When is the system of first order differential equations asymptotically stable? 
Given the system $$\mathbf{x}'=A\mathbf{x} $$
where $A=\begin{bmatrix}a &0 & 4 \\ -1 & -1 & 0 \\ -2-a & 0 & -3 \end{bmatrix}$
In what interval of $a$ is the system asymptotically stable, and for what value of $a$ is the system stable/unstable if such a case even exists?

Attempt
For a system to be asymptotically stable the real part of all eigenvalues must be negative. $$\Re(\lambda)<0, \: \: \: \text{for all} \: \: \lambda $$
Since we are dealing with a $3\times3$ matrix, I used maple to find the eigenvalues from the system matrix. I also tried to use Maple to solve the inequality case for the eigenvalues, but I don't think I'm getting correct results.
The results I get from Maple state that the system is asymptotically stable when $$-8 < a \leq5-4\sqrt{3} $$
I spoke with my peers, and they said that this interval of $a$ is wrong. I posted my Maple document below. Can anyone see what is going wrong?

 A: I agree with your peers. Simply because the range indicated means $a$ must be smaller than 0, while 0 is a feasible solution (roots at $-\frac{3}{2} \pm \frac{1}{2}\sqrt{23}i$). Since only the real part is indicating stability, it is important to figure out when the square root becomes imaginary:
$$a^2-10a-23=0$$
$$(a-5)^2=48$$
$$a = 5\pm 4\sqrt{3}$$
Knowing these roots, you can deduct that this is less than zero if:
$$5-4\sqrt{3}<a<5+4\sqrt{3}$$
Within this range, the stability is entirely relying on $ \frac{1}{2}a-\frac{3}{2}$. from here it can be seen that the upperbound of $a$ should be: $a<3$.
The lower bound is a bit tricky as it exceeds the range on which the root is imaginary, but luckily that has been already given by your calculator: $a>-8$, which is the value on which the second eigenvalue will yield 0. So the actual range of $a$ is:
$$-8<a<3$$
Why maple failed to give this answer might have something to do with the imaginary part. the given range is where the answer does not have an imaginary part and is strictly less than 0. I am not really familiar with Maple, but maybe make it try to solve real(your equation < 0){a}.
A: Below I use Maple 2020.1 (the current version).
restart;

A := <a,-1,-2-a|0,-1,0|4,0,-3>:

elist := convert(LinearAlgebra:-Eigenvalues(A),list):

map(lprint, elist):
   -1
   1/2*a-3/2+1/2*(a^2-10*a-23)^(1/2)
   1/2*a-3/2-1/2*(a^2-10*a-23)^(1/2)

W := map(L->evalc(Re(L)<0),elist):

map(lprint, W):
   -1 < 0
   1/2*a+1/4*abs(a^2-10*a-23)^(1/2)*(1+signum(a^2-10*a-23)) < 3/2
   1/2*a-1/4*abs(a^2-10*a-23)^(1/2)*(1+signum(a^2-10*a-23)) < 3/2

# solving for all three conditions together
solve(W,a);

        {-8 < a, a < 3}

plot(map(Re,elist),a=-9..4,thickness=1,
     view=-2.75..1/4,size=[500,200]);


# for fun (only), solve each separately

solve(W[1],a); # true for all `a`

                a

solve(W[2],a);

      RealRange(Open(-8), Open(3))

solve(W[3],a);

      RealRange(-infinity, Open(3))

