Recurrence series with matrix for $a_n=a_{n-1}-2a_{n-2}$ 
$a_0=0$, $a_1=1$, $a_n=a_{n-1} - 2a_{n-2}$ for $n\ge2$. Find a closed form for $a_n$.

So if I compute $a_2=1-0=1$, $a_3=1-2=-1$:
$$\begin{bmatrix} a_n\\a_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1} \\ a_{n-2}\end{bmatrix}$$
I think this matrix is true for $n\ge2$. But the idea is to find a matrix that satisfy $n\ge1$?
I can see that $\begin{bmatrix} 1 \\0\end{bmatrix}$ is base case? And if they ask to find general term of $a_n$, I should find expression that satisfy the recurrence or the base case?
I suppose that $\begin{bmatrix} a_n\\a_{n-1} \end{bmatrix}  $ = $x_n$
and $\begin{bmatrix} a_{n-1} \\ a_{n-2}\end{bmatrix}=x_{n-1}$
$$x_{n}=Ax_{n-1}$$
$$x_{n-1}=A^{-1}x_n$$
clearly do not satisfy the equation, and what is the purpose of finding a matrix for this problem?
The right pattern is $x_n=A^{n-1}x_1$, I was confused why it has always to be $x_1$ and how can I find it? thanks


So if I compute $a_2=1-0=1$, $a_3=1-2=-1$:
  $$\begin{bmatrix} a_n\\a_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1} \\ a_{n-2}\end{bmatrix}$$

Good start.  Now continue that....
$$\begin{align}
\begin{bmatrix} a_{n} \\ a_{n - 1} \end{bmatrix} &= 
\begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n - 1} \\ a_{n - 2} \end{bmatrix}
\\ &= \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n - 2} \\ a_{n - 3} \end{bmatrix}
\\ &= \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n - 3} \\ a_{n - 4} \end{bmatrix}
\\ &= \dots
\end{align}$$
In summary $$\begin{bmatrix} a_{n + 1} \\ a_{n} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} a_1 \\ a_0 \end{bmatrix}$$

I think this matrix is true for $n\ge2$. But the idea is to find a matrix that satisfy $n\ge1$?

Well lets look at $n=1$:
$$\begin{bmatrix} a_1 \\a_{0} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{-1}\end{bmatrix}$$
So the $n = 1$ case is the equation $a_1 = a_{0} - 2a_{-1}$.  Since we have no information about $a_{-1}$, this is neither provable nor useful.  So $n \ge 2$ is fine.

what is the purpose of finding a matrix for this problem?

Because it let's you reach $$\begin{bmatrix} a_{n + 1} \\ a_{n} \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} a_1 \\ a_0 \end{bmatrix}$$ after which you can use all kinds of nice matrix properties to say more things about the sequence.  For example, you could diagonalize $\begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix}$ to get
$$\begin{bmatrix} 1 & -2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix} \begin{bmatrix} \frac{1 - \sqrt{-7}}2 & 0 \\ 0 & \frac{1 + \sqrt{-7}}2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix}^{-1}$$
so
$$\begin{align}
\begin{bmatrix} a_{n + 1} \\ a_{n} \end{bmatrix} &= \left( \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix} \begin{bmatrix} \frac{1 - \sqrt{-7}}2 & 0 \\ 0 & \frac{1 + \sqrt{-7}}2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix}^{-1}\right)^n \begin{bmatrix} a_1 \\ a_0 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix} \begin{bmatrix} \frac{1 - \sqrt{-7}}2 & 0 \\ 0 & \frac{1 + \sqrt{-7}}2 \end{bmatrix}^n \begin{bmatrix} 1 & 1 \\ \frac{2}{1 - \sqrt{-7}} & \frac{2}{1 + \sqrt{-7}} \end{bmatrix}^{-1} \begin{bmatrix} a_1 \\ a_0 \end{bmatrix} 
\end{align}$$
Ugh, this sequence doesn't simplify into anything pretty because of all the complex numbers.  But it is a straightforward way of deriving a formula of the form $a_{n} = C_1 \phi^n + C_2 {\varphi}~^n$
Also matrices keep the problem only involving integers instead of complex numbers, so computation is so much nicer.

The right pattern is $x_n=A^{n-1}x_1$, I was confused why it has always to be $x_1$ and how can I find it? thanks

It is a bit cleaner to do $x_{n+1} = A^n x_1$.  You use $x_1$ as the base case because $a_1$ and $a_0$ is what you were given in the statement of the problem.  You can use any convenient $x$ that you can calculate as the base case, but why not just use the one given to you?
