Matrix A such that $A^{2} = 2 A - I,$ where I is the identity matrix. Then for $n \geq 2, A^{n}$ is equal to

n A - (n - 1) I

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n A - I

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

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2^{n - 1} A - I

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Solution

The correct option is A $n A - (n - 1) I$
Given , $A^{2} = 2 A - I$
Now, $A^{3} = A^{2} . A = 2 A^{2} - A I = 2 (2 A - I) - A = 4 A - 2 I - A = 3 A - (3 - 1) I$
$A^{4} = A^{3} A = (3 A - 2 I) A = 3 A^{2} - 2 A I = 3 (2 A - I) - 2 A = 4 A - 3 I = 4 A - (4 - 1) I$
$∴ A^{n} = n A - (n - 1) I$

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