Question:

Which of the following statements are correct?
If \[ A = \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix}, \] then

(A) \[ B^n = \begin{pmatrix} 1 & nq \\ 0 & 1 \end{pmatrix} \]

(B) \[ A^n = \begin{pmatrix} p^n & q\frac{p^n-1}{p-1} \\ 0 & 1 \end{pmatrix}, \; \text{if } p \neq 1 \]

(C) \[ AB = \begin{pmatrix} p & pq+q \\ 0 & 1 \end{pmatrix} \]

(D) \[ B^{n-1} = \begin{pmatrix} 1 & (n+1)q \\ 0 & 1 \end{pmatrix} \]

(E) \[ AB^n = \begin{pmatrix} p & (np+1)q \\ 0 & 1 \end{pmatrix} \]

Choose the correct answer from the options given below:

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When dealing with matrix powers, especially for 2x2 matrices, look for simple patterns. Proving by induction is the formal method, but for multiple-choice questions, testing for n=2 or n=3 is often sufficient to verify or disprove a given formula.
Updated On: Sep 29, 2025
  • (A), (B), (C) and (E) only
  • (B), (C) and (D) only
  • (C), (D) and (E) only
  • (A), (B), (D) and (E) only
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
This question requires verifying several statements involving matrix multiplication and exponentiation. We will evaluate each statement individually.

Step 2: Detailed Explanation:
Statement (A):
\[ B^n = \begin{pmatrix} 1 & nq \\ 0 & 1 \end{pmatrix} \] For \(n=2\): \[ B^2 = \begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2q \\ 0 & 1 \end{pmatrix}. \] This matches the formula. By induction, it holds true. Statement (A) is correct.

Statement (B):
\[ A^n = \begin{pmatrix} p^n & q\frac{p^n-1}{p-1} \\ 0 & 1 \end{pmatrix}, \; (p \neq 1) \] For \(n=2\): \[ A^2 = \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix} \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} p^2 & q(p+1) \\ 0 & 1 \end{pmatrix}. \] Formula gives: \[ q\frac{p^2-1}{p-1} = q(p+1). \] Matches perfectly. This comes from the geometric progression in the top-right entry. Statement (B) is correct.

Statement (C):
\[ AB = \begin{pmatrix} p & pq+q \\ 0 & 1 \end{pmatrix} \] \[ AB = \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} p & pq+q \\ 0 & 1 \end{pmatrix}. \] Statement (C) is correct.

Statement (D):
\[ B^{n-1} = \begin{pmatrix} 1 & (n+1)q \\ 0 & 1 \end{pmatrix} \] But from (A): \[ B^k = \begin{pmatrix} 1 & kq \\ 0 & 1 \end{pmatrix}. \] So, \[ B^{n-1} = \begin{pmatrix} 1 & (n-1)q \\ 0 & 1 \end{pmatrix}. \] The given version is wrong. Statement (D) is incorrect.

Statement (E):
\[ AB^n = \begin{pmatrix} p & (np+1)q \\ 0 & 1 \end{pmatrix} \] Using (A): \[ AB^n = \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & nq \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} p & npq+q \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} p & (np+1)q \\ 0 & 1 \end{pmatrix}. \] Statement (E) is correct.

Step 3: Final Answer:
The correct statements are: (A), (B), (C), and (E). Only (D) is incorrect.

Correct Choice: Option 1.
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