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

KonradZuse Group TitleBest 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.
 2 years ago

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

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

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

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I'm doing 6E
 2 years ago

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

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
so 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....
 2 years ago

TuringTest Group TitleBest 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?
 2 years ago

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

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
as shown in my pic
 2 years ago

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

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

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

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

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

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

TuringTest Group TitleBest 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
 2 years ago

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

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

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

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

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
oic yeah.
 2 years ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
deerrp :P
 2 years ago

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

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

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

KonradZuse Group TitleBest 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
 2 years ago

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

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

KonradZuse Group TitleBest 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
 2 years ago

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

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

TuringTest Group TitleBest 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...
 2 years ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
hehe, 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 :)
 2 years ago

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

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
http://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 :)
 2 years ago

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

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

TuringTest Group TitleBest 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!!!!!!!!!!!!
 2 years ago

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

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
COME BACK SOON SO I CAN ASK MOAR Q'S!
 2 years ago
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