If A=left[begin{array}{ll}1 & 1 0 & 1end{arr | vedclass

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(A) Given $A=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight]$ and $I=\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight]$.We calculate the

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

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(A) Given $A=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight]$ and $I=\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight]$.We calculate the powers of $A$:$A^2 = A \cdot A = \left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight] \left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight] = \left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight] = 2A - I = 2A - (2-1)I$.$A^3 = A^2 \cdot A = \left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight] \left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight] = \left[\begin{array}{ll}1 & 3 \ 0 & 1\end{array}ight] = 3A - 2I = 3A - (3-1)I$.By mathematical induction, we can generalize this pattern for any $n \in N$:$A^n = nA - (n-1)I$.

If $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then for all $n \in N$, find $A^n$.

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