$$\begin{align}
a_n &= \frac 12 \,a_{n-1} + \frac 12\, a_{n - 2}\\
a_{n-1} &= 1\, a_{n-1} + 0\, a_{n-2}
\end{align}$$
$$\begin{bmatrix}a_n \\ a_{n - 1}\end{bmatrix} = \begin{bmatrix} \frac 12 & \frac 12 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} a_{n-1} \\ a_{n - 2} \end{bmatrix}$$
$$\begin{bmatrix}a_{n + 1} \\ a_{n}\end{bmatrix} = \begin{bmatrix} \frac 12 & \frac 12 \\ 1 & 0 \end{bmatrix}^n\begin{bmatrix} a_{1} \\ a_{0} \end{bmatrix}$$
Now notice that your matrix $M = \begin{bmatrix} \frac 12 & \frac 12 \\ 1 & 0 \end{bmatrix}$ is a stochastic matrix. That is a very good thing. You could close this formula by finding eigenvalues and eigenvectors; but to find the steady state, you can actually skip that step with a stochastic matrix.
To find the steady state of your matrix $S = M^\infty$, which is where successive applications of $M$ don't change the state, which is $SM = S$, it comes down to solving $S(M-I) = 0$ which is $S = \text{nullspace}(M - I)$. Doing so you find:
$$\begin{bmatrix} \frac 12 & \frac 12 \\ 1 & 0 \end{bmatrix}^\infty = \begin{bmatrix} \frac 23 & \frac 13 \\ \frac 23 & \frac 13 \end{bmatrix}$$
So altogether:
$$\begin{bmatrix}a_{\infty + 1} \\ a_{\infty}\end{bmatrix} = \begin{bmatrix} \frac 12 & \frac 12 \\ 1 & 0 \end{bmatrix}^\infty\begin{bmatrix} a_{1} \\ a_{0} \end{bmatrix} = \begin{bmatrix} \frac 23 & \frac 13 \\ \frac 23 & \frac 13 \end{bmatrix}\begin{bmatrix}3000 \\ 1000 \end{bmatrix} = \begin{bmatrix}\frac{7000}{3} \\ \frac{7000}{3}\end{bmatrix}$$