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Question #40328

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all $4 \times 4$ matrices with real entries. For any matrix $M \in S$ , let $M^*$ be the matrix obtained by shifting the rows of $M$ cyclically downwards by one position (i.e., the $i$ -th row of $M$ becomes the $(i+1)$ -th row of $M^*$ for $i=1,2,3$ , and the $4$ -th row of $M$ becomes the $1$ -st row of $M^*$ ). A relation $R$ is defined on $S$ such that for $A, B \in S$ , $A \ R \ B$ if and only if $\det(A + B^*) = 0$ . Which of the following statements is true regarding the relation $R$ ?

Options

A.
R \text{ is reflexive only}
B.
R \text{ is symmetric only}
C.
R \text{ is reflexive and symmetric}
D.
R \text{ is an equivalence relation}

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

Relation as Subset|Factor Theorem in Determinants|Row/Column Exchange Parity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to translate a row operation (cyclic shift) into a matrix multiplication $A^* = PA$ and apply the determinant factorization property $\det(A+PA) = \det(I+P)\det(A)$ to prove reflexivity.

The core trap lies in the symmetry check: students often incorrectly assume that structural symmetry in the definition (like $A + B^*$ ) implies relational symmetry.

They fail to realize that matrix addition and permutation multiplication do not commute ( $PB \neq BP$ ), and falsely assume $\det(A+PB) = \det(B+PA)$ by incorrectly distributing transposes or inverses, falling into the parity trap of non-commutative determinant properties.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Relation as Subset
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free