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Question: Matrix A is such that \(A^{2} = 2A - I\)where I is the identity matrix. Then for \(n \geq 2,A^{n} =\...

Matrix A is such that A2=2A−IA^{2} = 2A - Iwhere I is the identity matrix. Then for n≥2,An=n \geq 2,A^{n} =

A

nA−(n−1)InA - (n - 1)I

B

nA−InA - I

C

2n−1A−(n−1)I2^{n - 1}A - (n - 1)I

D

2n−1A−I2^{n - 1}A - I

Answer

nA−(n−1)InA - (n - 1)I

Explanation

Solution

We have, A2=2A−IA^{2} = 2A - I ⇒ A2.A=(2A−I)AA^{2}.A = (2A - I)A; A3=2A2−IA=2[2A−I]−IAA^{3} = 2A^{2} - IA = 2\lbrack 2A - I\rbrack - IA ⇒ A3=3A−2IA^{3} = 3A - 2I

Similarly A4=4A−3IA^{4} = 4A - 3I and hence An=nA−(n−1)IA^{n} = nA - (n - 1)I