(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

(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
asked 2021-02-21
(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 \(\displaystyle{\left({A}^{{-{{1}}}}+{B}^{{-{{1}}}}\right)}\)
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