for n=2 ⇒ A2=3-41-1 3-41-1 =5-82-3 ∴ clearly option (4) is correct An=1+2n-4nn1-2n

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If $A = (\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}) then A^{n} =$

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$(\begin{matrix} 3 n & - 4 n \\ n & - n \end{matrix})$

$(\begin{matrix} 2 + n & 5 - n \\ n & - n \end{matrix})$

$(\begin{matrix} 3^{n} & {(- 4)}^{n} \\ 1^{n} & {(- 1)}^{n} \end{matrix})$

$(\begin{matrix} 1 + 2 n & - 4 n \\ n & 1 - 2 n \end{matrix})$

answer is D.

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Detailed Solution

$for n = 2 \Rightarrow A^{2} = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}] [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}] = [\begin{matrix} 5 & - 8 \\ 2 & - 3 \end{matrix}] ∴ clearly option (4) is correct A^{n} = [\begin{matrix} 1 + 2 n & - 4 n \\ n & 1 - 2 n \end{matrix}]$