Let A and B two non singular matrices of same order such that `(AB)^(k)=B^(k)A^(k)` for consecutive positive integral values of k, then `AB^(2)A^(-1)` is equal to

Options

A. `A^(2)`
B. B
C. A
D. `B^(2)`

Answer

To solve the problem, we need to find the value of \( AB^2A^{-1} \) given that \( (AB)^k = B^k A^k \) for consecutive positive integral values of \( k \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that \( (AB)^k = B^k A^k \) for consecutive positive integers \( k \). Let's first analyze this condition for \( k = 1 \):
\[ (AB)^1 = B^1 A^1 \implies AB = BA \]
This shows that matrices \( A \) and \( B \) commute.

2. **Using the Commutative Property**:
Since \( A \) and \( B \) commute, we can rearrange the products of \( A \) and \( B \) freely. Therefore, we can express \( AB^2A^{-1} \) as follows:
\[ AB^2A^{-1} = A(BB)A^{-1} = (AB)BA^{-1} \]
Since \( AB = BA \), we can replace \( AB \) with \( BA \):
\[ (AB)BA^{-1} = B(AB)A^{-1} = B(BA)A^{-1} \]

3. **Simplifying Further**:
Continuing with the simplification:
\[ B(BA)A^{-1} = B(B(A A^{-1})) = B(BI) = B^2 \]

4. **Conclusion**:
Therefore, we find that:
\[ AB^2A^{-1} = B^2 \]

### Final Answer:
\[ AB^2A^{-1} = B^2 \]