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Question #67355

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $3 \times 3$ matrix with real entries and $A = S + K$ be its decomposition into symmetric and skew-symmetric parts. Let $C$ be the cofactor matrix of $\alpha K$ , where $\alpha \in \mathbb{R}$ . If $t = \text{tr}(K^T K)$ , which of the following represents $\det(\text{cof}(C + \beta K))$ for any $\beta \in \mathbb{R}$ ?

Options

A.
$\frac{1}{16} \alpha^4 \beta^4 t^4$
B.
$\frac{1}{16} \alpha^6 \beta^4 t^4$
C.
$\frac{1}{64} \alpha^6 \beta^6 t^6$
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

Minors and Cofactors Sign Convention|Symmetric and Skew-Symmetric Decomposition|Determinant Scaling Property

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

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

Difficulty Boost : +4

Common Mistakes

The question tests the scaling properties of cofactor matrices and the determinant of sums involving skew-symmetric matrices.

A common trap is incorrectly applying $\text{cof}(kA) = k^n \text{cof}(A)$ instead of $k^{n-1}$ , leading to erroneous powers of $\alpha$ .

Furthermore, students often fail to recognize that for an odd-ordered skew-symmetric matrix $K$ , $\det(K)=0$ and $\text{adj}(K)$ is a symmetric rank-1 matrix.

By forcing the evaluation of $\det(C + \beta K)$ through the matrix determinant lemma (or $3 \times 3$ determinant addition formula), the problem collapses into a trace computation, punishing rote memorization of determinant properties.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Minors and Cofactors Sign Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free