A) \[\left[ \begin{matrix} -3 & 2 \\ 2 & -1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 3 & 2 \\ 2 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 2 \\ 2 & 3 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 3 & -2 \\ -2 & 1 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[A=\left[ \begin{matrix} 1 & 2 \\ 2 & 3 \\ \end{matrix} \right]\] \[|A|=\left| \begin{matrix} 1 & 2 \\ 2 & 3 \\ \end{matrix} \right|=3-4=-1\] adj \[(A)=\left[ \begin{matrix} 3 & -2 \\ -2 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{-1}}=\frac{1}{|A|}adj(A)=\frac{1}{-1}\left[ \begin{matrix} 3 & -2 \\ -2 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -3 & 2 \\ 2 & -1 \\ \end{matrix} \right]\]
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