 Solution

Let $B$ denote the zero-diagonal matrix with off-diagonal entries appearing in the iteration, and $G=-B/\alpha$ its iteration matrix. Its row sums are three, so $B1=31$. The other eigenvalues are $2\omega +{\omega }^2$ and $2{\omega }^2+\omega$, where $\omega =e^{2\pi i/3}$, namely $-3/2\pm i\sqrt{3}/2$, both of modulus $\sqrt{3}$. Thus the spectral radius is $\rho (G)=3/|\alpha |$.

For $|\alpha |>3$, the infinity-norm estimate $\Vert G{\Vert }_{\infty }=3/|\alpha |<1$ already gives a contraction, so the iterates converge from every initial vector to the unique solution of $(\alpha I+B)x=b$. Conversely, two iterates whose initial difference is a nonzero multiple of $1$ retain the difference $(-3/\alpha {)}^{k}1$. This fails to decay when $|\alpha |\le 3$, so convergence cannot be guaranteed for arbitrary starts. ThereforeAt $\alpha =3$ the offending mode oscillates; at $\alpha =-3$ it is stationary and the coefficient matrix is singular. A specially chosen starting vector can suppress a bad mode, but that is not guaranteed convergence.