A) \[{{k}^{2}}I\]
B) \[-{{k}^{3}}I\]
C) \[-{{k}^{2}}I\]
D) \[\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1-x & 1 \\ 1 & 1 & 1+y \\ \end{matrix} \right|\]
Correct Answer: A
Solution :
\[|A|=3,\]adj\[A=\left[ \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{-1}}=\frac{1}{3}\left[ \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right]\] \[\Rightarrow \] \[{{({{A}^{-1}})}^{3}}=\frac{1}{27}{{\left[ \begin{matrix} 1 & -2 \\ 0 & 3 \\ \end{matrix} \right]}^{3}}\] \[=\frac{1}{27}\left( \begin{matrix} 1 & -8 \\ 0 & 27 \\ \end{matrix} \right)\]
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