nubeer
  • nubeer
Anyone good at Eigen Vectors.
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

UnkleRhaukus
  • UnkleRhaukus
im good at eigen values
nubeer
  • nubeer
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.
zzr0ck3r
  • zzr0ck3r
do you have your eigen values ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

nubeer
  • nubeer
yes i have .. its 2
anonymous
  • anonymous
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)...
anonymous
  • anonymous
Where is lambda?? Oh my God...
UnkleRhaukus
  • UnkleRhaukus
\[\textbf Ax=\lambda x\]\[(\textbf A-\lambda \textbf I)x=0\]
anonymous
  • anonymous
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...
nubeer
  • nubeer
i have found lammida.. and its 2
UnkleRhaukus
  • UnkleRhaukus
there is repeated roots?
nubeer
  • nubeer
yes all 3 values of lamida are 2
anonymous
  • anonymous
I try to find eigen value now... You can carry on @UnkleRhaukus
UnkleRhaukus
  • UnkleRhaukus
\[\textbf A=\left[\begin{array}{ccc}6&5&2\\2&0&-8\\5&4&0\end{array}\right]\]\[x=\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]\]\[\lambda=2\] \[\left[\begin{array}{ccc}6&5&2\\2&0&-8\\5&4&0\end{array}\right]\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]=2\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]\]
UnkleRhaukus
  • UnkleRhaukus
\[\left[\begin{array}{ccc}6-2&5&2\\2&0-2&-8\\5&4&0-2\end{array}\right]\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{ccc} 0\\0\\0\end{array}\right]\]
nubeer
  • nubeer
ok.. got this far...
UnkleRhaukus
  • UnkleRhaukus
you have three simultaneous equations
nubeer
  • nubeer
ok so i have to just solve simultaneously to get first eigen vector?
UnkleRhaukus
  • UnkleRhaukus
something like that, i always go confused when it comes to eigenvectors
nubeer
  • nubeer
i think first eigen vector is x1= = 2 x2 = -2 x3 = 1
anonymous
  • anonymous
\[4x_1 + 5x_2 + 2x_3 = 0\] \[2x_1 -2x_2 - 8x_3 = 0\] \[5x_1 + 4x_2 - 2x_3 = 0\]
UnkleRhaukus
  • UnkleRhaukus
@nubeer , that certainly solves the three equations, how did you get there?
nubeer
  • nubeer
i have found x1 and x2 in terms of x3 while using subsitution method.
nubeer
  • nubeer
ok thanks guys.. i think i have done it... thank you very much espacially @UnkleRhaukus :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.