anonymous
  • anonymous
Find all the eigenvalues for the given matrices. if it is a 3x3. and a polynomial. how do i know what to break it down into?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
[ 4 0 1 -1 -6 -2 5 0 0 ] \[\lambda ^3+2 lambda ^2 - 29 lambda -30]/
UnkleRhaukus
  • UnkleRhaukus
\[\textbf Av=\lambda v\] \[\left(\textbf A-\lambda \textbf I\right)v =0\] \[\left|\textbf A-\lambda \textbf I\right|v =0\] \[\left|\left(\begin{array}\ 4&0&1\\-1&-6&-2\\5&0&0\end{array}\right)-\left(\begin{array}\ \lambda&0&0\\0&\lambda & 0\\0 &0 &\lambda \end{array}\right)\right|=0\] \[\left|\left(\begin{array}\ 4-\lambda&0&1\\-1&-6-\lambda&-2\\5&0&0-\lambda\end{array}\right)\right|=0\] \[(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]-0[-1\times(0-\lambda)]+1[-1\times0-(-2\times5)]=0\]
UnkleRhaukus
  • UnkleRhaukus
*\[(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]\]\[\qquad-0[-1\times(0-\lambda)-(5\times-2)]\]\[\qquad\qquad+1[-1\times0-(-6-\lambda)]=0\]

Looking for something else?

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

More answers

anonymous
  • anonymous
right, i know how to do all the calculations but when you get the the last step: when its all combined how do you know what lamda itself is equal to? ( like, you look at the constant and see its factors or something and do something..idk)_)
UnkleRhaukus
  • UnkleRhaukus
well at the last spet you'll have a bunch of factors =0 so to get zero at least one of the factors is zero
UnkleRhaukus
  • UnkleRhaukus
something like this only i have made mistakes \[(\lambda-4)[(6+\lambda)\lambda]+[6+\lambda]=0\]\[(6\lambda+\lambda^2-24-4\lambda)\lambda+6+\lambda=0\]\[\lambda^3+2\lambda^2-24\lambda+\lambda+6=0\]\[\lambda^3+3\lambda^2-24\lambda+6=0\]\[(\lambda-a)(\lambda-b)(\lambda-c)=0\] lambda would be a,b,c
anonymous
  • anonymous
but how do you know to break it into a b c?
UnkleRhaukus
  • UnkleRhaukus
how to factorize? you could try polynomial long division
anonymous
  • anonymous
ugh
anonymous
  • anonymous
\[\lambda^3 +2\lambda^2-29\lambda-30\]
anonymous
  • anonymous
how would you go about doing this? possibilities: +/- : 1,2,3,5,6,10,30,15
anonymous
  • anonymous
if you plug in -1, you get =0.then what?
UnkleRhaukus
  • UnkleRhaukus
your expression does factorizes nicely, try dividing by \((\lambda+1)\)
anonymous
  • anonymous
i dont know how... i knwo that sounds dumb but how do u do that?
UnkleRhaukus
  • UnkleRhaukus
|dw:1341885937030:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1341885983310:dw|
UnkleRhaukus
  • UnkleRhaukus
so\[\lambda^3 +2\lambda^2-29\lambda-30=(\lambda+1)(\lambda^2+\lambda-30)\]
UnkleRhaukus
  • UnkleRhaukus
now just factorize the quadratic factor...
anonymous
  • anonymous
thanks. also: do you know about diagnolization?

Looking for something else?

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