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Question #41255

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $3 \times 3$ matrix with complex entries such that $A A^\dagger = I$ and $\det(A) = i$ , where $A^\dagger$ is the conjugate transpose of $A$ . Let $B = \begin{pmatrix} 1 & 1+i & x \\ 1-i & -1 & i \\ x & -i & 1 \end{pmatrix}$ where $x$ is a real number. If $\det(A B A^T) = -3$ , then the product of all possible real values of $x$ is:

Options

A.
-7
B.
-1
C.
-3
D.
7

Check Your Answer

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

Properties of Transpose Determinant|Orthogonal Matrix Conditions|Algebraic Manipulation of i

Why This Question Is Tricky

Common Mistakes

This question tests the properties of determinants involving complex matrices and algebraic discipline.

A common trap is conflating the transpose $A^T$ with the conjugate transpose $A^\dagger$ .

Students often incorrectly assume $\det(A^T) = \overline{\det(A)} = -i$ , leading to $\det(A B A^T) = \det(B) = -3$ , which yields a product of $-1$ .

Another trap is miscalculating $i^2 = 1$ during the determinant expansion of $B$ , which leads to the incorrect quadratic $x^2 + 2x - 3 = 0$ and a product of $-3$ .

The condition $A A^\dagger = I$ acts as a distractor to trigger the unitary matrix conflation.

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	160 sec
Access	Free