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If $A^{2} = 2 A - I$ , then for $n \neq 2, A^{n}$ =

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nA - (n - 1)I

nA - I

$2^{n - 1} A - (n - 1) I$

$2^{n - 1} A - I$

answer is A.

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Detailed Solution

$A^{3} = A A^{2} = A (2 A - I) = 2 A^{2} - A I = 2 A^{2} - A = 2 (2 A - I) - A = 3 A - 2 I .$ $∴ A^{n} = n A - (n - 1) I$

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