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Question #40559

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, 3, 4, 5, 6\}$ . The set $S$ is partitioned into 3 unordered subsets of 2 elements each. For each such partition, the sums of the elements in the three subsets are assigned to the upper triangular entries $a_{12}, a_{13}, a_{23}$ of a $3 \times 3$ skew-symmetric matrix $A$ . Let $\mathcal{M}$ be the set of all distinct matrices $A$ that can be formed by considering all possible partitions of $S$ and all possible assignments of the sums. Find the value of $\sum_{A \in \mathcal{M}} \det(A \cdot \text{adj}(A) + 2I)$ where $I$ is the $3 \times 3$ identity matrix.

Options

A.
488
B.
720
C.
244
D.
122

Check Your Answer

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

Fundamental Principle of Counting (FPC)|Division into Groups|Skew-Symmetric Determinant|Adjoint-Determinant Relation

Why This Question Is Tricky

Common Mistakes

This question constructs a 'Lethal Collision' between the Fundamental Principle of Counting (FPC) and matrix properties.

The primary trap tests the Adjoint-Determinant relation: students often attempt to compute the determinant of an odd-order skew-symmetric matrix, failing to recognize it is always zero, which simplifies $A \cdot \text{adj}(A)$ to the zero matrix.

The secondary trap lies in the FPC Group Division: students typically calculate $15 \times 3! = 90$ matrices, ignoring that different partitions can yield identical multisets of sums (e.g., $\{5,7,9\}$ ) and that multisets with repeated elements (e.g., $\{3,9,9\}$ ) have fewer permutations.

This dual-layer design heavily penalizes rote calculation and rewards deep conceptual clarity.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Fundamental Principle of Counting (FPC)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free