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exraven
 2 years ago
Best ResponseYou've already chosen the best response.0\[\left( A\lambda I \right)x=0\]lamda is the eigenvalue, x is the eigenvector

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0i was able to find the eigenvalue \[\left[\begin{matrix}6 & 1 \\ 2 & 3\end{matrix}\right]\] to be \[\lambda _{1}=4, \lambda _{2}=5\] now what's the next step?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2@nightshade please don't post offtopic answers. @exraven and elica85 we want x to not be the zero vector also

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2Did you try to plug into exraven's formula to find the eigenvectors

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0i thought it's lambdaIA, not AlambdaI

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0idk what i'm doing, my school doesn't require us to take linear algebra to take diff eq so i'm suppose to know what to do but i don't

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0o if i just have to find x, i should be able to...

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2\[(A\lambda_n \cdot I)x=0\] For \[\lambda_1=4\] dw:1333932816786:dw

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2For \[\lambda_2=5\] dw:1333932896865:dw
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