# (7) If A and B are a square matrix of the same order. Prove that (ABA^−1)^3=AB^3A^−1 (8) If A, B and A+B are aach norsingular. Prove that A(A+B)^-1B=B(A+B)^-1A=(A^-1+B^-1)^-1

Question
Matrix transformations
(7) If A and B are a square matrix of the same order. Prove that $$\displaystyle{\left({A}{B}{A}^{ ## Answers (1) 2021-02-22 Let A and B are n*n matrix. Then we have PSK(ABA^-1)^3 = (ABA^-1) (ABA^-1) (AB^-1) = (AB) (A^-1A) B(A^-1A) (BA^-1)ZSK Matrix multiplication associative \(\displaystyle={\left({A}{B}\right)}{B}{\left({B}{A}^{{-{{1}}}}\right)}$$
$$\displaystyle{A}{B}^{{3}}{A}^{{-{{1}}}}$$
Let A, B and A + B are aach norsingular. We have to show $$\displaystyle{A}{\left({A}+{B}\right)}^{{-{{1}}}}{B}={B}{\left({A}+{B}\right)}^{{-{{1}}}}{A}={\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}^{{-{{1}}}}$$
Note that we need to show that A(A + B)^-1B is inverse of $$\displaystyle{\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}$$.
So, our first task is to show $$\displaystyle{A}{\left({A}+{B}\right)}^{{-{{1}}}}{B}{\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}={\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}{A}{\left({A}+{B}\right)}^{{-{{1}}}}{B}={1}$$ Notice that
PSKA(A+B)^-1B(A^-1+B^-1) = A(A+B)^-1(BA^-1 +1) =A(A+B)^-1(A+B)A^-1 =IZSK
Also we have
PSK(A^-1+B^-1)A(A+B)^-1B=(I+B^-1A)(A+B)^-1B =B^-1(B+A)(A+B)ZSK
This proves that $$\displaystyle{A}{\left({A}+{B}\right)}^{{-{{1}}}}{B}$$ is inverse of $$\displaystyle{\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}$$
Similarly, we can prove that $$\displaystyle{B}{\left({A}+{B}\right)}^{{-{{1}}}}{A}$$ is inverse of $$\displaystyle{\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}$$

### Relevant Questions

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