Calculati puterea unei matrice de ordinul 3

[turcas]

Sa se calculeze \( A^n , n \in \mathbb{N}^{*} \), unde \( A= \left( \begin{array}{ccc} 0 & 1& 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \end{array} \right) \).

Subiectul III, Etapa locala 24 Ianuarie 2009, Salaj, clasa a XI-a

[Tudor Micu]

\( A=B-I_{3} \), unde \( B=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right) \)
Prin inductie demonstram ca \( B^{k}=\left(\begin{array}{ccc} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{array}\right) \)
\( B \) si \( I_3 \) sunt comutative deci putem aplica binomul lui Newton:
\( A^{n}=(B-I_3)^n=\displaystyle\sum_{k=0}^{n-1}[(-1)^{k}\cdot C_n^k\cdot B^{n-k}]+(-1)^n\cdot I_n \)
Rezulta ca \( A^n=\left(\begin{array}{ccc} T+(-1)^n & T & T \\ T & T+(-1)^n & T \\ T & T & T+(-1)^n \end{array}\right) \), unde \( T=\displaystyle\sum_{k=0}^{n-1}(-1)^{k}\cdot C_n^k\cdot 3^{n-k-1}=\frac{1}{3}\cdot (3-1)^n-\frac{1}{3}\cdot (-1)^n=\frac{2^n-(-1)^n}{3} \)
\( A^n=\left(\begin{array}{ccc} \frac{2^n+2(-1)^n}{3} & \frac{2^n-(-1)^n}{3} & \frac{2^n-(-1)^n}{3} \\ \frac{2^n-(-1)^n}{3} & \frac{2^n+2(-1)^n}{3} & \frac{2^n-(-1)^n}{3} \\ \frac{2^n-(-1)^n}{3} & \frac{2^n-(-1)^n}{3} & \frac{2^n+2(-1)^n}{3} \end{array}\right) \)