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?
SolveValues[poly == 0, λ]
? It has all the solutions. $\endgroup$Times@@((1-#)&/@roots)
$\endgroup$Eigenvalues[m3, Quartics -> True]
shows explicit forms for these quartic roots. $\endgroup$