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Puterile unei matrice din M_3(R)

 
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Filip Chindea
Newton


Joined: 27 Sep 2007
Posts: 323
Location: Bucharest

PostPosted: Wed Jan 30, 2008 9:52 am    Post subject: Puterile unei matrice din M_3(R) Reply with quote

Pentru orice n \in \mathbb{N}^{\ast}, sa se calculeze A^n, unde
A = \left( \begin{array}{ccc} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \end{array} \right) \in \mathcal{M}_3(\mathbb{R}).
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Mateescu Constantin
Newton


Joined: 21 Apr 2009
Posts: 398
Location: Londra/Pitesti

PostPosted: Tue Aug 31, 2010 4:15 pm    Post subject: Reply with quote

Solutie. Se stie, sau se arata usor prin inductie ca daca A\in\mathcal{M}_3(\mathbb{R}) exista sirurile (x_n)_{\small n\ge 1} , (y_n)_{\small n\ge 1} si (z_n)_{\small n\ge 1}

astfel incat : \fbox{\ A^n=x_n\cdot A^2\ +\ y_n\cdot A\ +\ z_n\cdot I_3\ }\ unde \fbox{\underline{\overline{\begin{array}{|c|c|c|}
\ x_1=0\  &\ y_1=1\ &\ z_1=0\ \\\\\\\\\\    
\hline 
\ x_2=1\  &\ y_2=0\ &\ z_2=0\ \\\\\\\\\\   
\hline 
\ x_3=\tr\ (A)\ &\ y_3=\tr\ (A^{\ast})\ &\ z_3=\det\ (A)\ \\\\\\\\\\    
\hline 
\ x_{n+1}=x_3\cdot x_n+y_n\ &\ y_{n+1}=y_3\cdot x_n+z_n\ &\ z_{n+1}=z_3\cdot x_n\ \end{array}}}} .


In cazul nostru x_3=\tr\ (A)=7 , y_3=\tr\ (A^{\ast})=0 si z_3=\det\ (A)=-36. Prin urmare avem: \begin{array}{cccc} x_{n+2}=7x_{n+1} & - & 36x_{n-1}\end{array} ,

sir cu ecuatia caracteristica r^3=7r^2-36 (care-i totuna cu ecuatia caracteristica a matricei A) . Asadar, sirul (x_n)_{\small n\ge 1} este de forma :

x_n=\alpha\cdot (-2)^n+\beta\cdot 3^n+\gamma\cdot 6^n . Cunoscand primii termeni ai sirului se obtine usor ca : \fbox{\ \begin{array}x_n=\frac 1{40} & \cdot & (-2)^n & - & \frac 1{15} & \cdot & 3^n & + & \frac 1{24} & \cdot & 6^n\end{array}\ } .

Imediat gasim : \fbox{\ \begin{array}y_n=-\frac 9{10} & \cdot & (-2)^{n-2} & + & \frac {12}5 & \cdot & 3^{n-2} & - & \frac 32 & \cdot & 6^{n-2}\end{array}\ } si apoi \fbox{\ \begin{array}z_n=-\frac 9{10} & \cdot & (-2)^{n-1} & + & \frac {12}5 & \cdot & 3^{n-1} & - & \frac 32 & \cdot & 6^{n-1}\end{array}\ } .

In fine, \begin{array}{ccc} A^n=x_n & \cdot & \left\(\begin{array}{ccc} 11 & 9 & 7 \\ \\ 9 & 27 & 9 \\ \\ 7 & 9 & 11\end{array}\right\) & + & y_n & \cdot & \left\(\begin{array}{ccc} 1 & 1 & 3 \\ \\ 1 & 5 & 1 \\ \\ 3 & 1 & 1\end{array}\right\) & + & z_n & \cdot & \left\(\begin{array}{ccc} 1 & 0 & 0 \\ \\ 0 & 1 & 0 \\ \\ 0 & 0 & 1\end{array}\right\)=\left\(\begin{array}{ccccc} 11x_n+y_n+z_n & 9x_n+y_n & 7x_n+3y_n \\\\\\\\ 9x_n+y_n & 27x_n+5y_n+z_n & 9x_n+y_n \\\\\\\\ 7x_n+3y_n & 9x_n+y_n & 11x_n+y_n+z_n\end{array}\right\)\end{array}

unde sirurile (x_n)_{\small n\ge 1} , (y_n)_{\small n\ge 1} si (z_n)_{\small n\ge 1} au fost determinate mai sus .
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