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 \((A(A+B)^{-1}B)^{-1}(AB)\) given that \(A\) and \(B\) are two non-singular matrices that commute.

### Step-by-Step Solution

1. **Start with the expression**:
\[ (A(A+B)^{-1}B)^{-1}(AB) \]

2. **Apply the property of the inverse**:
The inverse of a product of matrices can be expressed as the product of their inverses in reverse order. Thus, we have:
\[ (A(A+B)^{-1}B)^{-1} = B^{-1}((A+B)^{-1})^{-1}A^{-1} \]
Since \(((A+B)^{-1})^{-1} = A+B\), we can rewrite it as:
\[ = B^{-1}(A+B)A^{-1} \]

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

4. **Simplify the expression**:
We can simplify \(A^{-1}(AB)\) to \(B\) because:
\[ A^{-1}(AB) = (A^{-1}A)B = IB = B \]
Therefore, the expression becomes:
\[ B^{-1}(A+B)B \]

5. **Distribute \(B\)**:
Now we distribute \(B\) inside the parentheses:
\[ B^{-1}(AB + B^2) \]

6. **Use the property of inverses**:
Since \(B^{-1}B = I\), we can simplify further:
\[ = B^{-1}AB + B^{-1}B^2 = B^{-1}AB + B \]

7. **Since \(A\) and \(B\) commute**:
We can replace \(AB\) with \(BA\):
\[ = B^{-1}BA + B = I + B = A + B \]

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