Here's the question you clicked on:
KonradZuse
Find the characteristic EQ of the following matrix.
@TuringTest pic to follow.
[7*lambda-8+(lambda-5)*(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.
which problem are you doing?
all of them in exercise 6?
I assume you know to subtract lambda from the diagonals, then find the determinant of that matrix?
MY book just shows for a 2x2 matrix that it's lambda - # on the main diagonal, and everything else is negated.
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....
\[A=\left[\begin{matrix}5&0&1\\1&1&0\\-7&1&0\end{matrix}\right]\]\[(\lambda I-A)=0=\left|\begin{matrix}\lambda-5&0&-1\\-1&\lambda-1&0\\7&-1&\lambda\end{matrix}\right|\]take the determinant and what do you get?
I got [7*lambda-8+(lambda-5)*(lambda(lambda)-1)]
I don't think you simplified correctly...
Idk check my pic, that's what maple gave me anyways....
[(lambda-5)*(lambda-1)(lambda)+7*(0*0)-(-1)*(-1)]+[(lambda-1)*7+(lambda-5)*0-(0*(-1))*lambda]
I have always done det(A - lambda*I)=0
That's what the book shows us.
It comes from \[ Ax =\lambda x\] or \[ Ax -\lambda x=0\] \[ (A -\lambda I) x=0\]
\[(\lambda-5)(\lambda-1)(\lambda)+(-1)^3-[7(\lambda-1)(-1)]=0\]what @phi said is true, you can do it either way
Yeah the book says that, then goes into normal determinants.e
Why is it only [7(λ−1)(−1)]?
the other parts have 0's so we can ignore them
and why do you only do it twice? I'm so confused. I thought the diagonal rule is 3 diags - 3 diags.
I don't like Maple's answer >( Wolfram doesn't either :P.
I use co-factors (L-5) * det (lower right 2x2) ignore the 0 -1 * det(lower left 2x2)
I hate co factors... From TT's example I get [-7*lambda(-1)+1+(lambda(lambda-1))(lambda)] = 0
[((lambda-5)(lambda-1))(lambda)+(-1)^3]+[-7*(lambda-1)(-1)] = 0; [-7 lambda(-1) + 1 + lambda(lambda - 1)(lambda)] = 0
I just took what you did and entered it into Maple :P.
oh I think I know what Id id :P.
> [((lambda-5)(lambda-1))(lambda)+(-1)^3]+[-7*(lambda-1)(-1)] = 0; [-7 lambda(-1) + 1 + lambda(lambda - 1)(lambda)] = 0
I frogot to add some extra brackets :). Does this shizzle look better >(
Same thing it looks like >(
\[(\lambda-5)(\lambda-1)(\lambda)+(-1)^3-[7(\lambda-1)(-1)]=0\]\[\lambda^3-6\lambda^2+5\lambda-1+7\lambda-7=0\]\[\lambda^3-6\lambda^2+12\lambda-8=0\]I could have messed up, but I did it by eye...
TT never messes up :)
hehe, that attitude has cost me plenty of credits, so I strongly suggest you double-check me, and everyone else for that matter, on everything. I appreciate the confidence though :)
Normally maple gives me the good answer, but nope it's stupid :P.
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 :)
yay :D always trust your brain first is the moral, I'd say Ok, dinner time, see ya!
NOOOOOOOOO!! 1 more quesitn plz.
\[\lambda^3-6\lambda^2+12\lambda-8=(\lambda-2)^3=0\]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!!!!!!!!!!!!
Thanks, diff Q on the eigenvalues though :)
COME BACK SOON SO I CAN ASK MOAR Q'S!