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
⇒ 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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