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Question #41355

Difficulty: 8/10 | 🟢 Free

The Question

Let AA be a 3×33 \times 3 matrix with real entries such that all its eigenvalues are real. Let S={x∈R:tr(adj(xI−A))=0}S = \{x \in \mathbb{R} : \text{tr}(\text{adj}(xI - A)) = 0\}. Which of the following statements is \textbf{INCORRECT}?

Options

  • A.

    If AA has three distinct eigenvalues, then SS has exactly two elements, and for all x∈Sx \in S, xI−AxI - A is non-singular.

  • B.

    If SS contains exactly one element x0x_0, then x0I−Ax_0I - A is a singular matrix.

  • C.

    If AA has exactly two distinct eigenvalues, then SS has exactly two elements, and for exactly one x∈Sx \in S, xI−AxI - A is singular.

  • D.

    If SS contains exactly two elements, then AA must have three distinct eigenvalues.

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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

Rolle's Theorem Anchors|Characteristic Equation|Singular vs Non-Singular Matrices

Why This Question Is Tricky

Common Mistakes

The question tests the application of Rolle's Theorem to the characteristic polynomial P(x)=det⁡(xI−A)P(x) = \det(xI - A).

The primary trap is failing to recognize that tr(adj(xI−A))\text{tr}(\text{adj}(xI - A)) is exactly the derivative P′(x)P'(x) via Jacobi's formula.

Students often incorrectly assume the converse of Rolle's theorem: if the derivative of a cubic has two distinct roots, the cubic itself must have three distinct roots.

This ignores the case of algebraic multiplicity 2 where one of the local extrema touches the x-axis, making the matrix singular at that specific point while still yielding two distinct roots for the derivative.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Derivatives - Class 12
ConceptRolle's Theorem Anchors
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree