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Question #41786

Difficulty: 8/10 | 🟢 Free

The Question

Let AA and BB be 3×33 \times 3 matrices with real entries. Define the matrix C=AB−BA+5I3C = AB - BA + 5I_3, where I3I_3 is the 3×33 \times 3 identity matrix. Let the matrix CC be given by C=(α21−1β430γ)C = \begin{pmatrix} \alpha & 2 & 1 \\ -1 & \beta & 4 \\ 3 & 0 & \gamma \end{pmatrix} Consider the following system of linear equations in variables x,y,zx, y, z: x+2y+3z=αx + 2y + 3z = \alpha 2x+3y+5z=β2x + 3y + 5z = \beta 7x+10y+17z=γ7x + 10y + 17z = \gamma If this system of equations is consistent, what is the value of β\beta?

Options

  • A.

    2

  • B.

    3

  • C.

    4

  • D.

    5

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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

Inconsistency Detection|Matrix Trace Properties|Matrix Multiplication Non-Commutativity

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of matrix trace properties and the conditions for consistency in a system of linear equations.

A common trap is for students to incorrectly assume that matrix multiplication is commutative (AB=BAAB = BA), which would lead them to conclude that C=5I3C = 5I_3 and thus β=5\beta = 5.

However, this assumption makes the linear system inconsistent, contradicting the problem statement.

The correct approach requires recognizing that the trace of the commutator [A,B][A, B] is always zero, yielding α+β+γ=15\alpha + \beta + \gamma = 15.

By identifying the linear dependence in the system's coefficients (4L2−L1=L34L_2 - L_1 = L_3), students can derive the consistency constraint α+γ=4β\alpha + \gamma = 4\beta, which elegantly resolves to the correct value of β\beta without needing to know the individual matrices.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLinear Programming - Class 12
ConceptInconsistency Detection
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree