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mathavraj Group Title

\[ \left( \begin{array}\ 1 & 0 &0 \\ 1 &0& 1\\ 0& 1 & 0 \end{array}\right)\] =A by cayley hamilton transformation method prove that \[A^n=A^n-2 +A^2-1\] and find \[A^50\]

  • 2 years ago
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  1. mathavraj Group Title
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    Sorry my question is not appearing...

    • 2 years ago
  2. mathavraj Group Title
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    In a 3*3 matrix with a11=1,a12=0,a13=0,a21=1,a22=0,a23=1,a31=0,a32=1,a33=0...prove by cayley hamilton theorem that A^n= A^n-2+A^2-1.....also find A^50

    • 2 years ago
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    @Jemurray3 @

    • 2 years ago
  4. Jemurray3 Group Title
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    The Cayley-Hamilton theorem says that matrices obey their own characteristic equations. So, \[det (A-\lambda I) = det\left(\begin{matrix}1-\lambda & 0 & 0 \\ 1 & -\lambda & 1 \\ 0 & 1 & -\lambda \end{matrix}\right) = \lambda^2(1-\lambda) -(1-\lambda) \] \[ = (\lambda^2 -1)(1-\lambda) = -\lambda^3+ \lambda^2 + \lambda - 1 = 0 \] implies that \[ -A^3 + A^2 + A - I = 0 \implies A^3 = A^2+ A - I \] We can use a proof by induction. Assume \[A^n = A^{n-2} + A^2 - I \] Then \[A^{n+1} = A\cdot A^n = A^{n-1} + A^3- A = A^{n-1} + A^2 + A - I - A\] \[ = A^{(n+1)-2} + A^2 - I \] Since we've already proved this for the case n = 2, it must hold for all n >= 2.

    • 2 years ago
  5. Jemurray3 Group Title
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    Furthermore, by the above theorem, \[A^{50} = A^{48} + A^2 - I = A^{46} + 2 ( A^2- I ) = A^{44} + 3(A^2- I) \] etc etc etc. Which yields the general formula (assuming that k is even) \[ A^k = A^2 + \frac{k-2}{2} \left( A^2 - I \right) \] So \[ A^{50} = A^2 + 24( A^2 - I ) = 25 A^2 - 24I\] A^2 is fairly easy to calculate: \[A^2 = \left(\begin{matrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}\right)\] So finally, \[A^{50} = \left( \begin{matrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 &1\end{matrix} \right) \]

    • 2 years ago
  6. mathavraj Group Title
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    but how did \[A^{48}+A^2-I\] turn into \[A^{46}+2(A^2-I)\]

    • 2 years ago
  7. Jemurray3 Group Title
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    Because via the proof above, \[ A^k = A^{k-2} + A^2 - I \]

    • 2 years ago
  8. mathavraj Group Title
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    thank you:)

    • 2 years ago
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