Let A and B be n x n square matrices with A being invertible. Show that if AB=BA, then A−1B=BA−1. Also, give an example such that AB=BA, but ATB=BAT.
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Is this right?
If AB=BA, then A−1AB=A−1BA
⟹B=A−1BA
⟹BA−1=A−1B
For AB=BA and ATB=BAT, let \(\displaystyle A=\left\[\begin{array}{cc} 1&2\\3&4\end{array}\right\]\) and let \(\displaystyle B=A^{-1}=\left\[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\]\).
Then AB=AA−1=A−1A=BA.
\(\displaystyle A^TB=\left\[\begin{array}{cc}1&3\\2&4\end{array}\right\] \left\[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\] = \left\[\begin{array}{cc}\frac{5}{2}&-\frac{1}{2}\\2&0\end{array}\right\]\)
\(\displaystyle BA^T = \left[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\] \left\[\begin{array}{cc}1&3\\2&4\end{array}\right\] = \left\[\begin{array}{cc}0&-2\\\frac{1}{2}&\frac{5}{2}\end{array}\right\]\).
Therefore ATB=BAT.
______________________________________
Is this right?
If AB=BA, then A−1AB=A−1BA
⟹B=A−1BA
⟹BA−1=A−1B
For AB=BA and ATB=BAT, let \(\displaystyle A=\left\[\begin{array}{cc} 1&2\\3&4\end{array}\right\]\) and let \(\displaystyle B=A^{-1}=\left\[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\]\).
Then AB=AA−1=A−1A=BA.
\(\displaystyle A^TB=\left\[\begin{array}{cc}1&3\\2&4\end{array}\right\] \left\[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\] = \left\[\begin{array}{cc}\frac{5}{2}&-\frac{1}{2}\\2&0\end{array}\right\]\)
\(\displaystyle BA^T = \left[\begin{array}{cc}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{array}\right\] \left\[\begin{array}{cc}1&3\\2&4\end{array}\right\] = \left\[\begin{array}{cc}0&-2\\\frac{1}{2}&\frac{5}{2}\end{array}\right\]\).
Therefore ATB=BAT.