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 $$(A+B)(A-B)=A^2-B^2$$

=> $$2A—B—(A+B)^{−1}\left(A+B\right)\left(A-B\right)$$

=> $$2A—B—\left(A-B\right)$$        [$$(A+B)(A+B)^{-1}=I$$]

=> $$2A—B—A+B$$

=> $$A$$