Matrice la puterea n

Lauryca Joi Feb 07, 2013 8:46 pm

Fie $A,B\epsilon M_{2}(R)$ iar $A=\begin{pmatrix} 1 &1 \\ 0&1 \end{pmatrix}$ ; $B=\begin{pmatrix} a &b \\ b&a \end{pmatrix}$ si $C=ABA^{-1}$

SA SE CALCULEZE : $C^{n},n\geq 1$

**Admin** Vin Feb 08, 2013 2:28 am

Rezolvarea problemei este destul de lunga (nu cred ca exista vreo scurtătura).

Se determina (prin metoda standard) ${A^{ - 1}} = {\left( {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}} 1&{ - 1}\\ 0&1 \end{array}} \right)$
Se efectueze:
$\begin{array}{l} C = AB{A^{ - 1}} = \left( {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} a&b\\ b&a \end{array}} \right) \times {\left( {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right)^{ - 1}}\\ \\ = \left( {\begin{array}{*{20}{c}} {a + b}&{a + b}\\ b&a \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1&{ - 1}\\ 0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {a + b}&0\\ b&{a - b} \end{array}} \right) \end{array}$
Se determina C^n prin calcul progresiv (C^2, C^3, se stabilește C^n, se demonstrează prin inducție): ${\color{Red} {C^n} = \left( {\begin{array}{*{20}{c}} {{{\left( {a + b} \right)}^n}}&0\\ {{b^n}}&{{{\left( {a - b} \right)}^n}} \end{array}} \right)}$

Matrice la puterea n

Matrice la puterea n

Re: Matrice la puterea n