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Question #40673

Difficulty: 8/10 | 🟢 Free

The Question

Let P=(2305)P = \begin{pmatrix} 2 & 3 \\ 0 & 5 \end{pmatrix} and II be the 2×22 \times 2 identity matrix. Let tkt_k denote the element in the first row and second column of the matrix Pk−Pk−1P^k - P^{k-1} for k≥1k \ge 1, where P0=IP^0 = I. A 3×33 \times 3 matrix MM is constructed such that its elements are given by Mi,j=∑k=1i+jtkM_{i,j} = \sum_{k=1}^{i+j} t_k for 1≤i,j≤31 \le i, j \le 3. The value of det⁡(M)\det(M) is:

Options

  • A.

    0

  • B.

    30

  • C.

    117

  • D.

    210

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

The Telescopic Series Trap|Elementary Row Operation Invariance|Linear Dependency of Rows

Why This Question Is Tricky

Common Mistakes

This question tests the 'Telescopic Series Trap' combined with matrix exponentiation and linear dependency.

Students who fail to recognize the telescopic collapse of the sum will struggle to find the general term.

Even if they find Mi,j=5i+j−2i+jM_{i,j} = 5^{i+j} - 2^{i+j}, the resulting 3×33 \times 3 matrix contains massive numbers (up to 15561), making Sarrus' rule or cofactor expansion a severe time-trap.

The smart path recognizes that Mi,jM_{i,j} can be decoupled into a sum of separable products (5i5j−2i2j5^i 5^j - 2^i 2^j), meaning the matrix can be factored into a 3×23 \times 2 and a 2×32 \times 3 matrix.

This immediately implies the rank is at most 2, and the determinant is identically zero.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptThe Telescopic Series Trap
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree