Q37 / 56 · IAT (IISER) 2026 · Mathematics
For a $2 \times 2$ matrix $A$, whose elements are real numbers, denote by $A^m$ the product $\underbrace{AA \cdots A}_{m \text{ times}}$, where $m$ is a positive integer. Define $x_0 = 0$, $x_1 = 1$, $x_n = x_{n-1} + x_{n-2}$, for all $n \geq 2$ and
$A_n = \begin{bmatrix} x_{n+1} & x_n \\ x_n & x_{n-1} \end{bmatrix}$, for all $n \geq 1$.
Which of the following statements is TRUE for all $m \geq 3$?
$A_n = \begin{bmatrix} x_{n+1} & x_n \\ x_n & x_{n-1} \end{bmatrix}$, for all $n \geq 1$.
Which of the following statements is TRUE for all $m \geq 3$?
- A$A_m - A_{m-1} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
- Correct answerB$A_1^m = A_1^{m-1} + A_1^{m-2}$
- C$\det(A_m) = -1$
- D$A_1^m - A_1^{m-1} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Solution
Given $x_0 = 0$, $x_1 = 1$, and $x_n = x_{n-1} + x_{n-2}$ for $n \geq 2$.
For $n = 1$, the matrix $A_1$ is:
$A_1 = \begin{bmatrix} x_2 & x_1 \\ x_1 & x_0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$
The characteristic equation of $A_1$ is given by $|A_1 - \lambda I| = 0$:
$\begin{vmatrix} 1-\lambda & 1 \\ 1 & -\lambda \end{vmatrix} = 0$
$\Rightarrow -\lambda(1-\lambda) - 1 = 0$
$\Rightarrow \lambda^2 - \lambda - 1 = 0$
By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. Therefore:
$A_1^2 - A_1 - I = 0$
$\Rightarrow A_1^2 = A_1 + I$
Multiplying both sides by $A_1^{m-2}$ (for $m \geq 3$), we get:
$A_1^m = A_1^{m-1} + A_1^{m-2}$
Checking the other options:
For (b), $\det(A_m) = x_{m+1}x_{m-1} - x_m^2 = (-1)^m$ (Cassini's identity), which is not always $-1$.
For (c), $A_1^m - A_1^{m-1} + I = A_1^{m-2} + I \neq 0$.
For (d), $A_m - A_{m-1} = \begin{bmatrix} x_{m+1}-x_m & x_m-x_{m-1} \\ x_m-x_{m-1} & x_{m-1}-x_{m-2} \end{bmatrix} = \begin{bmatrix} x_{m-1} & x_{m-2} \\ x_{m-2} & x_{m-3} \end{bmatrix} = A_{m-2} \neq I$.
Thus, only the first statement is true.
Answer: $A_1^m = A_1^{m-1} + A_1^{m-2}$
For $n = 1$, the matrix $A_1$ is:
$A_1 = \begin{bmatrix} x_2 & x_1 \\ x_1 & x_0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$
The characteristic equation of $A_1$ is given by $|A_1 - \lambda I| = 0$:
$\begin{vmatrix} 1-\lambda & 1 \\ 1 & -\lambda \end{vmatrix} = 0$
$\Rightarrow -\lambda(1-\lambda) - 1 = 0$
$\Rightarrow \lambda^2 - \lambda - 1 = 0$
By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. Therefore:
$A_1^2 - A_1 - I = 0$
$\Rightarrow A_1^2 = A_1 + I$
Multiplying both sides by $A_1^{m-2}$ (for $m \geq 3$), we get:
$A_1^m = A_1^{m-1} + A_1^{m-2}$
Checking the other options:
For (b), $\det(A_m) = x_{m+1}x_{m-1} - x_m^2 = (-1)^m$ (Cassini's identity), which is not always $-1$.
For (c), $A_1^m - A_1^{m-1} + I = A_1^{m-2} + I \neq 0$.
For (d), $A_m - A_{m-1} = \begin{bmatrix} x_{m+1}-x_m & x_m-x_{m-1} \\ x_m-x_{m-1} & x_{m-1}-x_{m-2} \end{bmatrix} = \begin{bmatrix} x_{m-1} & x_{m-2} \\ x_{m-2} & x_{m-3} \end{bmatrix} = A_{m-2} \neq I$.
Thus, only the first statement is true.
Answer: $A_1^m = A_1^{m-1} + A_1^{m-2}$
Hence, the correct answer is option: 'B'.