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Va rugam explicate, pe scurt, de ce credeti ca aceasta intrebare trebuie raportata.
Motivul pentru care raportezi utilizatorul.
1.Se da matricea:
![\[ A = \left( {\begin{array} {\cos \alpha } & 0 & {\sin \alpha } \\ 0 & 1 & 0 \\ { - \sin \alpha } & 0 & {\cos \alpha } \\ \end{array}} \right)\;\quad ,cu\;\alpha \in R \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-c8991b7a2e716b373fd641ec1234541f_l3.png "Rendered by QuickLaTeX.com")
a).Sa se calculeze A^n ,pt n din N stelat.
b). Fie matricele
![\[ C = \left( {\begin{array} x \\ y \\ z \\ \end{array}} \right)\;si\;D = \left( {\begin{array} {\sin 3\alpha } \\ {tg\;3\alpha } \\ {\cos 3\alpha } \\ \end{array}} \right) \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-56a38d2c69ec7a7f1e7fdbd03dcd00fb_l3.png "Rendered by QuickLaTeX.com")
.Determinati x,y,z apartin lui R astfel incat AC=D.
2.Se considera determinantul
![\[ \Delta (x) = \left| {\begin{array} {e^{2x^2 } } & {e^{ - a} } & {e^{ - x} } \\ {e^{ - a} } & {e^{2x} } & {e^{ - x^2 } } \\ {e^{ - x} } & {e^{ - x^2 } } & {e^{2a} } \\ \end{array}} \right|\;,a \in R \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-1b8c795bd327e3c910fe9ae06e55e88d_l3.png "Rendered by QuickLaTeX.com")
a).Sa se calculeze
![\[ \Delta (x)\]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-48752be991f02d59787c461298f1d76d_l3.png "Rendered by QuickLaTeX.com")
.
b).Sa se calculeze valorile parametrului real a pentru care ecuatia
![\[ \Delta (x)=0 \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-d6506c991aa55ba4ead5bfcd47530ac0_l3.png "Rendered by QuickLaTeX.com")
are radacini reale strict negative.
La prima merge prin inductie:
*** QuickLaTeX cannot compile formula: \[ \begin{array}{l} \left( {\begin{array} \cos \alpha & 0 & {\sin \alpha } \\ 0 & 1 & 0 \\ { - \sin \alpha } & 0 & {\cos \alpha } \\ \end{array}} \right) = A \\ A^2 = \left( {\begin{array} \cos \alpha & 0 & {\sin \alpha } \\ 0 & 1 & 0 \\ { - \sin \alpha } & 0 & {\cos \alpha } \\ \end{array}} \right)\left( {\begin{array} \cos \alpha & 0 & {\sin \alpha } \\ 0 & 1 & 0 \\ { - \sin \alpha } & 0 & {\cos \alpha } \\ \end{array}} \right) = \left( {\begin{array} \cos ^2 \alpha - \sin ^2 \alpha & 0 & {2\sin \alpha \cos \alpha } \\ 0 & 1 & 0 \\ { - 2\sin \alpha \cos \alpha } & 0 & {\cos ^2 \alpha - \sin ^2 \alpha } \\ \end{array}} \right) \\ \end{array} = \left( {\begin{array} \cos 2\alpha & 0 & {\sin 2\alpha } \\ 0 & 1 & 0 \\ { - 2\sin 2\alpha } & 0 & {\cos 2\alpha } \\ \end{array}} \right) \] *** Error message: Illegal character in array arg. leading text: \cos Missing number, treated as zero. leading text: \cos Missing \endcsname inserted. leading text: \cos Extra \endcsname. leading text: \cos Undefined control sequence \@nil. (in macro \@nextchar) leading text: \cos Missing number, treated as zero. (in macro \@nextchar) leading text: \cos Illegal unit of measure (pt inserted). leading text: \cos Illegal character in array arg. leading text: \cos Illegal parameter number in definition of \@preamble. leading text: \cos Illegal parameter number in definition of \@preamble. leading text: \cos Illegal parameter number in definition of \@preamble. leading text: \cos Illegal parameter number in definition of \@preamble.Acum parcurgi pasii de inductie si vei ajunge la
la b)
