQR)$ .

Students often incorrectly assume full commutativity $\text{Tr}(ABC) = \text{Tr}(ACB)$ or guess that the trace of the product of three skew-symmetric matrices is zero.

By mapping the matrix product to the scalar triple product, the anti-commutative nature of the cross product ( $\vec{w} \times \vec{v} = -\vec{v} \times \vec{w}$ ) introduces a critical negative sign: $\text{Tr}(M_{\vec{u}}M_{\vec{v}}M_{\vec{w}}) = -[\vec{u}\vec{v}\vec{w}]$ .

Missing this sign leads directly to the trap option 36.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Matrix Trace Properties
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free