Q. If $Z$is an idempotent matrix, then $(I + Z)^{n}$
A) $I + 2^{n}Z$
B) $I + (2^{n} - 1)Z$
C) $I - (2^{n} - 1)Z$
D) None of these
Explanation
(b)
$Z$ is idempotent, then
$Z^{2} = Z \Rightarrow Z^{1},\ Z^{4},\ \ldots,Z^{n} = Z$
$\therefore(I + Z)^{n} = {{}^{n}C}_{0}I^{n} + {{}^{n}C}_{1}I^{n - 1}Z + {{}^{n}C}_{2}I^{n - 2}Z^{2} + \ldots + {{}^{n}C}_{n}Z^{n}$
$= {{}^{n}C}_{0}I^{} + {{}^{n}C}_{1}Z + {{}^{n}C}_{2}Z^{} + {{}^{n}C}_{3} + \ldots + {{}^{n}C}_{n}Z\$
$= I + ({{}^{n}C}_{1} + {{}^{n}C}_{2} + {{}^{n}C}_{3} + \ldots + {{}^{n}C}_{n})Z$
$= I + (2^{n} - 1)Z$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.