`A` and `B` are two square matrices such that `A^(2)B=BA` and if `(AB)^(10)=A^(k)B^(10)`, then `k` is

Options

A. `1001`
B. `1023`
C. `1042`
D. none of these

Explanation

To solve the problem, we need to find the value of \( k \) given the conditions about the matrices \( A \) and \( B \).

### Step-by-Step Solution:

1. **Given Condition**: We have the equation \( A^2 B = B A \). This indicates that \( A^2 \) and \( B \) commute.

2. **Expression for \( (AB)^{10} \)**: We start by expanding \( (AB)^{10} \):
\[ (AB)^{10} = (AB)(AB)(AB)(AB)(AB)(AB)(AB)(AB)(AB)(AB) \]
We can group the terms:
\[ (AB)^{10} = A^1B^1 A^1B^1 A^1B^1 A^1B^1 A^1B^1 A^1B^1 A^1B^1 A^1B^1 A^1B^1 A^1B^1 \]

3. **Using Commutativity**: Since \( A^2 B = B A \), we can rearrange the terms:
\[ (AB)^{10} = A^2B A^2B A^2B A^2B A^2B A^2B A^2B A^2B A^2B A^2B \]
This can be simplified using the property of commutativity:
\[ = A^2 A^2 A^2 A^2 A^2 A^2 A^2 A^2 A^2 A^2 B^{10} = A^{10} B^{10} \]

4. **Relating to the Given Condition**: We know from the problem statement that:
\[ (AB)^{10} = A^k B^{10} \]
Thus, we can equate:
\[ A^{10} B^{10} = A^k B^{10} \]

5. **Comparing Powers of \( A \)**: Since \( B^{10} \) is common on both sides, we can compare the powers of \( A \):
\[ A^{10} = A^k \]
This implies:
\[ k = 10 \]

6. **Final Value of \( k \)**: Therefore, the value of \( k \) is:
\[ k = 10 \]