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UnkleRhaukusBest ResponseYou've already chosen the best response.1
im good at eigen values
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
ohh finally.. ok i have a question.. i know how to solve.. just for oncae type of case i dont know how to solve.. i will post up the question [ 6 5 2 2 0 8 5 4 0 ] this is the matrix.
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.0
do you have your eigen values ?
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
If I am not getting it wrong, then I think we will do like this to find eigen values : A  IX = 0 Something like that... No knowledge... @UnkleRhaukus please help your friend (me)...
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Where is lambda?? Oh my God...
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
\[\textbf Ax=\lambda x\]\[(\textbf A\lambda \textbf I)x=0\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Oh yeah.... I forgot all the things... One day, I found a page thrown by someone on the road, I read that.. There on the page, it was shown how to find Eigen values but the question was not complete.. And I have forgot that too...
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
i have found lammida.. and its 2
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
there is repeated roots?
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
yes all 3 values of lamida are 2
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
I try to find eigen value now... You can carry on @UnkleRhaukus
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
\[\textbf A=\left[\begin{array}{ccc}6&5&2\\2&0&8\\5&4&0\end{array}\right]\]\[x=\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]\]\[\lambda=2\] \[\left[\begin{array}{ccc}6&5&2\\2&0&8\\5&4&0\end{array}\right]\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]=2\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
\[\left[\begin{array}{ccc}62&5&2\\2&02&8\\5&4&02\end{array}\right]\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{ccc} 0\\0\\0\end{array}\right]\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
you have three simultaneous equations
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
ok so i have to just solve simultaneously to get first eigen vector?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
something like that, i always go confused when it comes to eigenvectors
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
i think first eigen vector is x1= = 2 x2 = 2 x3 = 1
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
\[4x_1 + 5x_2 + 2x_3 = 0\] \[2x_1 2x_2  8x_3 = 0\] \[5x_1 + 4x_2  2x_3 = 0\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
@nubeer , that certainly solves the three equations, how did you get there?
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
i have found x1 and x2 in terms of x3 while using subsitution method.
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
ok thanks guys.. i think i have done it... thank you very much espacially @UnkleRhaukus :)
 one year ago
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