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Question #41781

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $3 \times 3$ matrix with non-negative real entries such that its symmetric part is $S = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{pmatrix}$ . If $A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 7 \\ 6 \\ 8 \end{pmatrix}$ , then the maximum possible value of $\text{Tr}(A^T A)$ is:

Options

A.
105
B.
129
C.
143
D.
73

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

Non-Negativity Restriction|Symmetric and Skew-Symmetric Decomposition|Matrix Equality

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate matrix decomposition with linear programming constraints (non-negativity).

The core trap lies in the boundary check: students often find the parametric solution for the skew-symmetric components ( $y=1-x, z=x-1$ ) and attempt to maximize the quadratic objective function without applying the non-negativity restriction $a_{ij} \ge 0$ to ALL elements.

If they only check the constraints for elements involving $y$ or $z$ , they obtain wider, incorrect intervals ( $[-2, 4]$ or $[-3, 5]$ ), leading to the trap answers 129 and 143.

The true feasible region is strictly bounded by the intersection of all constraints, forcing $x \in [-2, 2]$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Non-Negativity Restriction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free