Question

Find the general solution of:\[\Large \vec {x}\;'\quad=\quad A\;\vec{x}\]
Where,\[\Large A\quad=\quad \left[\begin{matrix}0 & 1 & 0\\ 4 & 3 & -4\\ 1 & 2 & -1\end{matrix}\right]\]

So I found the eigenvalues:\[\Large \det(A-\lambda I)\quad=\det \left[\begin{matrix}-\lambda & 1 & 0 \\ 4 & 3-\lambda & -4\\ 1 & 2 & -1-\lambda\end{matrix}\right]\]
\[\Large -\lambda\left[(3-\lambda)(-1-\lambda)+8\right]-(1)\left[4(-1-\lambda)+4\right]=0\]Assuming I didn't make any mistakes ~ this simplifies to:\[\Large \lambda(\lambda-5)(\lambda+3)\quad=\quad0\]

Answer 1

For \(\Large \lambda=5\) I can't seem to find a vector :[
Maybe my brain isn't working correctly..\[\Large \left[\begin{matrix}-5 & 1 & 0\\ 4 & -2 & -4\\ 1 & 2 & -6\end{matrix}\right]\vec v\quad=\quad \vec 0\]

Answer 2

\(\Large \lambda(\lambda-1)^2=\quad0\)

Answer 3

Oh did I multiply that big mess incorrectly? :) Grrr lol
I better check my math :3

Answer 4

ive never worked before redundant roots ...

Answer 5

good luck !

Answer 6

Bahh thanks for the help :( yah I see what I did wrong there...
Grr I've messed up the multiplication on 3 of these 12 problems now.. so frustrating.

Answer 7

see if the answer helps to arrive at solution :)

http://www.wolframalpha.com/input/?i=eigen+vectors+%28%280%2C1%2C0%29%2C%284%2C3%2C-4%29%2C%281%2C2%2C-1%29%29

Answer 8

ya thanks guys c: seems to have worked out after fixing that up.

I have another one I'm stuck on :c i'll post it in a minute.
maybe one of you smarty pants can figure it out.

Answer 9

hey, friend, in what way , you find out characteristic equation for 3x3 matrix?

Answer 10

Umm I guess the Rule of Sarrus is kind of effective, but I prefer to use the other method, I can't remember what it's called.

Where you take the top row or column, and take each point and multiply it by the minor determinate 2x2 thing... and swap the sign between each one.. and blah blah blah..
I dunno XD I forget what it's called lol

Answer 11

that's why you always make mistake at it, there another simple way you don't have to take determinant of the cofactor matrices, why not? right?

Answer 12

hmm? 0_o

Answer 13

for 3x3 matrix, characteristic equation is constructed by :