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Question #41856

Difficulty: 8/10 | 🟢 Free

The Question

Let A(x)=(exx0ex)A(x) = \begin{pmatrix} e^x & x \\ 0 & e^x \end{pmatrix} and B(x)=(e−x0xe−x)B(x) = \begin{pmatrix} e^{-x} & 0 \\ x & e^{-x} \end{pmatrix} be two matrix-valued functions. Define U(x)=A(x)B(x)U(x) = A(x)B(x) and V(x)=B(x)A(x)V(x) = B(x)A(x). If the Wronskian matrix of UU and VV is defined as W(x)=U(x)V′(x)−U′(x)V(x)W(x) = U(x)V'(x) - U'(x)V(x), find the value of det⁡(W(2))\det(W(2)).

Options

  • A.

    −16-16

  • B.

    00

  • C.

    1616

  • D.

    −112-112

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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

Algebra of Functions|Linear Independence vs Dependence|Matrix Multiplication Non-Commutativity

Why This Question Is Tricky

Common Mistakes

This question tests the non-commutativity of matrix multiplication within the context of differential operators, specifically the Wronskian.

A common trap for students is to assume that algebraic products of functions behave commutatively (AB=BAAB = BA), which would lead to U(x)=V(x)U(x) = V(x) and incorrectly evaluating the Wronskian as UU′−U′UU U' - U' U, yielding a determinant of −112-112.

An even more severe conceptual error is assuming matrices commute with their own derivatives, which would result in a Wronskian of 00.

By forcing the explicit calculation of UV′−U′VU V' - U' V, the problem rigorously evaluates the student's ability to handle order-dependent product rules in matrix calculus.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 11
ConceptAlgebra of Functions
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree