Question

Se considera matricea A
$\left[\begin{array}{ccc}1&3&2\\3&9&6\\2&6&4\end{array}\right]$

aratati ca $A^{n} = 14^{n-1}* A$ , pt orice $n \geq 2$

Answer 1

[tex]A^2= \left[\begin{array}{ccc}1&3&2\\3&9&6\\2&6&4\end{array}\right] \left[\begin{array}{ccc}1&3&2\\3&9&6\\2&6&4\end{array}\right]=\left[\begin{array}{ccc}14&42&28\\42&126&84\\28&84&56\end{array}\right]=\\ \\ =14 \left[\begin{array}{ccc}1&3&2\\3&9&6\\2&6&4\end{array}\right]=14A\\ A^3=A^2A=14AA=14A^2=14*14A=14^2A\\ \text{Se demonstreaza prin inductie prin procedeul explicat mai sus:}\\ A^{n+1}=A^nA=14^nAA=14^nA^2=14^n*14A=14^{n+1}A [/tex]