Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

I got my questions answered at in under 10 minutes. Go to 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


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

[ 4 0 1 -1 -6 -2 5 0 0 ] \[\lambda ^3+2 lambda ^2 - 29 lambda -30]/
\[\textbf Av=\lambda v\] \[\left(\textbf A-\lambda \textbf I\right)v =0\] \[\left|\textbf A-\lambda \textbf I\right|v =0\] \[\left|\left(\begin{array}\ 4&0&1\\-1&-6&-2\\5&0&0\end{array}\right)-\left(\begin{array}\ \lambda&0&0\\0&\lambda & 0\\0 &0 &\lambda \end{array}\right)\right|=0\] \[\left|\left(\begin{array}\ 4-\lambda&0&1\\-1&-6-\lambda&-2\\5&0&0-\lambda\end{array}\right)\right|=0\] \[(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]-0[-1\times(0-\lambda)]+1[-1\times0-(-2\times5)]=0\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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)_)
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
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
but how do you know to break it into a b c?
how to factorize? you could try polynomial long division
\[\lambda^3 +2\lambda^2-29\lambda-30\]
how would you go about doing this? possibilities: +/- : 1,2,3,5,6,10,30,15
if you plug in -1, you get =0.then what?
your expression does factorizes nicely, try dividing by \((\lambda+1)\)
i dont know how... i knwo that sounds dumb but how do u do that?
so\[\lambda^3 +2\lambda^2-29\lambda-30=(\lambda+1)(\lambda^2+\lambda-30)\]
now just factorize the quadratic factor...
thanks. also: do you know about diagnolization?

Not the answer you are looking for?

Search for more explanations.

Ask your own question