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xor_NTG
maestru (V)
Pe: 2012-06-27T14:55:05+03:00 In: In:

[Matrice]

Salut!
Am urmatoarea matrice:

    \[ A = \left( {\begin{array} 1 & 1 & 5 \\ 0 & 1 & { - 2} \\ 0 & 0 & 1 \\ \end{array}} \right) \in M_3 \left( Q \right) \]

Se pune problema calcularii lui A^n, n \in N*.
Avem:

    \[ \begin{array}{l} A = \left( {\begin{array} 1 & 1 & 5 \\ 0 & 1 & { - 2} \\ 0 & 0 & 1 \\ \end{array}} \right) \in M_3 \left( Q \right) \\ A^2 = A \cdot A = \left( {\begin{array} 1 & 1 & 5 \\ 0 & 1 & { - 2} \\ 0 & 0 & 1 \\ \end{array}} \right)\left( {\begin{array} 1 & 1 & 5 \\ 0 & 1 & { - 2} \\ 0 & 0 & 1 \\ \end{array}} \right) = \left( {\begin{array} 1 & 2 & 8 \\ 0 & 1 & { - 4} \\ 0 & 0 & 1 \\ \end{array}} \right) \\ A^3 = A^2 \cdot A = \left( {\begin{array} 1 & 2 & 8 \\ 0 & 1 & { - 4} \\ 0 & 0 & 1 \\ \end{array}} \right)\left( {\begin{array} 1 & 1 & 5 \\ 0 & 1 & { - 2} \\ 0 & 0 & 1 \\ \end{array}} \right) = \left( {\begin{array} 1 & 3 & 9 \\ 0 & 1 & { - 6} \\ 0 & 0 & 1 \\ \end{array}} \right) \\ A^4 = A^3 \cdot A = \left( {\begin{array} 1 & 3 & 9 \\ 0 & 1 & { - 6} \\ 0 & 0 & 1 \\ \end{array}} \right)\left( {\begin{array} 1 & 1 & 5 \\ 0 & 1 & { - 2} \\ 0 & 0 & 1 \\ \end{array}} \right) = \left( {\begin{array} 1 & 4 & 8 \\ 0 & 1 & { - 8} \\ 0 & 0 & 1 \\ \end{array}} \right) \\ A^5 = A^4 \cdot A = \left( {\begin{array} 1 & 4 & 8 \\ 0 & 1 & { - 8} \\ 0 & 0 & 1 \\ \end{array}} \right)\left( {\begin{array} 1 & 1 & 5 \\ 0 & 1 & { - 2} \\ 0 & 0 & 1 \\ \end{array}} \right) = \left( {\begin{array} 1 & 5 & 5 \\ 0 & 1 & { - 10} \\ 0 & 0 & 1 \\ \end{array}} \right) \\ \end{array} \]

Am mers pana la A^5 pentru ca ultimul element de pe prima linie nu variaza regulat in functie de puterea matricei precum ceilalti membri:
Deci pana acum putem observa ca:

    \[ A^n = \left( {\begin{array} 1 & n & ? \\ 0 & 1 & { - 2n} \\ 0 & 0 & 1 \\ \end{array}} \right) \]

Elementul de pe prima linie nu are nicio ordine. Ce pot face in acest caz?
Multumesc.

Similare

2 raspunsuri

  1. DD
    profesor

    Draga colega, pune ; A=I3+B , determina B^2 si B^3. Iti va da ;B^3=03,
    ridica apoi A^n=(I+B)^n si este simplu.

  2. xor_NTG
    maestru (V)

    Am inteles. Deci rezolvarea corecta ar fi asa:

        A=I_3+B, \[ B = \left( {\begin{array} 0 & 1 & 5 \\ 0 & 0 & { - 2} \\ 0 & 0 & 0 \\ \end{array}} \right) \]

        \[ \begin{array}{l} \\ B^2 = \left( {\begin{array} 0 & 1 & 5 \\ 0 & 0 & { - 2} \\ 0 & 0 & 0 \\ \end{array}} \right) \cdot \left( {\begin{array} 0 & 1 & 5 \\ 0 & 0 & { - 2} \\ 0 & 0 & 0 \\ \end{array}} \right) = \left( {\begin{array} 0 & 0 & { - 2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}} \right) \\ B^3 = \left( {\begin{array} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}} \right) = O_3 \\ \end{array} \]

    Deci pentru

        \[ \begin{array}{l} n \ge 3,\,B^n = O_3 \\ A^n = \left( {I_3 + B} \right)^n = \sum\limits_{k = 0}^n {C_n^k } I_3^{n - k} B^k = C_n^0 I_3^n + C_n^1 I_3^{n - 1} B + C_n^2 I_3^{n - 2} B^2 + ... + C_n^n B^n \\ \end{array} \]

    Ne ramane doar termenii care nu contin pe B la o putere mai mare ca 3, adica cei de la inceput:

        I_3^n=I_3; \[ C_n^1 I_3^{n - 1} B ; C_n^2 I_3^{n - 2} B^2 \]

    , adica

        \[ I_3 ,nB,\frac{{\left( {n - 1} \right)n}}{2}B^2 \]

        \[ A^n = \left( {\begin{array} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}} \right) + \left( {\begin{array} 0 & n & {5n} \\ 0 & 0 & { - 2n} \\ 0 & 0 & 0 \\ \end{array}} \right) + \left( {\begin{array} 0 & 0 & { - n\left( {n - 1} \right)} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}} \right) = \left( {\begin{array} 1 & n & { - n^2 + 6n} \\ 0 & 1 & { - 2n} \\ 0 & 0 & 1 \\ \end{array}} \right) \]

    Observ ca se verifica si pe puterile mai mici, deci e corect. Multumesc DD.

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