Formula for $A^n$ where $n \in \{1, \ 2, \ \cdots \ \}$ for the matrix $A = \begin{bmatrix} 1 && b \\ 0 && 1 \end{bmatrix}$

Question

Formula for $A^n$ where $n \in \{1, \ 2, \ \cdots \ \}$ for the matrix $A = \begin{bmatrix} 1 && b \\ 0 && 1 \end{bmatrix}.$

Please help with the question if you can, it is for my Linear Algebra class and I cannot find anything close to an answer.

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EDIT: As advised in comments, I have tried this for the first few $n$'s:

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

$A^2= \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 2b \\ 0 & 1 \end{bmatrix}$

$A^3=A^2\cdot A= \begin{bmatrix} 1 & 2b \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & b+2b \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 3b \\ 0 & 1 \end{bmatrix}$

Answer 1

$$A^n=\begin{pmatrix}1&nb\\0&1\end{pmatrix}$$ Suppose this is true until $n$. Then $$A^{n+1}=A.A^n$$ Computing the right side almost completes the proof.

Answer 2

Without induction:

Write $A$ as $I+bE$, where $I$ is the unit matrix of rank $2$ and $E=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. As both matrices commute with each other, we can apply the binomial formula in the ring $M_2(\mathbf R)$, noting that $E^2=0$: $$A^n=\sum_{k=0}^n\binom nk I^{n-k}b^kE^k=I+nbE=\begin{bmatrix}1&nb\\0&1\end{bmatrix}.$$

Answer 3

Note that $A$ is the elementary matrix that adds $b$ times the second row to the first row. Then left multiplying by $A$ amounts to doing that operation, and the solution is now obvious.