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Question #41244

Difficulty: 8/10 | 🟢 Free

The Question

Let A=(034005000)A = \begin{pmatrix} 0 & 3 & 4 \\ 0 & 0 & 5 \\ 0 & 0 & 0 \end{pmatrix} and II be the 3×33 \times 3 identity matrix. Consider the infinite series of matrices S=∑k=1∞(kI−A)−1A((k+1)I−A)−1S = \sum_{k=1}^{\infty} (k I - A)^{-1} A ((k+1)I - A)^{-1} Then the sum of all the elements of the matrix SS is:

Options

  • A.

    27

  • B.

    12

  • C.

    15

  • D.

    30

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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

Nilpotent Matrix Logic|Inverse Existence Criterion|V-N Method (Partial Fractions)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize and manipulate telescopic matrix series (V-N method) alongside the properties of nilpotent matrices.

The primary trap lies in attempting to compute the inverse (kI−A)−1(kI - A)^{-1} algebraically using the adjoint method, which becomes computationally paralyzing.

Instead, students must identify the hidden difference of inverses X−1−Y−1=X−1(Y−X)Y−1X^{-1} - Y^{-1} = X^{-1}(Y-X)Y^{-1} to collapse the infinite sum.

Furthermore, they must apply the inverse existence criterion for nilpotent matrices, recognizing that since A3=0A^3 = 0, the Neumann series (I−A)−1=I+A+A2(I-A)^{-1} = I + A + A^2 terminates finitely, allowing for an exact evaluation of the boundary terms without convergence issues.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterMatrices - Class 12
ConceptNilpotent Matrix Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree