DeterminantHard

Question

Suppose A is a matrix such that A2 = A and (I + A)10 = I + kA, then k is

Options

A.127
B.511
C.1023
D.1024

Solution

A2 = A
⇒  A-1 A2 = A-1A   ⇒    A = I
∴  (I + A)10 = (I + I)10   = (2I)10
= 1024 I   = I + KI  = (K + 1) I
∴ K + 1 = 1024   ⇒    K = 1023

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