If A = begin{bmatrix} 1 & 2 0 & 1 end{bmatri | vedclass

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(A) Given $A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$.Calculate ${A^2} = A 	imes A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatr

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$\begin{bmatrix} 1 & 2n \ 0 & 1 \end{bmatrix}$

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(A) Given $A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$.Calculate ${A^2} = A 	imes A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+2(0) & 1(2)+2(1) \ 0(1)+1(0) & 0(2)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 4 \ 0 & 1 \end{bmatrix}$.Calculate ${A^3} = {A^2} 	imes A = \begin{bmatrix} 1 & 4 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+4(0) & 1(2)+4(1) \ 0(1)+1(0) & 0(2)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 6 \ 0 & 1 \end{bmatrix}$.Observing the pattern,we see that for any positive integer $n$,the matrix follows the form ${A^n} = \begin{bmatrix} 1 & 2n \ 0 & 1 \end{bmatrix}$.Thus,the correct option is $A$.

If $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$,then ${A^n} = $

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