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KonradZuse
 4 years ago
Find the characteristic EQ of the following matrix.
KonradZuse
 4 years ago
Find the characteristic EQ of the following matrix.

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

KonradZuse
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1which problem are you doing?

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

phi
 4 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
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0I got [7*lambda8+(lambda5)*(lambda(lambda)1)]

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

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

KonradZuse
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1I have always done det(A  lambda*I)=0

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

phi
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah the book says that, then goes into normal determinants.e

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

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

KonradZuse
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0I don't like Maple's answer >( Wolfram doesn't either :P.

phi
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0I just took what you did and entered it into Maple :P.

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

KonradZuse
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0I frogot to add some extra brackets :). Does this shizzle look better >(

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

TuringTest
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0TT never messes up :)

TuringTest
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0Normally maple gives me the good answer, but nope it's stupid :P.

KonradZuse
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0NOOOOOOOOO!! 1 more quesitn plz.

TuringTest
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks, diff Q on the eigenvalues though :)

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