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Question #41113

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all 3×33 \times 3 matrices with real entries. A relation RR on SS is defined by ARB  ⟺  det⁡(A)=det⁡(B)A R B \iff \det(A) = \det(B). Let [M][M] denote the equivalence class of a matrix M∈SM \in S. Consider the matrix P=(123246369)P = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix}. If a matrix A∈[P]A \in [P] satisfies the equation A⋅adj(A)+3A=A2+kIA \cdot \text{adj}(A) + 3A = A^2 + k I for some real number kk, which of the following statements is ALWAYS true?

Options

  • A.

    k=0k = 0 and the maximum possible trace of AA is 66

  • B.

    k=0k = 0 and the maximum possible trace of AA is 99

  • C.

    k=27k = 27 and the maximum possible trace of AA is 99

  • D.

    k=−27k = -27 and the maximum possible trace of AA is 66

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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

Equivalence Class Partitioning|Inversion and Determinant Product|Inverse Existence Criterion

Why This Question Is Tricky

Common Mistakes

This question tests the conceptual boundary of equivalence classes and matrix properties.

The trap lies in the 'Inverse Existence Criterion'.

Students often ignore the equivalence class constraint (A∈[P]  ⟹  det⁡(A)=0A \in [P] \implies \det(A) = 0) and assume AA can be any matrix satisfying the equation.

If they assume AA is invertible (e.g., A=3IA = 3I), they will incorrectly calculate k=27k = 27 and a maximum trace of 99 (Option C).

Even if they correctly find k=0k=0, forgetting that the singular nature of AA prevents the eigenvalues from being {3,3,3}\{3,3,3\} leads to the trap of maximum trace 99 (Option B).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 12
ConceptEquivalence Class Partitioning
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree