Question

Find the general solution of:\[\Large \vec x\;'\quad=\quad A\;\vec x\]Where,\[\Large A\quad=\quad \left[\begin{matrix}-3 & 4 \\ -2 & 1\end{matrix}\right]\]
So looking for eigenvalues:\[\Large \det(A-\lambda I)\quad=\quad \det \left[\begin{matrix}-3-\lambda & 4 \\ -2 & 1-\lambda\end{matrix}\right]\quad=\quad0\]

Giving us ummmm,\[\Large (-3-\lambda)(1-\lambda)+8\quad=\quad 0\]Simplifies to (maybe someone can check my work here, this seems to be the spot I make the most mistakes in):\[\Large \lambda^2+2\lambda+5\quad=\quad 0\]

Answer 1

Which gives us complex e-values:\[\Large \lambda\quad=\quad -1\pm2i\]

Answer 2

Whichhhh doesn't seem to be working out very nicely when I try to find the e-vectors :b
So maybe I made a mistake earlier somewhere...

For \(\Large \lambda\quad=\quad-1-2i\),

\[\Large \left[\begin{matrix}-2+2i & 4 \\ -2 & 2i\end{matrix}\right]\vec v\quad=\quad \vec 0\]

Answer 3

This is for ODE's, yes?

Answer 4

yes ODE2 :o

Answer 5

Hmm....I don't remember eigenvectors, but it's supposed to mean something.

Answer 6

lol dat abs :3 i think it's time you updated to an even prettier picture, like Ron Jeremy or something

Answer 7

Or Steve Buscemi!

Answer 8

I think we should focus our attention on the problem rather than my aesthetics :)

Answer 9

Anyblaze it 420, according to definition: \(\sf \color{red}{Ax=cx,~x}\) is nonzero vector,then you have \(\sf \color{}{Ax=0}\) which means it has nonzero solutions, and also means A is singlear. and that the eigenvector is not uniquely determined.

Answer 10

you have formula to find it out, right? just apply as if it is real eigenvalue,

Answer 11

complex number ? so what? no different

Answer 12

Oh I did my matrix incorrectly -_- grr
So it should be this... which is... maybe solvable now :p hmm
\[\Large \left[\begin{matrix}-2+2i & 4 \\ -2 & \color{teal}{2+2i}\end{matrix}\right]\vec v\quad=\quad \vec 0\]

Answer 13

hihihi.... I am dizzy in switching between your 2 posts. I am not back there again