A) \[{{k}^{3}}I\]
B) \[{{k}^{2}}I\]
C) \[-{{k}^{3}}I\]
D) \[-{{k}^{2}}I\]
Correct Answer: B
Solution :
Let\[I=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\], then \[kI=\left[ \begin{matrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \\ \end{matrix} \right]\] \[\Rightarrow \] \[adj\,(kI)=\left[ \begin{matrix} {{k}^{2}} & 0 & 0 \\ 0 & {{k}^{2}} & 0 \\ 0 & 0 & {{k}^{2}} \\ \end{matrix} \right]={{k}^{2}}I\].
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