= Solution
At a positive-rate state the <jump chain> moves to either other state with probability $1/2$, independently of the holding rate. At a zero-rate state there is no subsequent jump; extend the chain as permanently absorbed there, giving an identity row. This convention is needed to speak of all powers of its transition <matrix> in the absorbing cases.
When all rates are positive,
$$
\widehat P=\frac12(\mathbf1\mathbf1^T-I),\qquad
\widehat P^n=\frac13\mathbf1\mathbf1^T+
(-1/2)^n\left(I-\frac13\mathbf1\mathbf1^T\right).
$$
Its unique equilibrium is uniform and its powers tend to the <matrix> with all rows $(1/3,1/3,1/3)$. This differs from the holding-time-weighted equilibrium of the continuous-time chain unless the rates coincide.
With one zero-rate state the unique jump-chain equilibrium is its point mass and the limit has all rows that point mass. With two zero-rate states the equilibria are all mixtures on them, and the limiting rows are exactly the absorbing rows and half-half transient row found in part (d).
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