If $A = [\begin{array}{ll} x x \\ x x \end{array}]$ , then $A^{n} = \dots \dots \dots \dots, n \in N$

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$[\begin{array}{ll} 2^{n - 2} x^{n} 2^{n - 2} x^{n} \\ 2^{n - 2} x^{n} 2^{n - 2} x^{n} \end{array}]$

$[\begin{array}{cc} 2^{n} x^{n} & 2^{n} x^{n} \\ 2^{n} x^{n} & 2^{n} x^{n} \end{array}]$

$[\begin{array}{ll} 2^{n - 1} x^{n} 2^{n - 1} x^{n} \\ 2^{n - 1} x^{n} 2^{n - 1} x^{n} \end{array}]$

$[\begin{array}{ll} 2^{n - 1} x^{n - 1} 2^{n - 1} x^{n - 1} \\ 2^{n - 1} x^{n - 1} 2^{n - 1} x^{n - 1} \end{array}]$

answer is B.

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Detailed Solution

$\Rightarrow A^{2} = [\begin{array}{ll} x x \\ x x \end{array}] [\begin{array}{ll} x x \\ x x \end{array}] = [\begin{array}{ll} 2 x^{2} 2 x^{2} \\ 2 x^{2} 2 x^{2} \end{array}] \Rightarrow A^{3} = [\begin{array}{cc} 4 x^{3} & 4 x^{3} \\ 4 x^{3} & 4 x^{3} \end{array}] = [\begin{array}{cc} 2^{3 - 1} x^{3} & 2^{3 - 1} x^{3} \\ 2^{3 - 1} x^{3} & 2^{3 - 1} x^{3} \end{array}]$
$\Rightarrow A^{n} = [\begin{array}{ll} 2^{n - 1} x^{n} 2^{n - 1} x^{n} \\ 2^{n - 1} x^{n} 2^{n - 1} x^{n} \end{array}]$