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Question #41047

Difficulty: 8/10 | 🟢 Free

The Question

Let P=[p1p2p3]P = \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix} be a probability vector such that pi≥0p_i \ge 0 for i=1,2,3i=1,2,3 and ∑i=13pi=1\sum_{i=1}^3 p_i = 1. Consider the skew-symmetric matrix A=[02−3−2043−40]A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 4 \\ 3 & -4 & 0 \end{bmatrix} and the column vector B=[−12−1]B = \begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}. If PP satisfies the system of linear equations AP=BAP = B, then the value of 81(p12+p22+p32)81(p_1^2 + p_2^2 + p_3^2) is:

Options

  • A.

    35

  • B.

    45

  • C.

    25

  • D.

    55

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

Axiomatic Probability Constraints|Skew-Symmetric Determinant|System of Linear Equations (Cramer/Inverse)

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of linear algebra and axiomatic probability.

The trap lies in the system of equations AP=BAP = B.

Because AA is an odd-order skew-symmetric matrix, its determinant is strictly zero, making the system either inconsistent or dependent (having infinitely many solutions).

Students who blindly attempt to use Cramer's rule or matrix inversion will fail.

The correct path requires recognizing the linear dependence, expressing the variables parametrically (finding the Null Space translation), and utilizing the fundamental probability constraint ∑pi=1\sum p_i = 1 as the final independent equation to lock in the unique valid distribution.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 11
ConceptAxiomatic Probability Constraints
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree