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

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 $n A - (n - 1) I$
$A^{2} = 2 A - I$
$\Rightarrow A^{3} = A (2 A - I) = 2 A^{2} - A = 2 (2 A - I) - A = 3 A - 2 I$ -----(1)
$\Rightarrow A^{4} = A (3 A - 2 I) = 3 A^{2} - 2 A = 3 (2 A - I) - 2 A = 4 A - 3 I$ ----(2)
clearly,
$∴ A^{n} = n A - (n - 1) I$
Hence option A.

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Similar questions

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

A^{2}

(I + A)^{n}

A

A^{2} = 2 A - I

I

n \geq 2

A^{n}

If A = (\begin{matrix} 1 & 0 1 & 1 \end{matrix}] and I = (\begin{matrix} 1 & 0 0 & 1 \end{matrix}], then which of the following holds for all n \in N ?

A

A^{2} = 2 A - I

I

n \geq 2

n \in N; A^{n}

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