anonymous
  • anonymous
Let A = [-3,2,6 0,-1,6 0,0,-3] Find an invertible matrix P and a diagonal matrix D such that D = P^{-1}AP. find P and D I worked out the polynomial -7x^3-7x^2-15x-9 to give x= -1,-3,-3 My answer for P= [1,0,0 1,-3,-3 0,1,1] and D= -1,0,0 0,-3,0 0,0,-3 something's wrong!
Linear Algebra
  • 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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what type of decomposition are you doing here ?
anonymous
  • anonymous
I think you get something wrong at P, since eigenvalue -3 has 2 dimensions and they are not yours! recheck
anonymous
  • anonymous
you should get D=dia(-3,-1,-3)

Looking for something else?

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

More answers

anonymous
  • anonymous
yes, surely, because they are eigenvalues of A, what I need is checking the order this guy arrange in P to have corresponding of numbers, like (-3,-1,-3) or (-1,-3,-3)
anonymous
  • anonymous
ok I will try!
anonymous
  • anonymous
We cannot guess, right, kid? if we mess the order up, although we know it is the Diagonal matrix of A but the P is in different order of eigenvectors, We will be crazy when checking what wrong is, I had that experience. It made me crazy
anonymous
  • anonymous
tr(A)=-3-1-3=-7 A11+A22+A33=3+9+3=15 |A|=-3(3)=-9 so, the characteristic polynomial would be\[\Delta(t)=t^3+7t^2+15t+9\]
anonymous
  • anonymous
well, as long as the eigen vector columns are arranged in proper order, you should be good.
anonymous
  • anonymous
still no luck...
anonymous
  • anonymous
no
anonymous
  • anonymous
eigen vectors=? for t=-1: \[-2x+2y+6z=0\\6z=0\\-3z=0\implies v_1=(1,1,0)^T\]
anonymous
  • anonymous
yup
anonymous
  • anonymous
and v2 and v3 I put at (0,-3,1)
anonymous
  • anonymous
for t= -3: \[ 2y+6z=0\\ 2y+6z=0\\ v_{2,3}=(\alpha,-3\beta,\beta)^T \]
anonymous
  • anonymous
@thdrbird you cannot have them two identical
anonymous
  • anonymous
oh,ok..
anonymous
  • anonymous
you can use v2 when a=0,b=1 v3 when a=1,b=1
anonymous
  • anonymous
or v3 when a=1,b=0 ANY combination of alpha and beta that do not give the same vectors..
anonymous
  • anonymous
can you give me an example?
anonymous
  • anonymous
got it!
anonymous
  • anonymous
I forgot to set alpha!
anonymous
  • anonymous
thanks @electrokid, great help!
anonymous
  • anonymous
@Hoa as well! Thanks!
anonymous
  • anonymous
I did nothing guy!!!

Looking for something else?

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