To show that \(A(A+B)^{-1} B=B(A+B)^{-1} A\), we will use the following property of the inverse of a matrix: Given an invertible matrix \(X\), if there exists a matrix \(Y\) such that \(X Y=\) \(I\), then \(Y=X^{-1}\).

Let's denote the matrix \((A+B)^{-1}\) as \(C\). Then, we can rewrite the expression we need to prove as: \(A C B=B C A\).

Now let's compute the product of the matrices \(A C B\) and \(B C A\) :

\((A C B)(B C A)=A(C B)(C A)=A((C B) C) A\) (associativity of matrix multiplication)

Since matrix multiplication is associative, we can rewrite the expression inside the parentheses as:

\[

A((C B) C) A=A(C(B C)) A

\]

Now, notice that we can use the property of the inverse we mentioned earlier:

Since \(A+B=A C+B C\) and \(C\) is the inverse of \(A+B\), we have:

\[

(A+B) C=(A C+B C) C=I

\]

Thus,

\[

B C=(A+B) C-A C=I-A C

\]

Now, we can substitute this back into our expression:

\[

A(C(B C)) A=A(C(I-A C)) A

\]

Using the distributive property of matrix multiplication, we have:

\[

A(C I-C A C) A=A(C-C A C) A

\]

Since \(C I=C\), we get:

\[

A(C-C A C) A=A(C-C A C) A

\]

This shows that:

\[

A C B=B C A

\]

Therefore, we have proven that \(A(A+B)^{-1} B=B(A+B)^{-1} A\).