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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 \(A B=B A\) ].
in Mathematics by Platinum (108k points) | 570 views

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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
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:

Therefore, we have proven that \(A(A+B)^{-1} B=B(A+B)^{-1} A\).
by Platinum (108k points)

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