Finding Stability Given a Simple Characteristic Equation-FE/EIT Review

[@mikeb3789]

correct me if i'm wrong but we can determine the system is stable due to no variable K in the characteristic equation

[@TheUltimate1]

I am not sure if that is viable way to determine stability. I have linked another excellent video that explains:kzbin.info/www/bejne/jXOmi3WFd5h_eaM

[@e734127803]

sorry but you had a mistake when you made the second matrix marked with green it is (8*2-6*0)/2 while it should be only the first with third raw (4*2-6*0)/2=2 1 2 4 0 but the answer is still the same :)

[@RaiyaAcademy]

Hamad, yeah we're both right. Thanks for the check.

[@e734127803]

hahahah It is Emad Good luck :)

[@chenxiaomaoify]

Sorry I didn't get it. I have the same answer as Raiya's. Can you put you full matrix here?Thanks.

[@e734127803]

yes ,u will get the same answer but the correct approach is different u will use: 1 2 4 0 for the second product and not: 6 2 8 0 try it for q(s)=[2 5 6 4 3] the two approaches will have a different answer lu chen

[@user-is4qu9re3l]

31-8/4=6 Genius 😎 By the way Mam Your Voice❤️😌

[@HassanAQ-y2t]

it will be stable because u avoid to multiply the big number to a negative u always put the negative to the small value!

[@furkanyasar9186]

you are wrong 4*2- 1*0/4 must be

[@AhmedAli-ti7vc]

8*2 can never be 16.

[@englishwithvivek3602]

do a better practice on multiplication.....