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Data publicarii: 06 Noiembrie, 2014

EXERCITIUL 16

Suport teoretic:

Operatii cu matrice,termen general sir,inductia matematica.

Enunt:

Sa se calculeze

A^n,\;unde\;A=\begin{pmatrix}1&-2\\2&-3\end{pmatrix},\;n\in{\mathbb{R^*}}.A^n,\;unde\;A=\begin{pmatrix}1&-2\\2&-3\end{pmatrix},\;n\in{\mathbb{R^*}}.

Raspuns:

A^n={(-1)^n}\cdot{\begin{pmatrix}(-2n+1)&2n\\-2n&(2n+1)\end{pmatrix}},\;unde\;n\in{\mathbb{N^*}}.A^n={(-1)^n}\cdot{\begin{pmatrix}(-2n+1)&2n\\-2n&(2n+1)\end{pmatrix}},\;unde\;n\in{\mathbb{N^*}}.

Rezolvare:

Se calculeaza  A², A³ ,..., se obtine un sir de matrice de acelasi tip, elementele

acestora formand, ele insele, 4 siruri de numere reale, urmand a "descoperi", pe baze intuitive, termenii generali corespunzatori. Deci:

A^2={\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}\cdot{\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}=\begin{pmatrix}-3&4\\-4&5\end{pmatrix},A^2={\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}\cdot{\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}=\begin{pmatrix}-3&4\\-4&5\end{pmatrix},

A^3={\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}\cdot{\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}\cdot{\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}={\begin{pmatrix}-3&4\\-4&5\end{pmatrix}}\cdot{\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}=\begin{pmatrix}5&-6\\6&-7\end{pmatrix}.A^3={\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}\cdot{\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}\cdot{\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}={\begin{pmatrix}-3&4\\-4&5\end{pmatrix}}\cdot{\begin{pmatrix}1&-2\\2&-3\end{pmatrix}}=\begin{pmatrix}5&-6\\6&-7\end{pmatrix}.

Daca notam  A^n=\begin{pmatrix}a_n&b_n\\c_n&d_n\end{pmatrix},\;unde\;n\in{\mathbb{N^*}},A^n=\begin{pmatrix}a_n&b_n\\c_n&d_n\end{pmatrix},\;unde\;n\in{\mathbb{N^*}},   observam ca:

(an) este sirul   1, -3,   5, ... , a, ...

(bn) este sirul  -2,   4, -6, ... , bn, ...

(cn) este sirul    2, -4,   6, ... , cn, ...

si

(dn) este sirul  -3,   5, -7, ... , dn, ... . 

Se poate usor intui ca termenii generali ai celor 4 siruri sunt:

a_n=(-1)^{n+1}\cdot(2n-1),a_n=(-1)^{n+1}\cdot(2n-1),

b_n=(-1)^{n}\cdot(2n),b_n=(-1)^{n}\cdot(2n),

c_n=(-1)^{n+1}\cdot(2n),c_n=(-1)^{n+1}\cdot(2n),

d_n=(-1)^{n}\cdot(2n+1),d_n=(-1)^{n}\cdot(2n+1),

Se deduce imediat:

A^n=\begin{pmatrix}(-1)^{n+1}\cdot(2n-1)&(-1)^{n}\cdot(2n)\\(-1)^{n+1}\cdot(2n)&(-1)^{n}\cdot(2n+1)\end{pmatrix}={(-1)^n}\cdot{\begin{pmatrix}(-2n+1)&2n\\-2n&(2n+1)\end{pmatrix}}.A^n=\begin{pmatrix}(-1)^{n+1}\cdot(2n-1)&(-1)^{n}\cdot(2n)\\(-1)^{n+1}\cdot(2n)&(-1)^{n}\cdot(2n+1)\end{pmatrix}={(-1)^n}\cdot{\begin{pmatrix}(-2n+1)&2n\\-2n&(2n+1)\end{pmatrix}}.

Rezultatul se verifica apoi prin inductie matematica (vezi aici).


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