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Consider the following matrix:

m3 = {{-\[Beta]1 i2 - \[Beta]2 j - \[Beta]3 a - \[Mu], 
     0, -\[Beta]1 s, -\[Beta]2 s, -\[Beta]3 s}, {p \[Beta]1  i2 + 
      p \[Beta]2 j + p \[Beta]3 a, -b1, p \[Beta]1 s, 
     p \[Beta]2 s + \[Xi]1, 
     p \[Beta]3 s}, {(1 - p) \[Beta]1 i2 + (1 - p) \[Beta]2 j + (1 - 
         p) \[Beta]3 a, \[Epsilon], (1 - p) \[Beta]1 s - 
      b2, (1 - p) \[Beta]2 s + \[Xi]2, (1 - p) \[Beta]3 s}, {0, 0, 
     p1, -b3, 0}, {0, 0, 0, p2, -b4}} /. {s -> \[Nu]/\[Mu], i1 -> 0, 
    i2 -> 0, j -> 0, a -> 0};

Generates the characteristic polynomial:

-(1/(\[Mu]^2))(\[Lambda] + \[Mu]) (p1 p2 (-p \[Beta]3 \[Epsilon] \
\[Mu] \[Nu] + (-1 + 
          p) \[Beta]3 (b1 + \[Lambda]) \[Mu] \[Nu]) + (-b4 - \
\[Lambda]) ((-b3 - \[Lambda]) (-p \[Beta]1 \[Epsilon] \[Mu] \[Nu] - \
(b1 + \[Lambda]) \[Mu] (-b2 \[Mu] - \[Lambda] \[Mu] + \[Beta]1 \[Nu] -
              p \[Beta]1 \[Nu])) - 
       p1 (-\[Epsilon] \[Mu] (p \[Beta]2 \[Nu] + \[Mu] \[Xi]1) - (b1 \
+ \[Lambda]) \[Mu] (\[Beta]2 \[Nu] - 
             p \[Beta]2 \[Nu] + \[Mu] \[Xi]2))))

We see immediately that one solution is $\lambda=-\mu $. We are left with a quartic, but I don't know how to proceed. Maybe some manipulations of the remaining characteristic polynomial will help using $R_0$:

ss = FullSimplify[
   Together[
    Expand[(-b1 b2 b3 b4 + b4 p1 \[Epsilon] \[Xi]1 + 
      b1 b4 p1 \[Xi]2)/(-b1 b3 b4 \[Beta]1 + b1 b3 b4 p \[Beta]1 - 
      b1 b4 p1 \[Beta]2 + b1 b4 p1 q \[Beta]2 - b1 p1 p2 \[Beta]3 + 
      b1 p1 p2 r \[Beta]3 - b3 b4 p \[Beta]1 \[Epsilon] - 
      b4 p1 q \[Beta]2 \[Epsilon] - p1 p2 r \[Beta]3 \[Epsilon])]]];

r0 = (\[Nu]/\[Mu]) (1/ss);

EDIT:

as per the comments and using Mathematica, we solved it but Mathematica gives solutions but surely not in its simplest form. Is there a way to make them simpler?

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  • $\begingroup$ What's wrong with SolveValues[poly == 0, λ] ? It has all the solutions. $\endgroup$
    – flinty
    Jan 12, 2022 at 14:20
  • $\begingroup$ @flinty Ok is there a way to simplify the $\lambda$'s? Or even better, is there a way to simplify the characteristic polynomial in a multiplicative way? $\endgroup$
    – Math
    Jan 12, 2022 at 14:30
  • $\begingroup$ If you have the roots (in a list), you can write the the characteristic polynomial as product (up to a multiplicative factor) by Times@@((1-#)&/@roots) $\endgroup$ Jan 12, 2022 at 14:47
  • $\begingroup$ Eigenvalues[m3, Quartics -> True] shows explicit forms for these quartic roots. $\endgroup$
    – Roman
    Jan 12, 2022 at 14:48
  • $\begingroup$ What I mean is, is there a way to simplify these roots and/or the CP using $R_0$? $\endgroup$
    – Math
    Jan 12, 2022 at 15:30

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