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
ANSWERED
asked 2021-02-21

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

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

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