Question
If $$Z$$ is an idempotent matrix, then $${\left( {I + Z} \right)^n}$$
A.
$$I + {2^n}Z$$
B.
$$I + \left( {{2^n} - 1} \right)Z$$
C.
$$I - \left( {{2^n} - 1} \right)Z$$
D.
None of these
Answer :
$$I + \left( {{2^n} - 1} \right)Z$$
Solution :
$$Z$$ is idempotent then $${Z^2} = Z$$
$$\eqalign{
& \Rightarrow {Z^3},{Z^4},.....,{Z^n} = Z \cr
& {\left( {I + Z} \right)^n} = {\,^n}{C_0}{I^n} + {\,^n}{C_1}{I^{n - 1}}Z + {\,^n}{C_2}{I^{n - 2}}{Z^2} + ..... + {\,^n}{C_n}{Z^n} \cr
& = {\,^n}{C_0}I + {\,^n}{C_1}Z + {\,^n}{C_2}Z + {\,^n}{C_3}Z + ..... + {\,^n}{C_n}Z \cr
& = I + \left( {^n{C_1} + {\,^n}{C_2} + {\,^n}{C_3} + ..... + {\,^n}{C_n}} \right)Z \cr
& = I + \left( {{2^n} - 1} \right)Z \cr} $$