Question

I need help with math stuffs :c

I was able to find the eigenvalues of this matrix, I can't seem to figure out the eigenvectors.

Maybe someone can refresh my memory or provide a good link to some info.

Answer 1

\[\Large A\quad=\quad \left[\begin{matrix}2 & 1 \\ -3 & 6\end{matrix}\right]\]

Answer 2

I came up with these eigenvalues:\[\Large \lambda_1 \quad=\quad 4\qquad\qquad \lambda_2\quad=\quad4\]
Which I ... think is correct. :o

Answer 3

For the eigenvectors I have this:\[\Large \left[\begin{matrix}-2 & 1 \\ -3 & 2\end{matrix}\right]\vec v\quad=\quad \vec 0\]

Answer 4

I can't seem to figure this one out :c grr

Answer 5

So it works out to a system of equations,\[\Large \bf -2v_1+v_2\quad=\quad0\]\[\Large\bf -3v_1+2v_2\quad=\quad0\]And I need to find a vector:\[\Large \vec v\quad=\quad \left(\begin{matrix}\bf{v_1} \\ \bf{v_2}\end{matrix}\right)\]That satisfies the system, right?
Something like that?? +_+
grrr maf so hard!

Answer 6

Oh goodness ty loser :C
-3 times 1 is NOT 4.. man im such a doofus....

Answer 7

:)

Answer 8

\[\Large (2-\lambda)(6-\lambda)+3=0\]\[\Large \lambda^2-8\lambda+15=0\]\[\Large (\lambda-3)(\lambda-5)=0\]So something like that? +_+

Answer 9

hihihi... I think so.

Answer 10

For \(\Large \lambda=3\):\[\Large \left[\begin{matrix}-1 & 1 \\ -3 & 3\end{matrix}\right]\vec v\quad=\quad \vec 0\]Bahhh ok that would make a lot more sense D:::

Answer 11

you are the best

Answer 12

Giving us something like*\[\Large \vec v_1\quad=\quad \left(\begin{matrix}1 \\ 1\end{matrix}\right)\]

Answer 13

Ok ok ok I think I've got it from here \c:/ yayyy loser! thanks!