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mathavraj
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\[ \left( \begin{array}\ 1 & 0 &0 \\ 1 &0& 1\\ 0& 1 & 0 \end{array}\right)\] =A by cayley hamilton transformation method prove that \[A^n=A^n2 +A^21\] and find \[A^50\]
 2 years ago
 2 years ago
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 \[A^n=A^n2 +A^21\] and find \[A^50\]
 2 years ago
 2 years ago

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mathavraj Group TitleBest ResponseYou've already chosen the best response.0
Sorry my question is not appearing...
 2 years ago

mathavraj Group TitleBest ResponseYou've already chosen the best response.0
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^n2+A^21.....also find A^50
 2 years ago

mathavraj Group TitleBest ResponseYou've already chosen the best response.0
@Jemurray3 @
 2 years ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
The CayleyHamilton 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 \[A^n = A^{n2} + A^2  I \] Then \[A^{n+1} = A\cdot A^n = A^{n1} + A^3 A = A^{n1} + A^2 + A  I  A\] \[ = A^{(n+1)2} + A^2  I \] Since we've already proved this for the case n = 2, it must hold for all n >= 2.
 2 years ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
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) \[ A^k = A^2 + \frac{k2}{2} \left( A^2  I \right) \] 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

mathavraj Group TitleBest ResponseYou've already chosen the best response.0
but how did \[A^{48}+A^2I\] turn into \[A^{46}+2(A^2I)\]
 2 years ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Because via the proof above, \[ A^k = A^{k2} + A^2  I \]
 2 years ago

mathavraj Group TitleBest ResponseYou've already chosen the best response.0
thank you:)
 2 years ago
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