# Given the system <mi mathvariant="bold">x &#x2032; </msup> = A <mi mathvarian

Given the system
${\mathbf{x}}^{\prime }=A\mathbf{x}$
where
$A=\left[\begin{array}{ccc}a& 0& 4\\ -1& -1& 0\\ -2-a& 0& -3\end{array}\right]$
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.
$\mathrm{\Re }\left(\lambda \right)<0,\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\text{for all}\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\lambda$
Since we are dealing with a $3×3$ 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
I spoke with my peers, and they said that this interval of a is wrong. Can anyone see what is going wrong?
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Haylie Cherry
I agree with you. Simply because the range indicated means a must be smaller than 0, while 0 is a feasible solution (roots at $-\frac{3}{2}±\frac{1}{2}\sqrt{23}i$). Since only the real part is indicating stability, it is important to figure out when the square root becomes imaginary:
${a}^{2}-10a-23=0$
$\left(a-5{\right)}^{2}=48$
$a=5±4\sqrt{3}$
Knowing these roots, you can deduct that this is less than zero if:
$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