*** QuickLaTeX cannot compile formula: \[ \begin{array}{l} \left( {\begin{array}{*{20}c} {\cos \alpha } & 0 & {\sin \alpha } \\ 0 & 1 & 0 \\ { - \sin \alpha } & 0 & {\cos \alpha } \\ \end{array}} \right) = A \\ C = \left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right) \\ AC = \left( {\begin{array} \cos \alpha & 0 & {\sin \alpha } \\ 0 & 1 & 0 \\ { - \sin \alpha } & 0 & {\cos \alpha } \\ \end{array}} \right)\left( {\begin{array} x \\ y \\ z \\ \end{array}} \right) = \left( {\begin{array} {x\cos \alpha + z\sin \alpha } \\ y \\ { - x\sin \alpha + z\cos \alpha } \\ \end{array}} \right) = \left( {\begin{array} \sin 3\alpha \\ {tg3\alpha } \\ {\cos 3\alpha } \\ \end{array}} \right) \Rightarrow \left\{ \begin{array}{l} x\cos \alpha + z\sin \alpha = \sin 3\alpha \\ tg3\alpha = y \\ - x\sin \alpha + z\cos \alpha = \cos 3\alpha \\ \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} x\cos \alpha + z\sin \alpha = \sin 3\alpha \\ - x\sin \alpha + z\cos \alpha = \cos 3\alpha \\ \end{array} \right. \\ \left( 1 \right) \Rightarrow z = \frac{{\sin 3\alpha - x\cos \alpha }}{{\sin \alpha }} \\ \Downarrow \left( 2 \right) \Rightarrow - x\sin \alpha + \frac{{\sin 3\alpha - x\cos \alpha }}{{\sin \alpha }}\cos \alpha = \cos 3\alpha \Rightarrow \left( {amplificare} \right) \Rightarrow - x\sin ^2 \alpha + (\sin 3\alpha - x\cos \alpha )\cos \alpha = \cos 3\alpha \sin \alpha \\ \Rightarrow - x\sin ^2 \alpha - \cos ^2 \alpha x + \cos \alpha \sin 3\alpha = \cos 3\alpha \sin \alpha = - x\left( {\sin ^2 \alpha + \cos ^2 \alpha } \right) = \cos 3\alpha \sin \alpha - \cos \alpha \sin 3\alpha \Rightarrow - x = \sin \left( { - 2\alpha } \right) \Rightarrow x = \sin 2\alpha \\ \end{array} \] *** Error message: Illegal character in array arg. leading text: \cos Missing number, treated as zero. leading text: \cos Missing \endcsname inserted. leading text: \cos Extra \endcsname. leading text: \cos Undefined control sequence \@nil. (in macro \@nextchar) leading text: \cos Missing number, treated as zero. (in macro \@nextchar) leading text: \cos Illegal unit of measure (pt inserted). leading text: \cos Illegal character in array arg. leading text: \cos Illegal parameter number in definition of \@preamble. leading text: \cos Illegal parameter number in definition of \@preamble. leading text: \cos Illegal parameter number in definition of \@preamble. leading text: \cos Illegal parameter number in definition of \@preamble.Inlocuiesti, si scoti ce a mai ramas. Nu mai am timp acum. Revin mai tarziu.
Ultima:
Avem asa:
Deci am aplicat formula si am ajuns la acea forma, nu e mare inginerie.
la b)
*** QuickLaTeX cannot compile formula: \[ \begin{array}{l} \Delta \left( x \right) = \left( {e^{x^2 + x + a} } \right)^2 + \frac{2}{{e^{x^2 + x + a} }} - 3 \\ Notam] *** 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.Acum se pun conditiile
si 
. 
[tex]\[
\ne
\]0 \Rightarrow a \in \[
\emptyset
\]
[/tex]
asta daca nu am gresit pe undeva.
Se pare ca ai gresit valorile parametrului a..
Continui de la notarea ecuatiei…
Da, ai dreptate. Multumesc pentru observatie.
1b]. Se cere C, as ca ; A.C=D sau ; A^(-1).A.C=C=A^(-1).D.
Det(A)=1 si A^(-1)=
…………..(cos(a)……0…..-sin(a))
..A^(-1)=(0………….1………….0)
…………..(sin(a)…….0……cos(a)) si A^(-1).D=C, de unde ;
…………..(.sin(2.a).)
……….C=(.tg(3.a).)
…………..(cos(2.a).) -> x=sin(2.a) , y=tg(3.a) , z=cos(2.a) (obs. cele 3 paranteze de pe verticala ,inlocuesc paranteza ce identifica o matrice)
Multumesc mult la toata lumea!🙂
Mai am o mica rugaminte,inca un ex:
Se considera matricele
a).Calculati det X(a), a din R.
b).Sa se arate ca X(a)X(b)=X(a+b-ab), oricare ar fi a,b din R.
c).Aflati t apartine lui Z astfel incat X(-10)X(-9)….X(9)X(10)=X(t).