= Solution
{c}
For the <linear three-reservoir carbon model>, with concentrations as a column <vector>, the linear <ordinary differential equation> is
$$
\boxed{\dot{\mathbf x}=J\mathbf x+\mathbf f,\qquad
J=\begin{pmatrix}
-r/R_a&r/R_a&0\\
r/R_b&-(r+s)/R_b&s/R_b\\
0&s/R_d&-s/R_d
\end{pmatrix},\qquad
\mathbf f=\begin{pmatrix}E/R_a\\0\\0\end{pmatrix}.}
$$
A homogeneous <normal mode> has $\mathbf x=\mathbf v e^{\lambda t}$, where $J\mathbf v=\lambda\mathbf v$. For negative real <eigenvalues> the e-folding relaxation time is $\tau=-1/\lambda$. The row sums vanish, so $\mathbf v_3=(1,1,1)^T$ has $\lambda_3=0$ and
$$
\boxed{\tau_3=\infty.}
$$
Equal effective concentrations generate no exchange. Summing the reservoir budgets gives
$$
\frac{d}{dt}(R_aC_a+R_bC_b+R_dC_d)=E.
$$
Thus injected <carbon dioxide> is conserved after emissions cease: the model redistributes it but contains no permanent removal. The zero <normal mode> records this conserved excess <mass>, rather than a long but finite decay.
To check the other <normal modes> are damped, set $D=\operatorname{diag}(R_a,R_b,R_d)$ and $L=-DJ$. Then
$$
\mathbf z^TL\mathbf z=r(z_a-z_b)^2+s(z_b-z_d)^2\geq0.
$$
The <similarity transformation> $D^{1/2}JD^{-1/2}=-D^{-1/2}LD^{-1/2}$ is a symmetric negative-semidefinite <matrix> with a one-dimensional <kernel>, so the remaining two <eigenvalues> are strictly negative.
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