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Question #41678

Difficulty: 8/10 | 🟢 Free

The Question

Let $M = \begin{pmatrix} 5 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 5 \end{pmatrix}$ and $J = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$ . A function $f: \mathbb{R} \to \mathbb{R}$ is defined as $f(x) = \det(M + xJ)$ . Find the local maximum value of $f(x)$ .

Options

A.
44
B.
64
C.
-64
D.
10

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

Objective Function Gradient|Characteristic Equation|Row/Column Exchange Parity

Why This Question Is Tricky

Common Mistakes

This question deploys the 'Row/Column Exchange Parity' trap within an optimization context.

Students frequently attempt to simplify matrices involving an anti-diagonal variable $x$ by multiplying by the exchange matrix $J$ (or swapping rows) to form a standard characteristic polynomial structure $xI + A$ .

However, for a $3 \times 3$ matrix, $\det(J) = -1$ .

Neglecting this parity flip inverts the sign of the entire objective function $f(x)$ .

Consequently, the gradient and Hessian signs are reversed, causing the student to identify the local minimum of the true function as the local maximum of the altered function, leading directly to the trap option 64 instead of the correct maximum 44.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Objective Function Gradient
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free