A) \[\left[ \begin{matrix} 2 & -2 \\ 2 & 3 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 3 & -2 \\ 2 & 2 \\ \end{matrix} \right]\]
C) \[\frac{1}{10}\left[ \begin{matrix} 2 & 2 \\ -2 & 3 \\ \end{matrix} \right]\]
D) \[\frac{1}{10}\left[ \begin{matrix} 3 & 2 \\ -2 & 2 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[{{({{B}^{-1}}{{A}^{-1}})}^{-1}}={{({{A}^{-1}})}^{-1}}{{({{B}^{-1}})}^{-1}}=AB\] (Reversal law of inverses) \[=\,\left[ \,\begin{matrix} 2 & 2 \\ -3 & 2 \\ \end{matrix}\, \right]\,\left[ \,\begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix}\, \right]=\left[ \,\begin{matrix} 2 & -2 \\ 2 & 3 \\ \end{matrix}\, \right]\].
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