If A is an `3xx3` non-singular matrix such that `A A^'=A^' A` and `B""=""A^(-1)A^'` , then BB' equals (1) `I""+""B` (2) `I` (3) `B^(-1)` (4) `(B^(-1))^'`

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Define \( B \)**:
\[ B = A^{-1} A' \]

2. **Calculate \( B' \)**:
The transpose of \( B \) can be calculated using the property of transposes:
\[ B' = (A^{-1} A')' = (A')' (A^{-1})' = A (A^{-1})' \]
Since \( (A^{-1})' = (A')^{-1} \), we have:
\[ B' = A (A')^{-1} \]

3. **Calculate \( BB' \)**:
Now we compute \( BB' \):
\[ BB' = (A^{-1} A')(A (A')^{-1}) \]
Using the associative property of matrix multiplication:
\[ BB' = A^{-1} (A' A) (A')^{-1} \]

4. **Use the property \( AA' = A'A \)**:
Since \( AA' = A'A \), we can replace \( A' A \) with \( AA' \):
\[ BB' = A^{-1} (AA') (A')^{-1} \]

5. **Simplify \( BB' \)**:
Since \( A^{-1} A = I \), we can simplify:
\[ BB' = I (A') (A')^{-1} = I \]

Thus, we conclude that:
\[ BB' = I \]

### Final Answer:
The value of \( BB' \) is \( I \), which corresponds to option (2).