A) \[2A\]
B) \[A\]
C) \[-A\]
D) \[I\]
Correct Answer: B
Solution :
Given, \[A=\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] \[|A|=-1\] and \[C'=\left[ \begin{matrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \\ \end{matrix} \right]\] Now, \[{{A}^{-1}}=\frac{1}{|A|}.C'=\frac{1}{-1}\left[ \begin{matrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{A}^{-1}}=\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=A\]
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