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Question #41140

Difficulty: 8/10 | 🟢 Free

The Question

Let M=(010001000)M = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} and S={aM+bM2∣a,b∈R}S = \{ aM + bM^2 \mid a, b \in \mathbb{R} \}. A binary operation ∗* on SS is defined by A∗B=A+B+cABA * B = A + B + c AB, where cc is a non-zero real constant. For any A∈SA \in S, let A∘n=A∗A∗⋯∗AA^{\circ n} = A * A * \dots * A (nn times). If X∈SX \in S is the inverse of M∘5M^{\circ 5} under the operation ∗*, then XX is equal to:

Options

  • A.

    −5M+15cM2-5M + 15c M^2

  • B.

    −5M+10cM2-5M + 10c M^2

  • C.

    −5M−15cM2-5M - 15c M^2

  • D.

    5M−10cM25M - 10c M^2

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

Binary Operation Closure|Inverse Existence Criterion|Nilpotent Matrix Logic

Why This Question Is Tricky

Common Mistakes

This question tests the group structure of singular matrices under a custom binary operation.

The primary trap is a 'Logical Deadlock': students typically associate the existence of an inverse with non-zero determinants.

Here, every matrix in SS is nilpotent (and thus singular with det=0), yet the set forms a closed group under ∗*.

To solve it efficiently, students must avoid brute-force algebra (which leads to massive calculation errors for n=5n=5) and recognize the isomorphism A↦I+cAA \mapsto I+cA, transforming the operation into standard multiplication.

They must then apply the binomial expansion for negative indices, using the nilpotent property M3=OM^3=O to naturally terminate the infinite series.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 12
ConceptBinary Operation Closure
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree