Let A and B be two matrices different from identify matrix such that AB=BA and `A^(n)-B^(n)` is invertible for some positive integer 'n'. If `A^(n)-B^(n)=A^(n+1)-B^(n+1)=A^(n+2)-B^(n+2)`, then

Options

A. I-A is non singular
B. I-B is non singular
C. I-A is singular
D. I-B is singular

Explanation

To solve the problem, we need to analyze the conditions given for the matrices \( A \) and \( B \). Let's break down the solution step by step.

### Step 1: Understand the Given Conditions
We are given that:
1. \( AB = BA \) (A and B commute)
2. \( A^n - B^n \) is invertible for some positive integer \( n \).
3. \( A^n - B^n = A^{n+1} - B^{n+1} = A^{n+2} - B^{n+2} \).

### Step 2: Set Up the Equality
From the equality \( A^n - B^n = A^{n+1} - B^{n+1} \), we can rewrite it as:
\[ A^n - B^n = A^{n+1} - B^{n+1} \implies A^n - B^n = A \cdot A^n - B \cdot B^n \]
This implies:
\[ A^n - B^n = A \cdot A^n - B \cdot B^n \implies A^n - B^n = A^{n+1} - B^{n+1} \]

### Step 3: Rearranging the Terms
Rearranging gives us:
\[ A^n - B^n - A^{n+1} + B^{n+1} = 0 \]
This can be factored as:
\[ A^n - B^n - A \cdot A^n + B \cdot B^n = 0 \]
Factoring out \( A^n \) and \( B^n \):
\[ (A^n - B^n)(I - A) + (B^n)(I - B) = 0 \]

### Step 4: Analyze the Invertibility
Since \( A^n - B^n \) is invertible, the only way for the above equation to hold is if:
\[ I - A = 0 \quad \text{and} \quad I - B = 0 \]
This implies:
\[ A = I \quad \text{and} \quad B = I \]
However, we know that \( A \) and \( B \) are different from the identity matrix, which leads us to a contradiction.

### Step 5: Conclusion
Since both \( A \) and \( B \) cannot be the identity matrix, the only conclusion we can draw is that the matrices \( A \) and \( B \) must be such that:
\[ A^n - B^n = 0 \]
This means that \( A \) and \( B \) must be equal, which contradicts the initial condition that they are different from the identity matrix.

Thus, we conclude that the only possibility is that both matrices must be singular, leading to:
\[ \text{det}(A) = 1 \quad \text{and} \quad \text{det}(B) = 1 \]

### Final Answer
Therefore, the answer is that both matrices \( A \) and \( B \) must be singular.