If A is a `3xx3` non-singular matrix such that `"AA"'=A'A` and `B=A^(-1)A'`, then `"BB"'` equals:

Options

A. `I`
B. `B^(-1)`
C. `(B^(-1))`'
D. `I+B`

Answer

To solve the problem, we need to find the value of \( BB' \) where \( B = A^{-1} A' \) and \( A \) is a non-singular \( 3 \times 3 \) matrix satisfying \( AA' = A'A \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that \( A \) is a \( 3 \times 3 \) non-singular matrix and that \( AA' = A'A \). This implies that \( A \) is a normal matrix.

2. **Define Matrix B**:
We have \( B = A^{-1} A' \).

3. **Calculate \( BB' \)**:
To find \( BB' \), we first need to compute \( B' \):
\[ B' = (A^{-1} A')' = (A')' (A^{-1})' = A A^{-1} = I \]
where \( I \) is the identity matrix.

4. **Multiply B and B'**:
Now, we can calculate \( BB' \):
\[ BB' = B B = (A^{-1} A')(I) = A^{-1} A' I = A^{-1} A' = B \]

5. **Final Calculation**:
Since \( B = A^{-1} A' \), we need to find \( BB' \):
\[ BB' = A^{-1} A' A^{-1} A' = A^{-1} (A' A^{-1}) A' = A^{-1} I A' = A^{-1} A' = B \]

6. **Conclusion**:
Since \( B B' = I \), we conclude that:
\[ BB' = I \]

### Final Answer:
Thus, \( BB' = I \).