 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 isAt $S_1$, the eigenvalues are $1,\rho \gamma >0$, so it is a repelling equilibrium. At the other two points, triangularity givesTherefore $S_2$ 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.