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KonradZuse Group Title Find the characteristic EQ of the following matrix. one year ago one year ago

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1. KonradZuse Group Title

@TuringTest pic to follow.

2. KonradZuse Group Title

3. KonradZuse Group Title

[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 Group Title

which problem are you doing?

5. TuringTest Group Title

all of them in exercise 6?

6. phi Group Title

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

7. KonradZuse Group Title

I'm doing 6E

8. KonradZuse Group Title

MY book just shows for a 2x2 matrix that it's lambda - # on the main diagonal, and everything else is negated.

9. KonradZuse Group Title

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 Group Title

$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?

11. KonradZuse Group Title

I got [7*lambda-8+(lambda-5)*(lambda(lambda)-1)]

12. KonradZuse Group Title

as shown in my pic

13. TuringTest Group Title

I don't think you simplified correctly...

14. KonradZuse Group Title

Idk check my pic, that's what maple gave me anyways....

15. KonradZuse Group Title

[(lambda-5)*(lambda-1)(lambda)+7*(0*0)-(-1)*(-1)]+[(lambda-1)*7+(lambda-5)*0-(0*(-1))*lambda]

16. phi Group Title

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

17. KonradZuse Group Title

That's what the book shows us.

18. phi Group Title

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

19. TuringTest Group Title

$(\lambda-5)(\lambda-1)(\lambda)+(-1)^3-[7(\lambda-1)(-1)]=0$what @phi said is true, you can do it either way

20. KonradZuse Group Title

Yeah the book says that, then goes into normal determinants.e

21. KonradZuse Group Title

Why is it only [7(λ−1)(−1)]?

22. TuringTest Group Title

the other parts have 0's so we can ignore them

23. KonradZuse Group Title

and why do you only do it twice? I'm so confused. I thought the diagonal rule is 3 diags - 3 diags.

24. KonradZuse Group Title

oic yeah.

25. KonradZuse Group Title

deerrp :P

26. KonradZuse Group Title

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

27. phi Group Title

I use co-factors (L-5) * det (lower right 2x2) ignore the 0 -1 * det(lower left 2x2)

28. KonradZuse Group Title

I hate co factors... From TT's example I get [-7*lambda(-1)+1+(lambda(lambda-1))(lambda)] = 0

29. KonradZuse Group Title

[((lambda-5)(lambda-1))(lambda)+(-1)^3]+[-7*(lambda-1)(-1)] = 0; [-7 lambda(-1) + 1 + lambda(lambda - 1)(lambda)] = 0

30. KonradZuse Group Title

I just took what you did and entered it into Maple :P.

31. KonradZuse Group Title

oh I think I know what Id id :P.

32. KonradZuse Group Title

> [((lambda-5)(lambda-1))(lambda)+(-1)^3]+[-7*(lambda-1)(-1)] = 0; [-7 lambda(-1) + 1 + lambda(lambda - 1)(lambda)] = 0

33. KonradZuse Group Title

I frogot to add some extra brackets :). Does this shizzle look better >(

34. KonradZuse Group Title

Same thing it looks like >(

35. TuringTest Group Title

$(\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...

36. KonradZuse Group Title

TT never messes up :)

37. TuringTest Group Title

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 :)

38. KonradZuse Group Title

Normally maple gives me the good answer, but nope it's stupid :P.

39. KonradZuse Group Title
40. TuringTest Group Title

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

41. KonradZuse Group Title

NOOOOOOOOO!! 1 more quesitn plz.

42. TuringTest Group Title

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

43. KonradZuse Group Title

Thanks, diff Q on the eigenvalues though :)

44. KonradZuse Group Title

COME BACK SOON SO I CAN ASK MOAR Q'S!