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If A, B and A + B are non-singular matrices and AB = BA, then $$2A — B — A(A + B)^{−1}A + B(A + B)^{−1}B$$ equals
We are give $$BA=AB$$. We will post-multiply this by $$A^{-1}$$
=> $$BAA^{-1}=ABA^{-1}$$
=> $$B=ABA^{-1}$$ [$$AA^{-1}=I$$]
Now, we will pre-mutliply the equation with $$A^{-1}$$
=> $$A^{-1}B=A^{-1}ABA^{-1}$$
=> $$A^{-1}B=BA^{-1}$$
Adding $$A^{-1}A$$ both sides -
=> $$A^{-1}B+A^{-1}A=BA^{-1}+AA^{-1}$$
=> $$A^{-1}\left(B+A\right)=\left(B+A\right)A^{-1}$$
Taking the inverse on both sides [$$(AB)^{-1}=B^{-1}A^{-1}$$]
=> $$\left(A+B\right)^{-1}A=A\left(A+B\right)^{-1}\rightarrow1$$
Similarly, we will get => $$\left(A+B\right)^{-1}B=B\left(A+B\right)^{-1}\rightarrow2$$
Now, we need to find the value of $$2A — B — A(A + B)^{−1}A + B(A + B)^{−1}B$$
Substituting the value of $$\left(A+B\right)^{-1}A=A\left(A+B\right)^{-1}$$ from eq. 1, and the value of $$\left(A+B\right)^{-1}B=B\left(A+B\right)^{-1}$$ from eq. 2 -
=>$$2A—B—AA(A+B)^{−1}+BB(A+B)^{−1}$$
=> $$2A—B—A^2(A+B)^{−1}+B^2(A+B)^{−1}$$
=> $$2A—B—(A+B)^{−1}\left(A^2-B^2\right)$$
Now, $$(A+B)(A-B)=A^2+AB-BA-B^2$$, and since $$AB=BA$$, thus $$(A+B)(A-B)=A^2-B^2$$
=> $$2A—B—(A+B)^{−1}\left(A+B\right)\left(A-B\right)$$
=> $$2A—B—\left(A-B\right)$$ [$$(A+B)(A+B)^{-1}=I$$]
=> $$2A—B—A+B$$
=> $$A$$
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