Răspuns:
Explicație pas cu pas:
[tex]A^{2} = \left[\begin{array}{cc}3&3\\-1&-1\end{array}\right] *\left[\begin{array}{cc}3&3\\-1&-1\end{array}\right] = \left[\begin{array}{cc}6&6\\-2&-2\end{array}\right] = 2*\left[\begin{array}{cc}3&3\\-1&-1\end{array}\right] = 2*A[/tex]
[tex]A^{3} = A^2*A = (2*A)*A=2*(A*A) = 2*A^2 = 2*(2*A) = (2*2)*A=2^2*A[/tex]
[tex]A^{4} = A^3*A = (2^2*A)*A=2^2*(A*A) = 2^2*A^2 = 2^2*(2*A) = (2^2*2)*A = 2^3*A[/tex]
Prin inductie se demonstreaza ca :
[tex]A^n} = 2^{(n-1)}*A[/tex] , ∀ n ∈ N, n>0
Intradevar:
[tex]A ^1= 1*A =2^0*A= 2^{1-1}*A[/tex]
[tex]A^2 = 2*A = 2^{2-1}*A[/tex]
Daca [tex]A^n} = 2^{(n-1)}*A[/tex], atunci aratam ca [tex]A^{n+1} = 2^{[(n+1)-1]}*A[/tex]
[tex]A^{n+1} = A^n*A = [2^{(n-1)}*A]*A = 2^{(n-1)}*[A*A] = 2^{(n-1)}*A^2 =[/tex]
[tex]=2^{(n-1)}*(2*A) = [2^{(n-1)}*2]*A) = 2^{[(n-1)+1]}*A = 2^{[(n+1)-1]}*A[/tex]
Asadar:
[tex]A^n} = 2^{(n-1)}*A = 64 A[/tex]
[tex]2^{(n-1)} = 64 = 2^6[/tex]
[tex]n-1 = 6\\n = 7[/tex]
