If `A ` and `B` are two non-singular matrices which commute, then `(A(A+B)^(-1)B)^(-1)(AB)=`

Options

A. `A+B`
B. `A^(-1)+B^(-1)`
C. `A^(-1)+B`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \((A(A+B)^{-1}B)^{-1}(AB)\).

### Step-by-Step Solution:

1. **Identify the expression**: We start with the expression \((A(A+B)^{-1}B)^{-1}(AB)\).

2. **Use the property of inverses**: Recall that for any two matrices \(P\) and \(Q\), the inverse of their product is given by \((PQ)^{-1} = Q^{-1}P^{-1}\). Thus, we can rewrite:
\[ (A(A+B)^{-1}B)^{-1} = B^{-1}((A+B)^{-1})^{-1}A^{-1} \]
This simplifies to:
\[ B^{-1}(A+B)A^{-1} \]

3. **Substituting back into the expression**: Now substituting this back into our original expression gives:
\[ (B^{-1}(A+B)A^{-1})(AB) \]

4. **Distributing the multiplication**: We can now distribute the multiplication:
\[ B^{-1}(A+B)(A^{-1}AB) \]

5. **Simplifying \(A^{-1}AB\)**: Notice that \(A^{-1}AB = B\) (since \(A^{-1}A = I\), the identity matrix). Therefore, we can rewrite the expression as:
\[ B^{-1}(A+B)B \]

6. **Distributing \(B\)**: Now, we can distribute \(B\) inside the parentheses:
\[ B^{-1}(AB + B^2) \]

7. **Using the property of inverses**: Since \(B^{-1}B = I\), we have:
\[ B^{-1}AB + B^{-1}B^2 = B^{-1}AB + B \]

8. **Final simplification**: Thus, the expression simplifies to:
\[ B^{-1}AB + B = A + B \]

### Final Result:
Therefore, the final result is:
\[ (A(A+B)^{-1}B)^{-1}(AB) = A + B \]