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In a group (G $^{*}$ ), then equation $x^{*} a=b$ has a
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Verified Answer
The correct answer is:
unique solution $b^{*} a^{-1}$
$$
\begin{array}{l}
\mathrm{x}^{*} \mathrm{a}=\mathrm{b} \Rightarrow\left(\mathrm{x}^{*} \mathrm{a}\right)^{*} \mathrm{a}^{-1}=\mathrm{b}^{*} \mathrm{a}^{-1} \\
\Rightarrow \mathrm{x}^{*}\left(\mathrm{a}^{*} \mathrm{a}^{-1}\right)=\mathrm{b}^{*} \mathrm{a}^{-1} \\
\Rightarrow \mathrm{x}^{*} \mathrm{e}=\mathrm{b}^{*} \mathrm{a}^{-1} \\
\Rightarrow \mathrm{x}=\mathrm{b}^{*} \mathrm{a}^{-1}
\end{array}
$$
\begin{array}{l}
\mathrm{x}^{*} \mathrm{a}=\mathrm{b} \Rightarrow\left(\mathrm{x}^{*} \mathrm{a}\right)^{*} \mathrm{a}^{-1}=\mathrm{b}^{*} \mathrm{a}^{-1} \\
\Rightarrow \mathrm{x}^{*}\left(\mathrm{a}^{*} \mathrm{a}^{-1}\right)=\mathrm{b}^{*} \mathrm{a}^{-1} \\
\Rightarrow \mathrm{x}^{*} \mathrm{e}=\mathrm{b}^{*} \mathrm{a}^{-1} \\
\Rightarrow \mathrm{x}=\mathrm{b}^{*} \mathrm{a}^{-1}
\end{array}
$$
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