You are correct for the spectral radius:
$$\rho (A) = max |\lambda|,$$
where $\lambda$ is an eigenvalue of $A$.
To write the Jacobi iteration, we solve each equation in the system as:
$E1: x_1 = -2x_2 + 1$
$E2: x_2 = -3x_1 + 0$
This is typically written as, $Ax = (D - L - U)x = b$,
where $D$ is the diagonal, $-L$ is the lower triangular and $-U$ is the upper triangular. Solving this system results in:
$x = D^{-1}(L + U)x + D^{-1}b$ and the matrix form of the Jacobi iterative technique is:
$x_{k} = D^{-1}(L + U)x_{k-1} + D^{-1}b, k = 1, 2, \ldots$
Writing these out, gives:
$$A = \begin{pmatrix} 1&2 \\ 3&1\end{pmatrix} = D - L - U = \begin{pmatrix} 1&0 \\ 1&0\end{pmatrix} - \begin{pmatrix} 0&0 \\ -3&0\end{pmatrix} -\begin{pmatrix} 0&-2 \\ 0&0\end{pmatrix}.$$
This results in an iteration formula of (compare this to what I started with with $E1$ and $E2$ above):
$$x_{k} = D^{-1}(L + U)x_{k-1} + D^{-1}b = \begin{pmatrix} 0&-2 \\ -3&0\end{pmatrix}x_{k-1} + \begin{pmatrix} 1 \\ 0\end{pmatrix}$$
This can also be written in a component-wise form.
We know the solution here is $\displaystyle x = (-\frac{1}{5}, \frac{3}{5})$, but no initial $x_{0}$ choice will give convergence here because $A$ is not diagonally dominant (it is easy to manually crank tables for different starting $x_0's$ and see what happens).
Note: See the nice comment below from Elmar Zander, which is an oversight on my part! Thanks Elmar!
Regards