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Question #67411

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be an $n \times n$ matrix (where $n \ge 3$ is an odd integer) such that $A^T + A = O$ . Let $P$ be an $n \times n$ permutation matrix formed by interchanging exactly two rows of the identity matrix $I_n$ . If $B = P A$ , which of the following expressions is identically equal to the null matrix $O$ ?

Options

A.
$\text{adj}(B^T) - P \cdot \text{adj}(A)$
B.
$\text{adj}(B^T) + P \cdot \text{adj}(A)$
C.
$\text{adj}(B^T) - \text{adj}(A) \cdot P$
D.
$\text{adj}(B^T) + \text{adj}(A) \cdot P$

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

Row/Column Exchange Parity|Adjoint-Determinant Relation|Singular vs Non-Singular Matrices

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question exploits the common student reflex to use the identity adj(M) = |M|M⁻¹ to simplify adjugate expressions.

By defining A as a skew-symmetric matrix of odd order, A is strictly singular (|A|=0), making A⁻¹ undefined.

If a student illegally assumes invertibility, they will incorrectly derive adj(Bᵀ) = |Bᵀ|(Bᵀ)⁻¹ = (-|A|)(-PA⁻¹) = P(|A|A⁻¹) = P·adj(A), leading them directly to the trap option adj(Bᵀ) - P·adj(A) = O.

The parity change (|B| = -|A|) acts as a distractor in this invalid derivation, perfectly canceling the negative sign from (-A)⁻¹ = -A⁻¹.

The correct path requires recognizing the null space constraint and using the universally valid polynomial identity adj(XY) = adj(Y)adj(X), along with the property adj(-M) = (-1)ⁿ⁻¹adj(M).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Row/Column Exchange Parity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free