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Let $$A$$ and $$B$$ be $$n \times n$$ matrices. If $$A+B$$ is invertible, show that $$A(A+B)^{-1} B=$$ $$B(A+B)^{-1} A$$. [Don't assume that $$A B=B A$$ ].
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A(A+B)-1B = (A+B - B)(A+B)-1B = (A+B)(A+B)-1B - B(A+B)-1B = B - B(A+B)-1B ;

B(A+B)-1A = B(A+B)-1(A+B - B) = B(A+B)-1(A+B) - B(A+B)-1B = B - B(A+B)-1B.

by (104 points)
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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 $$A(A+B)^{-1} B=B(A+B)^{-1} A$$.
by Platinum (108k points)

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