Solution
= Solution
At a <steady state>, $K\mathbf y_0=\mathbf f_0$. All <eigenvalues> of $K$ are nonzero, so it is invertible and
$$
\boxed{\mathbf y_0=K^{-1}\mathbf f_0,\qquad
K^{-1}=\frac14\begin{pmatrix}3&1&0\\1&3&0\\0&0&1\end{pmatrix}}.
$$
For the specified constant source this gives
$$
\boxed{\mathbf y_0=(1,1,1/4)^T}.
$$
Indeed the first two rows of $K\mathbf y_0$ each give one, and the third gives $4(1/4)=1$.