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Question #40501

Difficulty: 8/10 | 🟢 Free

The Question

Let AA be a 3×33 \times 3 matrix with positive integer eigenvalues. Let M1=(tr(A)−1∣A∣0)M_1 = \begin{pmatrix} \text{tr}(A) & -1 \\ |A| & 0 \end{pmatrix} and M2=(tr(adj(A))−1∣adj(A)∣0)M_2 = \begin{pmatrix} \text{tr}(\text{adj}(A)) & -1 \\ |\text{adj}(A)| & 0 \end{pmatrix}. If M1M_1 and M2M_2 share a common eigenvalue and ∣A∣≠1|A| \neq 1, find the value of tr(A)+∣A∣+tr(adj(A))\text{tr}(A) + |A| + \text{tr}(\text{adj}(A)).

Options

  • A.

    15

  • B.

    17

  • C.

    19

  • D.

    21

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Condition for Common Roots|Characteristic Equation|Adjoint-Determinant Relationship

Why This Question Is Tricky

Common Mistakes

The problem tests the synthesis of matrix properties and the theory of equations.

The trap lies in the Adjoint-Determinant relationship: students often correctly identify ∣adj(A)∣=∣A∣2|\text{adj}(A)| = |A|^2 but fail to relate tr(adj(A))\text{tr}(\text{adj}(A)) to the elementary symmetric polynomials of the eigenvalues.

A common mistake is assuming the matrices M1M_1 and M2M_2 must be identical for them to share an eigenvalue, leading to e1=e2e_1 = e_2 and e3=e32e_3 = e_3^2, which incorrectly forces ∣A∣=1|A|=1 (violating the given condition).

The correct path requires deriving the resultant condition for the two quadratics and solving the resulting Diophantine equation for the eigenvalues.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterComplex Numbers and Quadratic Equations - Class 11
ConceptCondition for Common Roots
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree