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 \( AB^2 A^{-1} \) 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 have the equation \( (AB)^k = B^k A^k \). Let's first analyze this for \( k = 1 \):
\[ (AB)^1 = B^1 A^1 \implies AB = BA \]
This shows that matrices \( A \) and \( B \) commute.

2. **Finding \( AB^2 A^{-1} \)**:
We want to evaluate \( AB^2 A^{-1} \). We can rewrite this expression as:
\[ AB^2 A^{-1} = A(BB)A^{-1} \]

3. **Using the Commutative Property**:
Since \( A \) and \( B \) commute, we can replace \( AB \) with \( BA \):
\[ AB^2 A^{-1} = A(BB)A^{-1} = (AB)B A^{-1} \]
Now, we can substitute \( AB \) with \( BA \):
\[ (AB)B A^{-1} = (BA)B A^{-1} \]

4. **Continuing with the Commutative Property**:
Again, we can use the commutative property:
\[ (BA)B A^{-1} = B(AB)A^{-1} \]
Now substituting \( AB \) with \( BA \) again:
\[ B(AB)A^{-1} = B(BA)A^{-1} \]

5. **Simplifying Further**:
Now, we can simplify \( (BA)A^{-1} \):
\[ B(BA)A^{-1} = B(B)I = B^2 \]
where \( I \) is the identity matrix, since \( AA^{-1} = I \).

6. **Final Result**:
Therefore, we conclude that:
\[ AB^2 A^{-1} = B^2 \]

### Conclusion:
The final answer is \( B^2 \).