= Solution
Substitute $\mathbf y=\mathbf v e^{-\gamma t}$ into the unforced system. It gives $(K-\gamma I)\mathbf v=0$, so a nonzero single-exponential solution must use one of the eigendirections in part (a):
$$
\boxed{\begin{aligned}
\mathbf y_1(t)&=A(1,1,0)^Te^{-t},\\
\mathbf y_2(t)&=B(1,-1,0)^Te^{-2t},\\
\mathbf y_3(t)&=C(0,0,1)^Te^{-4t}.
\end{aligned}}
$$
These are all nonzero solutions having one common decay constant. Their arbitrary <linear combination> is the general unforced solution; all decay because the three <eigenvalues> are positive.
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