A) \[-30\]
B) \[-30/13\]
C) \[-40/9\]
D) \[-35/12\]
Correct Answer: B
Solution :
[b] Let \[S=A-\frac{{{A}^{2}}}{3}+\frac{{{A}^{3}}}{9}-\frac{{{A}^{4}}}{27}+.....\] \[\therefore \,\,\,\,\,-\frac{AS}{3}=-\frac{{{A}^{2}}}{3}+\frac{{{A}^{3}}}{9}-\frac{{{A}^{4}}}{27}+....\] Subtracting, we get \[\Rightarrow \,\,\,\,\,S+\frac{AS}{3}=A\] \[\Rightarrow \,\,\,\,\,S\left( I+\frac{A}{3} \right)=A\] \[\Rightarrow \,\,\,\,\,S(3I+A)=3A\] \[\Rightarrow \,\,\,\,\,S\left[ \begin{matrix} 4 & -3 \\ -1 & 4 \\ \end{matrix} \right]\,\,=\,\,\left[ \begin{matrix} 3 & -9 \\ -3 & 3 \\ \end{matrix} \right]\] \[\Rightarrow \,\,\,\,\,S=\left[ \begin{matrix} 3 & -9 \\ -3 & 3 \\ \end{matrix} \right]\,\,{{\left[ \begin{matrix} 4 & -3 \\ -1 & 4 \\ \end{matrix} \right]}^{-1}}=\frac{1}{13}\left[ \begin{matrix} 3 & -27 \\ -9 & 3 \\ \end{matrix} \right]\]
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