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maca
user (0)
Pe: 2010-07-23T08:59:45+03:00 In: In:

matrici …. help plssssss.. :(

se da matricea A= ( 1 2 )
0 1
sa se calculeze suma elementelor matricei B= suma (deasupra lui sigma este 10 ..si jo k=1 ) din A la puterea k ……. [/code]

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5 raspunsuri

  1. sindex
    user (0)

        \[\begin{array}{l} A = \left( {\begin{array} 1 & 2 \\ 0 & 1 \\ \end{array}} \right) \\ B = \sum\limits_{k = 1}^{10} {{A^k}} = {A^1} + {A^2} + ... + {A^{10}} \\ \end{array}\]

    Observi ca e o progresie geometrica, cu ratia A
    Aplici formula sumei pentru progresia geometrica:

        \[{S_n} = {a_1} \cdot \frac{{{q^n} - 1}}{{q - 1}}\]

    Particularizezi pentru problema:
    \[\begin{array}{l} {a_1} = A \\ q = A \\ n = 10 \\ \end{array}\

        \[1 = {I_2}\]

    si de aici iti iese suma primilor 10 termeni :

        \[B = \sum\limits_{k = 1}^{10} {{A^k}} = {A^1} + {A^2} + ... + {A^{10}} = A \cdot \frac{{{A^{10}} - {I_2}}}{{A - {I_2}}}\]

        \[{A^{10}} = {({A^2})^5}\]

        \[\begin{array}{l} {A^2} = A \cdot A \\ {({A^2})^5} = {A^2} \cdot {A^2} \cdot {A^2} \cdot {A^2} \cdot {A^2} \\ \end{array}\]

    Calculele le faci tu, ca inmultirea matricelor trebuie sa o stii 🙂
    Succes.

    Daca nu iese asa, poti folosi metoda muncitoreasca:

        \[\begin{array}{l} {A^2} = A \cdot A \\ {A^3} = {A^2} \cdot A \\ {A^4} = {A^2} \cdot {A^2} \\ {A^5} = {A^3} \cdot {A^2} \\ {A^6} = {({A^3})^2} \\ {A^7} = {A^5} \cdot {A^2} \\ {A^8} = {({A^4})^2} \\ {A^9} = {({A^3})^3} \\ {A^{10}} = {({A^5})^2} \\ \end{array}\]

    Si apoi sa le aduni la sfarsit.

    Sper sa-ti dea, daca ai probleme, revino cu intrebari

    • 0
  2. ali
    maestru (V)

    E mult mai simplu decat atat:

        \[ \begin{array}{l} A = \left( {\begin{array} 1 & 2 \\ 0 & 1 \\ \end{array}} \right) = \left( {\begin{array} 1 & 0 \\ 0 & 1 \\ \end{array}} \right) + 2\underbrace {\left( {\begin{array} 0 & 1 \\ 0 & 0 \\ \end{array}} \right)}_{ = B} = {\rm{I}}_2 + 2B \\ \left\{ \begin{array}{l} \overline {\underline {\left| {B^n = B} \right|} } \\ \overline {\underline {\left| {{\rm{I}}_{_2 }^n = {\rm{I}}_2 } \right|} } \\ \overline {\underline {\left| {{\rm{I}}_{_2 }^n \cdot B^n = B^n } \right|} } \\ \end{array} \right. \Rightarrow A^n = \left( {{\rm{I}}_2 + 2B} \right)^n = ... = {\rm{I}}_2 + B\left[ {2 \cdot C_n^1 + \left( 2 \right)^2 \cdot C_n^2 + .....} \right] \\ \end{array} \]

    Cazul tau n=10, si tinand cont de observatiile din acolada ai sa vezi cat de usor se finalizeaza.
    😀

    • 0
  4. ali
    maestru (V)

    In caz ca te intrebi cu cine este egala suma:

    *** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 S = 2C_n^1  + 2^2 C_n^2  + .... + 2^k C_n^k  \\
	 S = \underbrace {C_n^0 }_{ = 1} + 2C_n^1  + 2^2 C_n^2  + .... + 2^k C_n^k  - 1 \\
	 Observam\;\;ca]
	
	
	

*** Error message:
\begin{array} on input line 10 ended by \end{document}.
leading text: \end{document}
Improper \prevdepth.
leading text: \end{document}
Missing $ inserted.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Missing \cr inserted.
leading text: \end{document}
Missing $ inserted.
leading text: \end{document}
You can't use `\end' in internal vertical mode.
leading text: \end{document}
\begin{array} on input line 10 ended by \end{document}.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Emergency stop.

    In cazul tau n=10

    • 0
  5. sindex
    user (0)

    Pai el tot trebuie sa calculeze suma elementelor matricei B (b11+..b22)
    Si pana la urma, e nevoit sa calculeze pe

        \[{A^{10}}\]

    😀

    • 0
  6. ali
    maestru (V)

    Fibonacci wrote: E mult mai simplu de cat atat:
    \[ \begin A = \left( {\begin{array} 1 & 2 \\ 0 & 1 \\ \end{array}} \right) = \left( {\begin{array} 1 & 0 \\ 0 & 1 \\ \end{array}} \right) + 2\underbrace {\left( {\begin{array} 0 & 1 \\ 0 & 0 \\ \end{array}} \right)}_{ = B} = {\rm{I}}_2 + 2B \\


    Am notat cu B matricea ( 0 1 0 0) fara sa stiu ca in enunti se preciza ca B= sum …
    Gresala mea, oricum tu poti s-o notez cu C sau orice alta litera in afara de B.
    !!! Odata determinata formula generala pentru matricea A^n se poate calcula orice putere a lui A economisind un timp valoros.

    ❗ Imi cer scuze pentru neatentia formularii,
    Cu bine,

    • 0
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