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

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

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(C) Given $A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}$.Calculate $A^2 = A 	imes A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix}$.Calculate $A^3 = A^2 	imes A = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix}$.By observation,$A^n = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$ for all $n \ge 1$.Now,evaluate $nA - (n - 1)I$:$nA = n \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} n & 0 \ n & n \end{bmatrix}$.$(n - 1)I = (n - 1) \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} n - 1 & 0 \ 0 & n - 1 \end{bmatrix}$.$nA - (n - 1)I = \begin{bmatrix} n - (n - 1) & 0 - 0 \ n - 0 & n - (n - 1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$.Since $A^n = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$,it follows that $A^n = nA - (n - 1)I$.

If $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then which one of the following holds for all $n \ge 1$ (by the principle of mathematical induction)?

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