## KonradZuse Group Title Find the characteristic EQ of the following matrix. 2 years ago 2 years ago

@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.

4. TuringTest

which problem are you doing?

5. TuringTest

all of them in exercise 6?

6. phi

I assume you know to subtract lambda from the diagonals, then find the determinant of that matrix?

I'm doing 6E

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....

10. TuringTest

$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)]

as shown in my pic

13. TuringTest

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]

16. phi

I have always done det(A - lambda*I)=0

That's what the book shows us.

18. phi

It comes from $Ax =\lambda x$ or $Ax -\lambda x=0$ $(A -\lambda I) x=0$

19. TuringTest

$(\lambda-5)(\lambda-1)(\lambda)+(-1)^3-[7(\lambda-1)(-1)]=0$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)]?

22. TuringTest

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.

oic yeah.

deerrp :P

I don't like Maple's answer >( Wolfram doesn't either :P.

27. phi

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 >(

35. TuringTest

$(\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 :)

37. TuringTest

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.

40. TuringTest

yay :D always trust your brain first is the moral, I'd say Ok, dinner time, see ya!

NOOOOOOOOO!! 1 more quesitn plz.

42. TuringTest

$\lambda^3-6\lambda^2+12\lambda-8=(\lambda-2)^3=0$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!!!!!!!!!!!!