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KonradZuse

  • 3 years ago

Find the characteristic EQ of the following matrix.

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  1. KonradZuse
    • 3 years ago
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    @TuringTest pic to follow.

  2. KonradZuse
    • 3 years ago
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  3. KonradZuse
    • 3 years ago
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    [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
    • 3 years ago
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    which problem are you doing?

  5. TuringTest
    • 3 years ago
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    all of them in exercise 6?

  6. phi
    • 3 years ago
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    I assume you know to subtract lambda from the diagonals, then find the determinant of that matrix?

  7. KonradZuse
    • 3 years ago
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    I'm doing 6E

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

  9. KonradZuse
    • 3 years ago
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    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
    • 3 years ago
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    \[A=\left[\begin{matrix}5&0&1\\1&1&0\\-7&1&0\end{matrix}\right]\]\[(\lambda I-A)=0=\left|\begin{matrix}\lambda-5&0&-1\\-1&\lambda-1&0\\7&-1&\lambda\end{matrix}\right|\]take the determinant and what do you get?

  11. KonradZuse
    • 3 years ago
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    I got [7*lambda-8+(lambda-5)*(lambda(lambda)-1)]

  12. KonradZuse
    • 3 years ago
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    as shown in my pic

  13. TuringTest
    • 3 years ago
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    I don't think you simplified correctly...

  14. KonradZuse
    • 3 years ago
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    Idk check my pic, that's what maple gave me anyways....

  15. KonradZuse
    • 3 years ago
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    [(lambda-5)*(lambda-1)(lambda)+7*(0*0)-(-1)*(-1)]+[(lambda-1)*7+(lambda-5)*0-(0*(-1))*lambda]

  16. phi
    • 3 years ago
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    I have always done det(A - lambda*I)=0

  17. KonradZuse
    • 3 years ago
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    That's what the book shows us.

  18. phi
    • 3 years ago
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    It comes from \[ Ax =\lambda x\] or \[ Ax -\lambda x=0\] \[ (A -\lambda I) x=0\]

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

  20. KonradZuse
    • 3 years ago
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    Yeah the book says that, then goes into normal determinants.e

  21. KonradZuse
    • 3 years ago
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    Why is it only [7(λ−1)(−1)]?

  22. TuringTest
    • 3 years ago
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    the other parts have 0's so we can ignore them

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

  24. KonradZuse
    • 3 years ago
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    oic yeah.

  25. KonradZuse
    • 3 years ago
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    deerrp :P

  26. KonradZuse
    • 3 years ago
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    I don't like Maple's answer >( Wolfram doesn't either :P.

  27. phi
    • 3 years ago
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    I use co-factors (L-5) * det (lower right 2x2) ignore the 0 -1 * det(lower left 2x2)

  28. KonradZuse
    • 3 years ago
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    I hate co factors... From TT's example I get [-7*lambda(-1)+1+(lambda(lambda-1))(lambda)] = 0

  29. KonradZuse
    • 3 years ago
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    [((lambda-5)(lambda-1))(lambda)+(-1)^3]+[-7*(lambda-1)(-1)] = 0; [-7 lambda(-1) + 1 + lambda(lambda - 1)(lambda)] = 0

  30. KonradZuse
    • 3 years ago
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    I just took what you did and entered it into Maple :P.

  31. KonradZuse
    • 3 years ago
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    oh I think I know what Id id :P.

  32. KonradZuse
    • 3 years ago
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    > [((lambda-5)(lambda-1))(lambda)+(-1)^3]+[-7*(lambda-1)(-1)] = 0; [-7 lambda(-1) + 1 + lambda(lambda - 1)(lambda)] = 0

  33. KonradZuse
    • 3 years ago
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    I frogot to add some extra brackets :). Does this shizzle look better >(

  34. KonradZuse
    • 3 years ago
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    Same thing it looks like >(

  35. TuringTest
    • 3 years ago
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    \[(\lambda-5)(\lambda-1)(\lambda)+(-1)^3-[7(\lambda-1)(-1)]=0\]\[\lambda^3-6\lambda^2+5\lambda-1+7\lambda-7=0\]\[\lambda^3-6\lambda^2+12\lambda-8=0\]I could have messed up, but I did it by eye...

  36. KonradZuse
    • 3 years ago
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    TT never messes up :)

  37. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    Normally maple gives me the good answer, but nope it's stupid :P.

  39. TuringTest
    • 3 years ago
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    yay :D always trust your brain first is the moral, I'd say Ok, dinner time, see ya!

  40. KonradZuse
    • 3 years ago
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    NOOOOOOOOO!! 1 more quesitn plz.

  41. TuringTest
    • 3 years ago
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    \[\lambda^3-6\lambda^2+12\lambda-8=(\lambda-2)^3=0\]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!!!!!!!!!!!!

  42. KonradZuse
    • 3 years ago
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    Thanks, diff Q on the eigenvalues though :)

  43. KonradZuse
    • 3 years ago
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    COME BACK SOON SO I CAN ASK MOAR Q'S!

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