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Given the matrix \begin{align} A = \begin{bmatrix}1&a\\0&2\end{bmatrix} , \end{align} prove that \begin{align} A^n=\begin{bmatrix}1&(2^n-1)a\\0&2^n\end{bmatrix} \end{align} for all $n \geq 0$.

I have no clue on how to prove it right. All I could do was to try with $A^2$, $A^3$ and so on but I can't prove it.

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closed as off-topic by Saad, José Carlos Santos, egreg, Cesareo, Lee David Chung Lin Mar 24 at 2:32

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    $\begingroup$ By induction... $\endgroup$ – Yu Ding Mar 23 at 9:28
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    $\begingroup$ Welcome to MSE. In the future please include your own thoughts, the effort made so far, and the specific difficulties that got you stuck. $\endgroup$ – Lee David Chung Lin Mar 24 at 2:32
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You can prove this by mathematical induction.

Base:

$$\begin{bmatrix}1&a\\0&2\end{bmatrix} = \begin{bmatrix}1&(2^1-1)a\\0&2^1\end{bmatrix}$$

Induction step:

Suppose $A^n = \begin{bmatrix}1&(2^n-1)a\\0&2^n\end{bmatrix}$. Then $$A^{n + 1} = A^nA = \begin{bmatrix}1&(2^n-1)a\\0&2^n\end{bmatrix}\begin{bmatrix}1&a\\0&2\end{bmatrix} = \begin{bmatrix}1&a + 2(2^n-1)a\\0&2^{n + 1}\end{bmatrix} = \begin{bmatrix}1&(2^{n+1}-1)a\\0&2^{n+1}\end{bmatrix}$$

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    $\begingroup$ The idea behind this proof is that the expression holds true for n=1, and for every n that it holds true, it still does for n+1. Therefore the expression is valid for n=1, therefore for n=2, therefore for n=3 and so on... $\endgroup$ – David Mar 23 at 9:33
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Note that $$A^n=\left( I+ \begin{bmatrix}0&a\\0&1\end{bmatrix} \right)^n=I+\sum_{k=1}^n\binom{n}{k}\begin{bmatrix}0&a\\0&1\end{bmatrix}^k= I+(2^n-1)\begin{bmatrix}0&a\\0&1\end{bmatrix}=\begin{bmatrix}1&(2^n-1)a\\0&2^n\end{bmatrix}$$ where we used the fact that $\begin{bmatrix}0&a\\0&1\end{bmatrix}^k=\begin{bmatrix}0&a\\0&1\end{bmatrix}$ and $\sum_{k=1}^n\binom{n}{k}=2^n-1$.

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$A$'s characteristic equation is $\lambda^2 -3\lambda +2=0$, whose roots are $1$ and $2$.
Hence, $A^n=B+2^n\cdot C$,where $B$ and $C$ are some matrices that you will find by substituting $n=0$ and $n=1$ into this relation.
Note: This is a method that works for any $2 \times 2$ matrix provided that the computations do not get messy.

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