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 \(\displaystyle{\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}\)

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 \(\displaystyle{\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}\)