System of sequences How can I solve this system of two sequences, 
$$
\begin{cases}
a_{n+1}=0.9a_n+0.3b_n \\
b_{n+1}=0.1a_n+0.7b_n
\end{cases}
$$
with $(a_1,b_1)=(1000,1000)$
using matrix ? 
I know the answer is $(1500,500)$ but it not gives me the method. I seek a bit on the net but I haven't find good key words to seek this. 
EDIT: What I get with the comments and the hint:
$$\binom{a_{n}}{b_{n}}=\begin{pmatrix} 0.9 & 0.3 \\ 0.1 & 0.7 \end{pmatrix}^{n-1} \binom{a_1}{b_1} $$
So because the eigenvalues are $1$ and $0.6$ then $$\binom{a_{n}}{b_{n}}=P\begin{pmatrix} 1 & 0 \\ 0 & 0.6 \end{pmatrix}^{n-1}P^{-1} \binom{a_1}{b_1} $$
I just need to get the matrix $P$ and take the limit. Is that correct?
 A: Hint:
You can set up the matrix like so:
$$\pmatrix{0.9 \;\;\; 0.3 \\ 0.1  \;\;\;  0.7}\pmatrix{a_n \\ b_n}=\pmatrix{a_{n+1} \\ b_{n+1}}$$
As you can see, going from $\pmatrix{a_n \\ b_n} \rightarrow \pmatrix{a_{n+k} \\ b_{n+k}}$ you need to apply the matrix $k$ times. However, this is - as it stands now - quite involved. But now remember that if you're dealing with a diagonal matrix, then multiplying together $k$ of those diagonal matrices is the same as raising it to the $k$th power. 
That should get you going. If not, let me know. 
EDIT: If you have a diagonal matrix $A$, applying it to some vector $v$ is the same as just multiplying the vector by some scalar $\lambda$ (one for each eigenvalue), like this: $Av=\lambda v$. So if you do it a second time you get $AAv=\lambda^2v$. This is the reason why if we want to apply a matrix $k$ times, we can just raise it to the power of $k$, so $AAA...k$ times $=A^k$, which means that the individual entries in the matrix are raised to the $k$th power. 
