A^n=? Va rog ................

[tex]A^2 = \left[\begin{array}{cc}8&4\\16&8\end{array}\right] = 4 \cdot A [/tex]

Aplicand succesiv aceasta identitate in produsul [tex]A^n = A \cdot A \cdot A \cdot... \cdot A[/tex], obtinem:

[tex]A^n = 4 \cdot A^{n-1} = 4^2 \cdot A^{n-2} =... = 4^{n-1} \cdot A. [/tex]

[tex]A^n = 4^{n-1}\left[\begin{array}{cc}2&1\\4&2\end{array}\right] = 2^{2n-2}\left[\begin{array}{cc}2&1\\4&2\end{array}\right] = \left[\begin{array}{cc}2^{2n-1}&2^{2n-2}\\2^{2n}&2^{2n-1}\end{array}\right] [/tex]

a^1= 2 1
         4 2

A²   =2 1   2 1  = 8   4 =         =4^1*A
        4 2   4 2     16  8


A³=A²*A= 8   4    2 1=   32  16  =16A=4²A
                 16 8    4 2     64  32


presupunem A^n=4^(n-1)*A      (1)
care a fost calculata si se verifica pt n=1,2,3

caklculam A^(n+1)
A^(n+1)=A^n*A=4^(n-1) A*A=4^(n-1) A²=
=4^(n-1) *4^1*A=4^n*A=4^((n+1)-1)*A
Pn⇒P(n+1)
deci formula (1) este adevarata, fiind demonbstrata prin inductie (paranteaz matematica paranteza)completa