Question

For a $2 \times 2$ matrix $A$, whose elements are real numbers, denote by $A^m$ the product $AA\dots A$ ($m$ times), where $m$ is a positive integer. Define $x_0 = 0$, $x_1 = 1$, $x_n = x_{n-1} + x_{n-2}$, for all $n \ge 2$ and \[ A_n = \begin{bmatrix} x_{n+1} & x_n\\ x_n & x_{n-1} \end{bmatrix}, \text{ for all } n \ge 1. \] Which of the following statements is TRUE for all $m \ge 3$?

• A. $A_1^m = A_1^{m-1} + A_1^{m-2}$
• B. $\det(A_m) = -1$
• C. $A_1^m - A_1^{m-1} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
• D. $A_m - A_{m-1} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Hint:

The matrix $A = \begin{bmatrix} 1 & 1
1 & 0 \end{bmatrix}$ is the famous Fibonacci generator matrix.
Its powers generate successive Fibonacci numbers according to $A^n = \begin{bmatrix} F_{n+1} & F_n
F_n & F_{n-1} \end{bmatrix}$.
Using the Cayley-Hamilton theorem on this matrix immediately yields the recurrence relation in matrix form.

Answer 1

The Correct Option is A

Step 1: Understanding the Question

The sequence \( x_n \) is the standard Fibonacci sequence starting with \( x_0 = 0 \) and \( x_1 = 1 \).

The matrix \( A_n \) contains terms of this sequence. We need to identify a matrix identity that holds true for all exponents \( m \ge 3 \).

Step 2: Key Formula or Approach

First, the matrix \( A_1 \) is:

\[ A_1 = \begin{bmatrix} x_2 & x_1 \\ x_1 & x_0 \end{bmatrix} \]

Since \( x_0 = 0, x_1 = 1, x_2 = 1 \), we have:

\[ A_1 = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \]

We will find the characteristic equation of \( A_1 \) using the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation.

Step 3: Detailed Explanation

Characteristic equation:

\[ \det(A_1 - \lambda I) = \det \begin{bmatrix} 1-\lambda & 1 \\ 1 & -\lambda \end{bmatrix} = 0 \]

\[ (1-\lambda)(-\lambda) - (1)(1) = 0 \implies \lambda^2 - \lambda - 1 = 0 \]

Applying Cayley-Hamilton theorem:

\( A_1^2 - A_1 - I = O \), where \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

\( \implies A_1^2 = A_1 + I \quad \text{--- (Equation 1)} \)

General power relation for \( m \ge 3 \):

Multiply both sides of Equation 1 by \( A_1^{m-2} \):

\[ A_1^2 \cdot A_1^{m-2} = (A_1 + I) \cdot A_1^{m-2} \implies A_1^m = A_1^{m-1} + A_1^{m-2} \]

This matches Option (A) directly.

Why other options are false:

Option (B) claims \( \det(A_1^m) = -1 \) for all \( m \ge 3 \). But \( \det(A_1) = (1)(0) - (1)(1) = -1 \). Then \( \det(A_1^m) = (\det A_1)^m = (-1)^m \), which is +1 for even \( m \) and -1 for odd \( m \). So it is not always -1.

Step 4: Final Answer

The true statement for all \( m \ge 3 \) is:

\( A_1^m = A_1^{m-1} + A_1^{m-2} \)