Finding $n$th power of a $3\times 3$ matrix

Question

Find the $A^n$ if $$A=\begin{bmatrix}1 & a & b \\0 & 1 &a\\0 &0 &1\end{bmatrix}$$ I tried inductive method to show $$A^n=\begin{bmatrix}1 & na & nb+\frac{n(n-1)}{2}a^2 \\0 & 1 &na\\0 &0 &1\end{bmatrix}$$ now : My question is : Is there other method (idea ) to find $A^n$ ?
Thanks in advance. Can the idea apply for $$A=\begin{bmatrix}1 & a & b \\0 & 1 &c\\0 &0 &1\end{bmatrix}$$ when $c \neq a$ ?

Answer 1

$$A=\begin{bmatrix}1 & a & b \\0 & 1 &c\\0 &0 &1\end{bmatrix}$$

Write $A = I + U$, where $U^2 = \begin{bmatrix} a \\ 0 \\ 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & c \end{bmatrix}$ and $U^3=0$.

$$ A^n=(I+U)^n=I+nU+{n\choose 2}U^2 = \pmatrix{1&na&nb+\frac{n(n-1)}{2}ac\\0&1&nc\\0&0&1}$$

Answer 2

If we define $$ N=\begin{bmatrix}0&a&b\\0&0&a\\0&0&0\end{bmatrix} $$ then $A=I+N$, $N^3=0$, and $I$ and $N$ commute. Therefore we can use to the binomial theorem to obtain $$ A^n=(I+N)^n=I+nN+{n\choose 2}N^2 $$ which matches your answer.

Answer 3

$$A=I+aJ+bJ^2$$ where $$J=\pmatrix{0&1&0\\0&0&1\\0&0&0}.$$ Then $J^3=O$. Now $$A^n=\sum_{k=0}^n{n\choose k}(aJ+bJ^2)^n =I+n(aJ+bJ^2)+{n\choose 2}(aJ+bJ^2)^2$$ as $(aJ+bJ^2)^3=O$. But $(aJ+bJ^2)^2=a^2J^2$ so $$A^n=I+naJ+nbJ^2+\frac{n(n-1)}2a^2J^2 =\pmatrix{1&na&nb+\frac{n(n-1)}{2}a^2\\0&1&na\\0&0&1}.$$