A) \[\frac{1}{27}\left( \begin{matrix} 1 & -26 \\ 0 & 27 \\ \end{matrix} \right)\]
B) \[\frac{1}{27}\left( \begin{matrix} -1 & 26 \\ 0 & 27 \\ \end{matrix} \right)\]
C) \[\frac{1}{27}\left( \begin{matrix} 1 & -26 \\ 0 & -27 \\ \end{matrix} \right)\]
D) \[\frac{1}{27}\left( \begin{matrix} -1 & -26 \\ 0 & -27 \\ \end{matrix} \right)\]
Correct Answer: A
Solution :
\[|A|=3,\,AdjA=\left( \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right)\]; \[\therefore \] \[{{A}^{-1}}=\frac{1}{3}\,\left( \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right)\] \[\Rightarrow \] \[{{({{A}^{-1}})}^{3}}=\frac{1}{27}\,{{\left( \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right)}^{3}}=\frac{1}{27}\,\left( \begin{matrix} 1 & -26 \\ 0 & 27 \\ \end{matrix} \right)\].
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