# Proving $A^n=\left[\begin{smallmatrix}1&(2^n-1)a\\0&2^n\end{smallmatrix}\right]$ [closed]

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.

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

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• By induction... – Yu Ding Mar 23 at 9:28
• Welcome to MSE. In the future please include your own thoughts, the effort made so far, and the specific difficulties that got you stuck. – Lee David Chung Lin Mar 24 at 2:32

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

• 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... – David Mar 23 at 9:33

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 $$\sum_{k=1}^n\binom{n}{k}=2^n-1$$.

$$A$$'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.