[tex]A^2=A\cdot A=\left(\begin{array}{ccc}1&3&2\\3&9&6\\2&6&4\end{array}\right)\cdot\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[/tex]
Vom demonstra prin inductie.
[tex]P(n):A^n=14^{n-1}A, n\geq2\\P(2):A^2=14A, \text{ adev\u arat, demonstrat mai sus}[/tex]
[tex]\text{Presupunem } P(k):A^k=14^{k-1}A\text{ adev\u arat\u a.}\\\text{Demonstr\u am } P(k+1):A^{k+1}=14^{(k+1)-1}A\\A^{k+1}=A^k\cdot A=14^{k-1}A\cdot A=14^{k-1}\cdot A^2=\\=14^{k-1}\cdot14A=14^{k-1+1}A=14^kA\Rightarrow P(k+1)\text{ adev\u arat\u a}\Rightarrow P(n)\text{ adev\u arat\u a, }\forall n\geq2[/tex]
