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

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(C) First,calculate the powers of $A$:$A^2 = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatri

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

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(C) First,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}$$A^3 = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix}$By induction,we can observe that $A^n = \begin{bmatrix} 1 & 0 \ n & 1 \end{bmatrix}$.Now,evaluate the expression in option $C$:$nA = n \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} n & 0 \ n & n \end{bmatrix}$$(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} = A^n$.Thus,the correct relation is $A^n = nA - (n - 1)I$.

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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