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Question #41256

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = \begin{pmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & 1 \end{pmatrix}$ and $B$ be a $3 \times 3$ real matrix such that $\det(B) = 2$ . If $x \in \mathbb{R}$ satisfies the equation $\det(A^T B A) = \det(\det(A) B^{-1})$ , then the sum of all distinct real values of $x$ is:

Options

A.
3
B.
2
C.
1
D.
4

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

Properties of Transpose Determinant|Determinant Product Rule|Singular vs. Non-Singular Threshold

Why This Question Is Tricky

Common Mistakes

This question tests the 'Singular Threshold' trap within determinant equations.

Students frequently divide both sides of the intermediate equation $2\det(A)^2 = \frac{1}{2}\det(A)^3$ by $\det(A)^2$ , implicitly assuming $A$ is non-singular ( $\det(A) \neq 0$ ).

This algebraic cancellation causes them to lose the valid solution $x=1$ , leading to an incorrect sum of 2.

Furthermore, it rigorously tests the scalar multiplication property of determinants $\det(kM) = k^n \det(M)$ , where failing to raise the scalar $\det(A)$ to the power of 3 (the matrix dimension) will completely derail the equation.

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