A is even ordered non singular symmetric matrix and B is even ordered non singular skew symmetric matrix such that AB = BA, then `A^3B^3(B'A)^(-1)(A^(-1)B^(-1))'` AB is equal to :

Options

A. `A^2B^2`
B. `B^2A^2`
C. `-A^2B^2`
D. `-B^2A^2`

Answer

To solve the problem step by step, we need to simplify the expression \( A^3 B^3 (B'A)^{-1} (A^{-1} B^{-1})' AB \).

### Step 1: Understand the properties of matrices
Given:
- \( A \) is a symmetric matrix: \( A' = A \)
- \( B \) is a skew-symmetric matrix: \( B' = -B \)
- \( AB = BA \)

### Step 2: Rewrite the expression
We start with the expression:
\[ A^3 B^3 (B'A)^{-1} (A^{-1} B^{-1})' AB \]
Substituting the properties of \( A \) and \( B \):
\[ = A^3 B^3 (-BA)^{-1} (A^{-1} (-B))' AB \]

### Step 3: Simplify \( (B'A)^{-1} \)
Since \( B' = -B \):
\[ (B'A)^{-1} = (-BA)^{-1} = -A^{-1} B^{-1} \]

### Step 4: Simplify \( (A^{-1} B^{-1})' \)
Using the property of transpose:
\[ (A^{-1} B^{-1})' = (B^{-1})' (A^{-1})' = (-B^{-1}) A^{-1} \]

### Step 5: Substitute back into the expression
Now substituting back:
\[ = A^3 B^3 (-A^{-1} B^{-1}) (-B^{-1} A^{-1}) AB \]
This simplifies to:
\[ = A^3 B^3 A^{-1} B^{-1} B^{-1} A^{-1} AB \]
\[ = A^3 B^3 A^{-1} B^{-2} A^{-1} AB \]

### Step 6: Simplify \( A^3 A^{-1} \)
Using the property of inverses:
\[ = A^2 B^3 B^{-2} AB \]

### Step 7: Simplify \( B^3 B^{-2} \)
This simplifies to:
\[ = A^2 B AB \]

### Step 8: Use the commutative property of \( A \) and \( B \)
Since \( AB = BA \):
\[ = A^2 B^2 A \]

### Step 9: Final simplification
Thus, we can write:
\[ = A^2 B^2 \]
or equivalently:
\[ = B^2 A^2 \]

### Final Answer
The expression simplifies to:
\[ A^2 B^2 \quad \text{or} \quad B^2 A^2 \]