Is the Jacobian Matrix for an Equilibrium Point Unstable?

[Somefantastik]

Hello,

I have a Jacobian matrix

$$A = \left( \begin{array}{cc} -2x+1-a_{1}y & -a_{1}x \\ -a_{2}ry & r-2ry-ra_{2}x \end{array} \right)\$$

where

$$a_{1} > 1 \ ; \ a_{2} > 1 \ ; \ and \ r>0$$

the trace is

$$Tr(A) = 1-2x-a_{1}y+r(1-y-a_{2}x)-ry = x(-2-ra_{2}) +y(-a_{1}-2r)+1+r$$;

and the determinant is

$$det(A) = r(1-2y-a_{2}x-2x+4yx+2a_{2}x^{2}-a_{1}y+2y^{2}a_{1})$$;

I need to show that Tr(A)>0 or det(A)<0. Neither one of them seem to be working out for me. Can anyone point me in the right direction? The point of this exercise is showing an equilibrium point is unstable.

Update: I can see from MATLAB that the determinant must be zero as the phase portrait shows a saddle point.

Please help me show det(A) < 0.

 

[gtoguy97]

To do this, you need to rearrange the equation for det(A) so that it is in a form such that all the terms are negative. This will require factoring and then multiplying the factors together. Once you have done this, it should be easy to see that det(A) will always be negative. Alternatively, you could use MATLAB or another computer program to plot the phase portrait of the system and determine that there is a saddle point, which implies that det(A) must be zero.