Question

solve \[\left(\begin{matrix}x_1 \\ x_2\\x_3\\x_4\end{matrix}\right)'=\left[\begin{matrix}0 & -1 & 0 & 0\\ -2 & -2 & 1 & 0\\ 0 & 0 & 0& 1\\1&0&-2&-2\end{matrix}\right]\left(\begin{matrix}x_1 \\ x_2\\x_3\\x_4\end{matrix}\right)\]

Answer 1

@satellite73 can you help?

Answer 2

@terenzreignz can you help?

Answer 3

That prime symbol means transpose?

Answer 4

no it means its derivative... it's an ODE problem

Answer 5

Okay... never seen these before D:

mercy T.T

I'll still look into it though, try to enlighten myself ^_^

Answer 6

ok thanks. i found the eigenvalues as such:\[\lambda=(-1+\sqrt{2}i),(-1-\sqrt{2}i),(-1),(-1)\]I don't know how to find the eigenvectors

Answer 7

I'm allergic to matrices -_-

But finding eigenvectors for a specific eigenvalue seems simple enough, in the case where the eigenvalue is -1, just look for the vectors which satisfy
\[\left[\begin{matrix}0 & -1 & 0 & 0\\ -2 & -2 & 1 & 0\\ 0 & 0 & 0& 1\\1&0&-2&-2\end{matrix}\right]\left(\begin{matrix}x_1 \\ x_2\\x_3\\x_4\end{matrix}\right)=-\left(\begin{matrix}x_1 \\ x_2\\x_3\\x_4\end{matrix}\right)\]

Answer 8

i don't have a clue how to do the 4x4 at all :(