Question

# (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 thatA(A+B)^-1B=B(A+B)^-1A=(A^-1+B^-1)^-1

Matrix transformations

(7) If A and B are a square matrix of the same order. Prove that $$(ABA^{−1})^3=AB^3A^{−1}$$

2021-02-22

Let A and B are n*n matrix. Then we have
$$(ABA^{-1})^3 = (ABA^{-1}) (ABA^{-1}) (AB^{-1}) = (AB) (A^{-1}A) B(A^{-1}A) (BA^{-1})$$ 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 $$A(A + B)^{-1}B$$ 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
$$A(A+B)^{-1}B(A^{-1}+B^{-1}) = A(A+B)^{-1}(BA^{-1} +1) =A(A+B)^{-1}(A+B)A^{-1} =I$$
Also we have
$$(A^{-1}+B^{-1})A(A+B)^{-1}B=(I+B^{-1}A)(A+B)^{-1}B =B^{-1}(B+A)(A+B)$$
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 $$\displaystyle{\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}$$