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

elica85Best ResponseYou've already chosen the best response.0
i 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?
 2 years ago

myininayaBest 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
 2 years ago

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

elica85Best ResponseYou've already chosen the best response.0
i thought it's lambdaIA, not AlambdaI
 2 years ago

elica85Best ResponseYou've already chosen the best response.0
idk 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
 2 years ago

elica85Best ResponseYou've already chosen the best response.0
o if i just have to find x, i should be able to...
 2 years ago

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

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