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(D) Given $A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}$.We calculate the powers of $A$:$A^2 = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} \begin

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$A^n = nA - (n-1)I$

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(D) Given $A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}$.We calculate the powers of $A$:$A^2 = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix}$$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix}$By induction,we can hypothesize that $A^n = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$.Now,let us check option $(D)$: $nA - (n-1)I = n \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} - (n-1) \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} n & 0 \ n & n \end{bmatrix} - \begin{bmatrix} n-1 & 0 \ 0 & n-1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$.This matches our derived form for $A^n$. Thus,option $(D)$ is correct.

If $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then which of the following holds for all $n \geq 2, n \in N$?

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