Efectueaza o cautare in website!

Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu.

Data publicarii: 13 Mai, 2011

EXERCITIUL 2

Suport teoretic:

Ecuatii matriciale,matrici inversabile,clase de resturi.

Enunt: 

Sa se rezolve, in multimea claselor de resturi modulo 5, ecuatia matriciala

A·X + B = O5, unde

A=\begin{pmatrix}{\hat{2}}&{\hat{3}}&{\hat{0}}\\{\hat{4}}&{\hat{1}}&{\hat{3}}\\{\hat{1}}&{\hat{2}}&{\hat{4}}\end{pmatrix},\;B=\begin{pmatrix}{\hat{0}}&{\hat{1}}&{\hat{1}}\\{\hat{1}}&{\hat{2}}&{\hat{0}}\\{\hat{3}}&{\hat{0}}&{\hat{2}}\end{pmatrix}.A=\begin{pmatrix}{\hat{2}}&{\hat{3}}&{\hat{0}}\\{\hat{4}}&{\hat{1}}&{\hat{3}}\\{\hat{1}}&{\hat{2}}&{\hat{4}}\end{pmatrix},\;B=\begin{pmatrix}{\hat{0}}&{\hat{1}}&{\hat{1}}\\{\hat{1}}&{\hat{2}}&{\hat{0}}\\{\hat{3}}&{\hat{0}}&{\hat{2}}\end{pmatrix}.

Raspuns:

X=\begin{pmatrix}{\hat{4}}&{\hat{4}}&{\hat{2}}\\{\hat{1}}&{\hat{4}}&{\hat{2}}\\{\hat{1}}&{\hat{2}}&{\hat{0}}\end{pmatrix}.X=\begin{pmatrix}{\hat{4}}&{\hat{4}}&{\hat{2}}\\{\hat{1}}&{\hat{4}}&{\hat{2}}\\{\hat{1}}&{\hat{2}}&{\hat{0}}\end{pmatrix}.

Rezolvare:

Cercetam daca matricea A este inversabila:

det(A)=\begin{vmatrix}{\hat{2}}&{\hat{3}}&{\hat{0}}\\{\hat{4}}&{\hat{1}}&{\hat{3}}\\{\hat{1}}&{\hat{2}}&{\hat{4}}\end{vmatrix}=\cdots=\hat{2}\not=\hat{0},det(A)=\begin{vmatrix}{\hat{2}}&{\hat{3}}&{\hat{0}}\\{\hat{4}}&{\hat{1}}&{\hat{3}}\\{\hat{1}}&{\hat{2}}&{\hat{4}}\end{vmatrix}=\cdots=\hat{2}\not=\hat{0},

deci matricea, nefiind nula, este inversabila (in multimea claselor de resturi modulo n,

cu n = 5, numar prim, singurul element care nu admite invers este clasa lui 0). 

Deci ecuatia admite solutia unica:

X={A^{-1}}\cdot{(-B)},X={A^{-1}}\cdot{(-B)},

unde (-B) reprezinta opusa matricei B.

Folosind formula cunoscuta de calcul al inversei unei matrice, obtinem succesiv:

X={{\hat{2}}^{-1}}\cdot{A^{*}}={\hat{3}}\cdot{\begin{pmatrix}{\hat{3}}&{\hat{3}}&{\hat{4}}\\{\hat{2}}&{\hat{3}}&{\hat{4}}\\{\hat{2}}&{\hat{4}}&{\hat{0}}\end{pmatrix}}=\begin{pmatrix}{\hat{4}}&{\hat{4}}&{\hat{2}}\\{\hat{1}}&{\hat{4}}&{\hat{2}}\\{\hat{1}}&{\hat{2}}&{\hat{0}}\end{pmatrix}.X={{\hat{2}}^{-1}}\cdot{A^{*}}={\hat{3}}\cdot{\begin{pmatrix}{\hat{3}}&{\hat{3}}&{\hat{4}}\\{\hat{2}}&{\hat{3}}&{\hat{4}}\\{\hat{2}}&{\hat{4}}&{\hat{0}}\end{pmatrix}}=\begin{pmatrix}{\hat{4}}&{\hat{4}}&{\hat{2}}\\{\hat{1}}&{\hat{4}}&{\hat{2}}\\{\hat{1}}&{\hat{2}}&{\hat{0}}\end{pmatrix}.

Observatie:

Matricea adjuncta A* se obtine inlocuind elementele matricei transpuse tA prin

complementii lor algebrici.


Adăugaţi un comentariu

Adăugaţi un comentariu
Introdu codul din imagine

Răspunsuri şi comentarii

Chelsia

rSvGm3a6Ww, 15.08.2016 21:37

That's an inenigous way of thinking about it.

erg

arr, 19.02.2016 17:54

Observatie: Matricea adjuncta A* se obtine inlocuind elementele matricei transpuse tA prin complementii lor algebrici. cum adica???? cum se obtin complementii aia?

Răspuns: Citeste aici: http://www.profesoronline.ro/inversa_unei_matrice_liceu/

sal

ciorap21, 08.08.2012 15:54

si unde e rezolvare.

Răspuns: Pentru a avea acces la rezolvare, trebuie selectat textul "CLICK AICI PENTRU...", aflat sub enuntul problemei.

 

CATEGORII :


Arhiva blog-ului

 

 

http:// www.supermatematic


Developed by Hagau Ioan