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Question #40481

Difficulty: 8/10 | 🟢 Free

The Question

Let B=(1−32+3312−3003)B = \begin{pmatrix} 1 & -\sqrt{3} & 2+\sqrt{3} \\ \sqrt{3} & 1 & 2-\sqrt{3} \\ 0 & 0 & 3 \end{pmatrix}. For each n∈Nn \in \mathbb{N}, define the matrix Mn=Bn−2nIM_n = B^n - 2^n I, where II is the 3×33 \times 3 identity matrix. Let SS be the set of all integers n∈{1,2,…,100}n \in \{1, 2, \dots, 100\} such that det⁡(adj(Mn))>0\det(\text{adj}(M_n)) > 0. Find the sum of all elements in the set SS.

Options

  • A.

    4234

  • B.

    5050

  • C.

    4134

  • D.

    4324

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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

De Moivre’s Theorem Applications|Characteristic Equation|Singular vs Non-Singular Matrices

Why This Question Is Tricky

Common Mistakes

This question tests the application of De Moivre's Theorem to find the eigenvalues of a matrix power, combined with the properties of the adjoint matrix.

The specific trap lies in the assumption that Bn−2nIB^n - 2^n I is always non-singular because BB has no eigenvalue equal to 2 (its eigenvalues are 3 and 1±i31 \pm i\sqrt{3}).

A student might falsely assume that if B−2IB - 2I is non-singular, then Bn−2nIB^n - 2^n I is also non-singular for all nn, leading them to include all integers from 1 to 100 (Option B).

However, because the complex eigenvalues of BB have a magnitude of exactly 2, their nn-th powers align with 2n2^n at specific intervals (n=6kn=6k), causing the matrix MnM_n to become singular and collapsing the determinant to zero.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterComplex Numbers and Quadratic Equations - Class 11
ConceptDe Moivre’s Theorem Applications
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree