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ureji1c8r1 2022-05-11 Answered
Given the system
x = A x
where
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?
Attempt
For a system to be asymptotically stable the real part of all eigenvalues must be negative.
( λ ) < 0 , for all λ
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 < 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?
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Answers (1)

Haylie Cherry
Answered 2022-05-12 Author has 18 answers
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 < a < 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 < 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
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Background
It is relatively easier to find a solution to a system of linear equations in the form of A v = b given the matrix A. But what systematic ways are there that allows us to obtain a matrix given a equation?
For example, consider the following equations with all terms existing in R
[ a b c d e f g h i ] [ 2 3 4 ] = [ 1 1 1 ]
Although it is easy to see that a = 1 2 , e = 1 3 , i = 1 4 with all other terms being 0 is a viable solution, I am curious if there is a more systematic way of finding a matrix that satisfies a equation. Even more importantly, how should these methods be adapted when there are added constraints on the properties of the matrix? For example, if we require that the matrix of interest should be invertible, or of rank = k?
Why I am interested in such question
Consider the vector space P 2 ( R ) , the problem of finding a basis β such that [ x 2 + x + 1 ] β = ( 2 , 3 , 4 ) T can be reduced to a problem that has been stated above.

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