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Question #68018

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $3 \times 3$ matrix with real entries such that its distinct eigenvalues are the roots of the equation $x^2 - 3x + 2 = 0$ . It is given that $\det(A A^T) = 4$ . If $\alpha = \text{Tr}(A)$ , find the value of $\det(A^T A - \alpha A^T)$ .

Options

A.
-36
B.
-8
C.
36
D.
-72

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

Properties of Transpose Determinant|Matrix Trace Properties|Even Multiplicity Roots

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 understanding of algebraic multiplicity and matrix invariants.

The primary trap lies in the calculation of the trace $\alpha$ .

Students often mistakenly assume that the trace is the sum of the distinct eigenvalues (calculating $\alpha = 1 + 2 = 3$ ), ignoring the fact that a $3 \times 3$ matrix must have 3 eigenvalues counting multiplicities.

This leads them to evaluate $\det(A - 3I)$ instead of $\det(A - 4I)$ , yielding the trap answer $-8$ .

Furthermore, the use of $\det(A A^T) = 4$ forces the student to correctly apply transpose determinant properties to deduce the exact algebraic multiplicities, nullifying any shortcuts that ignore the non-diagonalizable potential of the matrix.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Properties of Transpose Determinant
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free