= Solution
Abbreviate $A=\alpha_{12}$, $B=\alpha_{21}$. The coordinate factors immediately give the three boundary <equilibria> $S_1,S_2,S_3$. The <Jacobian matrix> is
$$
J(u)=\begin{pmatrix}1-2u_1-Au_2&-Au_1\\-\rho Bu_2&\rho(\gamma-2u_2-Bu_1)\end{pmatrix}.
$$
At $S_1$, the <eigenvalues> are $1,\rho\gamma>0$, so it is a repelling <equilibrium>. At the other two points, triangularity gives
$$
\boxed{\begin{aligned}S_2:&\quad-1,\ \rho(\gamma-B),\\S_3:&\quad1-A\gamma,\ -\rho\gamma.\end{aligned}}
$$
Therefore $S_2$ is asymptotically stable for $\gamma<B$ and a <saddle point> for $\gamma>B$; $S_3$ is asymptotically stable for $\gamma>1/A$ and a <saddle point> for $\gamma<1/A$. Equality gives a zero <eigenvalue>, so linearization alone is inconclusive there.
For later use, the coexistence point is physically positive exactly when $B<\gamma<1/A$ if $AB<1$, or $1/A<\gamma<B$ if $AB>1$. Its stability is as supplied in the question. If $AB=1$ the displayed formula for $S_4$ does not apply: when also $\gamma=B=1/A$, the coincident non-axis <nullclines> give a whole segment of <equilibria>.
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