Given the system x ′ = A xwhere A = [ a 0 4 − 1 − 1 0 − 2 − a 0 − 3 ] 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?AttemptFor a system to be asymptotically stable the real part of all eigenvalues must be negative. ℜ ( λ ) &lt; 0 , for all λSince 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 &lt; a ≤ 5 − 4 3 I spoke with my peers, and they said that this interval of a is wrong. Can anyone see what is going wrong?

Question

Given the system        x    ′    =  A  xwhere  A  =      [                            a                          0                          4                                      −          1                          −          1                          0                                      −          2          −          a                          0                          −          3                      ]  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?AttemptFor a system to be asymptotically stable the real part of all eigenvalues must be negative.  ℜ  (  λ  )  &amp;lt;  0  ,        for all      λSince 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&#039;t think I&#039;m getting correct results.The results I get from Maple state that the system is asymptotically stable when  −  8  &amp;lt;  a  ≤  5  −  4      3  I spoke with my peers, and they said that this interval of a is wrong. Can anyone see what is going wrong?

Haylie Cherry · Accepted Answer

I agree with you. Simply because the range indicated means a must be smaller than 0, while 0 is a feasible solution (roots at   −      3    2    ±      1    2        23    i). Since only the real part is indicating stability, it is important to figure out when the square root becomes imaginary:      a    2    −  10  a  −  23  =  0  (  a  −  5      )    2    =  48  a  =  5  ±  4      3  Knowing these roots, you can deduct that this is less than zero if:  5  −  4      3    &amp;lt;  a  &amp;lt;  5  +  4      3  Within this range, the stability is entirely relying on       1    2    a  −      3    2  . from here it can be seen that the upperbound of   a should be:   a  &amp;lt;  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  &amp;gt;  −  8, which is the value on which the second eigenvalue will yield 0. So the actual range of   a is:  −  8  &amp;lt;  a  &amp;lt;  3

Given the system <mi mathvariant="bold">x ′ </msup> = A <mi mathvarian

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