Q. If $k \in R_{0}$, then det{adj($kI_{n})\}$ is equal to
A) $k^{n - 1}$
B) $k^{n(n - 1)}$
C) $k^{n}$
D) $k$
Explanation
(b)
$\left( kI_{n} \right)$adj$\left( kI_{n} \right) = \left| kI_{n} \right|I_{n}\lbrack using\ A(adjA) = |A|I\rbrack$
adj$\left( kI_{n} \right) = k^{n - 1}I_{n}$
$\left| adj(kI_{n}) \right| = k^{n(n - 1)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.