A) \[\left( \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right)\]
B) \[\frac{1}{(ad-bc)}\left( \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right)\]
C) \[\frac{1}{|A|}\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right)\]
D) \[\left( \begin{matrix} b & -a \\ d & -c \\ \end{matrix} \right)\]
Correct Answer: B
Solution :
\[|A|\,=(ad-bc)\] \[\therefore \,\,{{A}^{-1}}=\frac{1}{(ad-bc)}\,\,\left[ \,\begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]\].
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