A community for students.
Here's the question you clicked on:
 0 viewing
KonradZuse
 2 years ago
Find the characteristic EQ of the following matrix.
KonradZuse
 2 years ago
Find the characteristic EQ of the following matrix.

This Question is Closed

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0@TuringTest pic to follow.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0[7*lambda8+(lambda5)*(lambda(lambda)1)] seems to be the answer, accordng to my book in the first question it gave the "characteristic polynomial" as the answer... So I wanted to do the same here.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1which problem are you doing?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1all of them in exercise 6?

phi
 2 years ago
Best ResponseYou've already chosen the best response.1I assume you know to subtract lambda from the diagonals, then find the determinant of that matrix?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0MY book just shows for a 2x2 matrix that it's lambda  # on the main diagonal, and everything else is negated.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0so I took the determinant for the 3x3, but my book wants the characteristic EQ, which is what I posted above, but the answer in the b ook showed the polynomial characteristic which was lambda^3 etc etc and then sifted out....

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1\[A=\left[\begin{matrix}5&0&1\\1&1&0\\7&1&0\end{matrix}\right]\]\[(\lambda IA)=0=\left\begin{matrix}\lambda5&0&1\\1&\lambda1&0\\7&1&\lambda\end{matrix}\right\]take the determinant and what do you get?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I got [7*lambda8+(lambda5)*(lambda(lambda)1)]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1I don't think you simplified correctly...

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Idk check my pic, that's what maple gave me anyways....

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0[(lambda5)*(lambda1)(lambda)+7*(0*0)(1)*(1)]+[(lambda1)*7+(lambda5)*0(0*(1))*lambda]

phi
 2 years ago
Best ResponseYou've already chosen the best response.1I have always done det(A  lambda*I)=0

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0That's what the book shows us.

phi
 2 years ago
Best ResponseYou've already chosen the best response.1It comes from \[ Ax =\lambda x\] or \[ Ax \lambda x=0\] \[ (A \lambda I) x=0\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1\[(\lambda5)(\lambda1)(\lambda)+(1)^3[7(\lambda1)(1)]=0\]what @phi said is true, you can do it either way

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah the book says that, then goes into normal determinants.e

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Why is it only [7(λ−1)(−1)]?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1the other parts have 0's so we can ignore them

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0and why do you only do it twice? I'm so confused. I thought the diagonal rule is 3 diags  3 diags.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I don't like Maple's answer >( Wolfram doesn't either :P.

phi
 2 years ago
Best ResponseYou've already chosen the best response.1I use cofactors (L5) * det (lower right 2x2) ignore the 0 1 * det(lower left 2x2)

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I hate co factors... From TT's example I get [7*lambda(1)+1+(lambda(lambda1))(lambda)] = 0

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0[((lambda5)(lambda1))(lambda)+(1)^3]+[7*(lambda1)(1)] = 0; [7 lambda(1) + 1 + lambda(lambda  1)(lambda)] = 0

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I just took what you did and entered it into Maple :P.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0oh I think I know what Id id :P.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0> [((lambda5)(lambda1))(lambda)+(1)^3]+[7*(lambda1)(1)] = 0; [7 lambda(1) + 1 + lambda(lambda  1)(lambda)] = 0

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I frogot to add some extra brackets :). Does this shizzle look better >(

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Same thing it looks like >(

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1\[(\lambda5)(\lambda1)(\lambda)+(1)^3[7(\lambda1)(1)]=0\]\[\lambda^36\lambda^2+5\lambda1+7\lambda7=0\]\[\lambda^36\lambda^2+12\lambda8=0\]I could have messed up, but I did it by eye...

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0TT never messes up :)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1hehe, that attitude has cost me plenty of credits, so I strongly suggest you doublecheck me, and everyone else for that matter, on everything. I appreciate the confidence though :)

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Normally maple gives me the good answer, but nope it's stupid :P.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%5B%28%28%CE%BB%E2%88%925%29%28%CE%BB%E2%88%921%29%28%CE%BB%29%29%2B%28%28%E2%88%921%29%5E3%29%5D%E2%88%92%5B7%28%CE%BB%E2%88%921%29%28%E2%88%921%29%5D wolfram says you're good :)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1yay :D always trust your brain first is the moral, I'd say Ok, dinner time, see ya!

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0NOOOOOOOOO!! 1 more quesitn plz.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1\[\lambda^36\lambda^2+12\lambda8=(\lambda2)^3=0\]triple Eigenvalue, you can manage that I think ;) I really have to go, food getting cold read the link I gave you and good luck!!!!!!!!!!!!

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks, diff Q on the eigenvalues though :)

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0COME BACK SOON SO I CAN ASK MOAR Q'S!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.