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OpenStudy (loser66):

find eigenvalue of \[\left(\begin{matrix}-2i&-4\\1&-2i\end{matrix}\right)\] Please, help

OpenStudy (anonymous):

Don't worry we will help

OpenStudy (loser66):

Thanks

OpenStudy (anonymous):

Is there any answer choices?

OpenStudy (loser66):

nope,

OpenStudy (anonymous):

Can you write it how you see it without using equation

OpenStudy (loser66):

ok, got what you mean. I have problem as \[x'=\left(\begin{matrix}-1&-4\\1&-1\end{matrix}\right)x\]

OpenStudy (anonymous):

Give me one second

OpenStudy (loser66):

oh, I got it now. I am sorry for bothering and thanks for response. XD

OpenStudy (anonymous):

Are you sure?

OpenStudy (loser66):

not exactly mine, but kind of, take a look at this http://www.youtube.com/watch?v=efpuBso8ZVw

OpenStudy (anonymous):

Also where was the question from?

OpenStudy (loser66):

nope, I am not sure just this part. I took linear last Spring and afraid of this part only. the rest is ok, I can handle it.

OpenStudy (anonymous):

Let me keep trying

OpenStudy (anonymous):

What did you get?

OpenStudy (loser66):

I got eigenvector \([2i,b]^T\)

OpenStudy (anonymous):

Hmm

OpenStudy (loser66):

what do you mean by "Hhm"?

OpenStudy (anonymous):

Just checking my answers

OpenStudy (loser66):

oh, I am sorry, it's \([2i, 1]\) not b, hihihi

OpenStudy (anonymous):

Correct

OpenStudy (anonymous):

Good job!

OpenStudy (loser66):

hehehe...thanks for being here with me.

OpenStudy (anonymous):

Yeah no problem!

OpenStudy (loser66):

hey, I have another question

OpenStudy (loser66):

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