1. Given information

We are given:

• $A$ is a $3 \times 3$ non-singular matrix
• $AA' = A'A$
• $B = A^{-1}A'$

Here $A'$ denotes the transpose of $A$.

We need to find $BB'$.

2. Find $B'$

Since $B = A^{-1}A',$ its transpose is $B' = (A^{-1}A')' = (A')'(A^{-1})'.$ Using $(XY)' = Y'X'$ and $(A')' = A$, we get $B' = A\,(A^{-1})'.$ Now, $(A^{-1})' = (A')^{-1},$ so $B' = A(A')^{-1}.$

3. Compute $BB'$

Now, $BB' = (A^{-1}A')\big(A(A')^{-1}\big).$ Rearrange using associativity: $BB' = A^{-1}(A'A)(A')^{-1}.$

Given that $AA' = A'A,$ we substitute $A'A = AA'$: $BB' = A^{-1}(AA')(A')^{-1}.$ Now simplify: $BB' = (A^{-1}A)A'(A')^{-1} = I \cdot I = I.$

Thus, $BB' = I.$

4. Check options

• A: $B^{-1}$ — not necessarily equal to $BB'$
• B: $(B^{-1})'$ — not equal in general
• C: $I+B$ — incorrect
• D: $I$ — correct

5. Conclusion

Therefore, $\boxed{BB' = I}$ so the correct option is D.