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:

• A. (A), (B), (C) and (E) only
• B. (B), (C) and (D) only
• C. (C), (D) and (E) only
• D. (A), (B), (D) and (E) only

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question requires verifying several statements involving matrix multiplication and exponentiation. Each statement will be evaluated 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):
\[A^n = \begin{pmatrix} p^n & q\frac{p^n-1}{p-1} \\ 0 & 1 \end{pmatrix}, \; (p eq 1)\] 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 evaluation: \[q\frac{p^2-1}{p-1} = q(p+1).\] This matches. The top-right entry corresponds to a geometric progression. Statement (B) is correct.

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

Statement (D):
\[B^{n-1} = \begin{pmatrix} 1 & (n+1)q \\ 0 & 1 \end{pmatrix}\] From statement (A), we have: \[B^k = \begin{pmatrix} 1 & kq \\ 0 & 1 \end{pmatrix}.\] Therefore, \[B^{n-1} = \begin{pmatrix} 1 & (n-1)q \\ 0 & 1 \end{pmatrix}.\] The provided statement is incorrect. Statement (D) is incorrect.

Statement (E):
\[AB^n = \begin{pmatrix} p & (np+1)q \\ 0 & 1 \end{pmatrix}\] Using the result from (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:
Statements (A), (B), (C), and (E) are correct. Statement (D) is incorrect.

Correct Choice: Option 1.