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SabahS
user (0)
Pe: 2009-02-14T14:32:07+02:00 In: In:

Matrice pt cei bravi ! Admitere ASe 2008

*** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 A = (\begin{array}{*c}
	   1 & 0 & {i\sqrt 3 }  \\
	   0 & 0 & 0  \\
	   {i\sqrt 3 } & 0 & 1  \\
	\end{array}) \\
	 B = A\nolimits^{2010}  \\
	 S = \sum {|el.matricea.B|}  \\
	 \end{array}
	\]
	

*** Error message:
Missing number, treated as zero.
leading text: 	 A = (\begin{array}{*c}
 Illegal character in array arg.
leading text: 	 A = (\begin{array}{*c}
Missing # inserted in alignment preamble.
leading text: 	 A = (\begin{array}{*c}
Extra alignment tab has been changed to \cr.
leading text: 	   1 &
Extra alignment tab has been changed to \cr.
leading text: 	   1 & 0 &
Missing $ inserted.
leading text: 	   1 & 0 & {i\sqrt 3
Extra }, or forgotten $.
leading text: 	   1 & 0 & {i\sqrt 3 }
Missing } inserted.
leading text: 	   0
Extra }, or forgotten $.
leading text: 	   0
Missing } inserted.
leading text: 	   0
Extra }, or forgotten $.
leading text: 	   0
Missing } inserted.
leading text: 	   0
Extra }, or forgotten $.
leading text: 	   0
Missing } inserted.

am facut si A patrat A la 3a nu gasesc regula si nici cu A(xy) nu reusesc poate vreunu dintre voi care vad ca va pricepeti reusiti plsss help .

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3 raspunsuri

  1. ali
    maestru (V)

    Uite si o indicatie ‘importanta’:
    Rescriem matricea in felul urmator:

        \[ \begin{array}{l} \left( {\begin{array}{*{20}c} 1 & 0 & {i\sqrt 3 } \\ 0 & 0 & 0 \\ {i\sqrt 3 } & 0 & 1 \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} 0 & 0 & {i\sqrt 3 } \\ 0 & { - 1} & 0 \\ {i\sqrt 3 } & 0 & 0 \\ \end{array}} \right) = I_3 + \left( {\begin{array}{*{20}c} 0 & 0 & {i\sqrt 3 } \\ 0 & { - 1} & 0 \\ {i\sqrt 3 } & 0 & 0 \\ \end{array}} \right) \\ B = A^{2010} \Rightarrow B = \left[ {I_3 + \left( {\begin{array}{*{20}c} 0 & 0 & {i\sqrt 3 } \\ 0 & { - 1} & 0 \\ {i\sqrt 3 } & 0 & 0 \\ \end{array}} \right)} \right]^{2010} \\ {\rm Observatii:} \\ {\rm I}_{\rm 3}^{\rm n} = I_3 \\ I_3 *A = A*I_3 = A. \\ \end{array} \]

    Deci tot ce mai ai de facut este sa gaseste o frmula pt A.

    • 0
  2. ali
    maestru (V)

    *20c$1=1
    Nu stiu dece asa aparut in ex.

    • 0
  4. omegatheo
    user (0)

        \[ \begin{array}{l} \left( {\begin{array}{*{20}c} 1 & 0 & {i\sqrt 3 } \\ 0 & 0 & 0 \\ {i\sqrt 3 } & 0 & 1 \\ \end{array}} \right) = 2\left( {\begin{array}{*{20}c} {\frac{1}{2}} & 0 & {i\frac{{\sqrt 3 }}{2}} \\ 0 & 0 & 0 \\ {i\frac{{\sqrt 3 }}{2}} & 0 & {\frac{1}{2}} \\ \end{array}} \right) = 2\left( {\begin{array}{*{20}c} {\cos \frac{\pi }{3}} & 0 & {i\sin \frac{\pi }{3}} \\ 0 & 0 & 0 \\ {i\sin \frac{\pi }{3}} & 0 & {\cos \frac{\pi }{3}} \\ \end{array}} \right) = 2C; \\ C^n = \left( {\begin{array}{*{20}c} {\cos \frac{{n\pi }}{3}} & 0 & {i\sin \frac{{n\pi }}{3}} \\ 0 & 0 & 0 \\ {i\sin \frac{{n\pi }}{3}} & 0 & {\cos \frac{{n\pi }}{3}} \\ \end{array}} \right),pr{\rm in inductie matematica;} \\ {\rm Deci A}^{{\rm 2010}} = 2^{2010} C^{2010} ;si{\rm S = 2}^{{\rm 2010}} *S(C^{2010} );\cos \frac{{2010\pi }}{3} = - 1;\sin \frac{{2010\pi }}{3} = 0; \\ Deci{\rm S}(C^{2010} ) = - 2*2^{2010} = - 2^{2011} ; \\ \end{array} \]

    • 0
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