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Equality of Matrices

Matrix A is...

Question

Matrix $A$ is such that $A^{2} = 2 A - 1$ , where $l$ is the identity matrix, then for $n \geq 2, A^{n}$ is equal to

n A - (n - 1) l

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n A - 1

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

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2^{n - 1} A - 1

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Solution

The correct option is A $n A - (n - 1) l$
Given that,
$A^{2} = 2 A - I$
Now,
$\begin{matrix} A^{3} = 2 A^{2} - I A = 2 (2 A - I) - A A^{3} = 3 A - 2 I A^{4} = 4 A - 3 I A^{5} = 5 A - 4 I A^{n} = n A - (n - 1) I \end{matrix}$
Hence,
Option $A$ is correct answer.

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