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KonradZuseBest ResponseYou've already chosen the best response.0
@TuringTest pic to follow.
 one year ago

KonradZuseBest 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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
which problem are you doing?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
all of them in exercise 6?
 one year ago

phiBest 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?
 one year ago

KonradZuseBest 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.
 one year ago

KonradZuseBest 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....
 one year ago

TuringTestBest 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?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I got [7*lambda8+(lambda5)*(lambda(lambda)1)]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I don't think you simplified correctly...
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Idk check my pic, that's what maple gave me anyways....
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
[(lambda5)*(lambda1)(lambda)+7*(0*0)(1)*(1)]+[(lambda1)*7+(lambda5)*0(0*(1))*lambda]
 one year ago

phiBest ResponseYou've already chosen the best response.1
I have always done det(A  lambda*I)=0
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
That's what the book shows us.
 one year ago

phiBest ResponseYou've already chosen the best response.1
It comes from \[ Ax =\lambda x\] or \[ Ax \lambda x=0\] \[ (A \lambda I) x=0\]
 one year ago

TuringTestBest 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
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Yeah the book says that, then goes into normal determinants.e
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Why is it only [7(λ−1)(−1)]?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the other parts have 0's so we can ignore them
 one year ago

KonradZuseBest 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.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I don't like Maple's answer >( Wolfram doesn't either :P.
 one year ago

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

KonradZuseBest 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
 one year ago

KonradZuseBest 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
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I just took what you did and entered it into Maple :P.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
oh I think I know what Id id :P.
 one year ago

KonradZuseBest 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
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I frogot to add some extra brackets :). Does this shizzle look better >(
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Same thing it looks like >(
 one year ago

TuringTestBest 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...
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
TT never messes up :)
 one year ago

TuringTestBest 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 :)
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Normally maple gives me the good answer, but nope it's stupid :P.
 one year ago

KonradZuseBest 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 :)
 one year ago

TuringTestBest 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!
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
NOOOOOOOOO!! 1 more quesitn plz.
 one year ago

TuringTestBest 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!!!!!!!!!!!!
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Thanks, diff Q on the eigenvalues though :)
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
COME BACK SOON SO I CAN ASK MOAR Q'S!
 one year ago
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