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Question #67894

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $3 \times 3$ matrix with complex entries such that $\text{tr}(A) = 0$ , $\text{tr}(A^2) = -6$ , and $\det(A) = 2i$ . It is given that the matrix $B = A^2 + \alpha A + \beta I$ (where $\alpha, \beta \in \mathbb{C}$ ) is nilpotent. Find the value of $|\det(A + \alpha I) + \beta^2|^2$ .

Options

A.
16
B.
32
C.
20
D.
64

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

Nilpotent Matrix Logic|Matrix Trace Properties|Vieta’s Formulas in Complex Domain

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the intersection of matrix invariants and nilpotent operators.

The primary trap lies in the complex domain: students often assume that real trace values imply real coefficients, but here the determinant is imaginary, forcing the use of Vieta's formulas (Newton's sums) to construct a complex characteristic polynomial $P(\lambda) = \lambda^3 + 3\lambda - 2i$ .

A common fatal mistake is attempting to compute the eigenvalues of $B$ via brute-force matrix multiplication or assuming $A$ must be diagonalizable, leading to algebraic exhaustion.

The rigorous path requires recognizing that nilpotency mandates all eigenvalues of $B$ to be strictly zero, which translates to a simple linear system in $\alpha$ and $\beta$ when evaluated at the roots of $P(\lambda)$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Nilpotent Matrix Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free