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I have the following Jacobian matrix for an equilibrium of an SIR model $$J=\left( \begin{array}{cccc} -\text{$\alpha $N} & 0 & \zeta & 0 \\ \text{$\alpha $N} & -\beta -\rho & 0 & 0 \\ 0 & \beta & -\zeta & 0 \\ 0 & \rho & 0 & 0 \\ \end{array} \right)$$

However, when using Mathematica to calculate $\det{(J- \lambda I)}$, I get the following

$$\lambda ^3 \left(\lambda -4 \left( \begin{array}{cccc} -\text{$\alpha $N} & 0 & \zeta & 0 \\ \text{$\alpha $N} & -\beta -\rho & 0 & 0 \\ 0 & \beta & -\zeta & 0 \\ 0 & \rho & 0 & 0 \\ \end{array} \right)\right)$$

What does this mean? Why is it giving me a matrix back in the determinant? I need to find the determinant in order to analyse the stability of an equilibrium.

Here is my code

$In[1]:= \text{J2}\text{:=}\left( \begin{array}{cccc} -\text{$\alpha $N} & 0 & \zeta & 0 \\ \text{$\alpha $N} & -\beta -\rho & 0 & 0 \\ 0 & \beta & -\zeta & 0 \\ 0 & \rho & 0 & 0 \\ \end{array} \right)$

$In[2]:= \text{FullSimplify}[\left| \text{J2}-\lambda \text{IdentityMatrix}[4]\right| ]$

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  • $\begingroup$ can you show us your code $\endgroup$ – Alex R. Nov 27 '14 at 23:04
  • $\begingroup$ @AlexR. - Edited the post with my code $\endgroup$ – user860374 Nov 27 '14 at 23:08
  • $\begingroup$ depending on the typset, try $\lambda$*IdentityMatrix[4], and also try Det[ ] instead of $||$. $\endgroup$ – Alex R. Nov 27 '14 at 23:11
  • $\begingroup$ @AlexR. I tried that as well, it gives the same :( $\endgroup$ – user860374 Nov 27 '14 at 23:11
  • $\begingroup$ What happens if you type your matrix J2 in terms as $\{\{-\alpha N, 0, \zeta, 0\},\{\alpha N, -\beta-\rho, 0 , 0\},\cdots\}\}$ $\endgroup$ – Alex R. Nov 27 '14 at 23:13
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What happens if you type your matrix J2 in terms as $\{\{-\alpha N, 0, \zeta, 0\},\{\alpha N, -\beta-\rho, 0 , 0\},\cdots\}\}$

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  • $\begingroup$ Thank you! :). It seems as if Mathematica is not too fond of Matrixform when calculating determinants :) $\endgroup$ – user860374 Nov 27 '14 at 23:19

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