Matrix $A$ is 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, $A^{2} = 2 A - I$ ...(i)
On multiplying by $A$ both sides, we get
$A^{2} \cdot A = (2 A - I) A$
$\Rightarrow A^{3} = 2 A^{2} - I A$
$= 2 (2 A - I) - I A$ [from equation (i)]
$= 4 A - 2 I - A$ ..... $(∵ I A = A)$
$= 3 A - 2 I$
Similarly, $A^{4} = 4 A - 3 I$
$\Rightarrow A^{5} = 5 A - 4 I$
Hence, $A^{n} = n A - (n - 1) I$

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