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Question #40584

Difficulty: 8/10 | 🟢 Free

The Question

Let $n \ge 4$ be an integer. Let $A = [a_{ij}]$ be an $n \times n$ matrix where $a_{ij}$ is the number of derangements of the set $\{1, 2, \dots, n\}$ that map $i$ to $j$ . Let $D_n$ be the total number of derangements of $n$ objects. If $I$ is the $n \times n$ identity matrix, which of the following represents the value of $\det(A) - \det(A - D_n I)$ ?

Options

A.
$D_n \left( \frac{-D_n}{n-1} \right)^{n-1}$
B.
$D_n \left( \frac{D_n}{n-1} \right)^{n-1}$
C.
$D_n \left( \frac{-D_n}{n} \right)^{n-1}$
D.
$0$

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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

Derangements (Inclusion-Exclusion)|Triangular/Diagonal Matrix Shortcut|Singular vs Non-Singular Matrices

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of combinatorics and linear algebra.

The primary trap is failing to recognize that the row sums of the derangement matrix are constant ( $D_n$ ), which immediately makes $D_n$ an eigenvalue and $\det(A - D_n I) = 0$ .

Students who miss this will attempt a brute-force determinant expansion of $A - D_n I$ , which is computationally prohibitive.

Furthermore, calculating $\det(A)$ requires recognizing $A$ as a scaled translation of the all-ones matrix $J$ , utilizing the eigenvalues of $J$ as a shortcut rather than performing complex row operations.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Derangements (Inclusion-Exclusion)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free