Question

Determinati cel mai mic numar natural n pentru care matricea A= $\left(\begin{array}{ccc}3&3\\-1&-1\end{array}\right)$ verifica relatia $A^{n}$ =64A

Answer 1

Răspuns:

Explicație pas cu pas:

$A^{2} = \left[\begin{array}{cc}3&3\\-1&-1\end{array}\right] *\left[\begin{array}{cc}3&3\\-1&-1\end{array}\right] = \left[\begin{array}{cc}6&6\\-2&-2\end{array}\right] = 2*\left[\begin{array}{cc}3&3\\-1&-1\end{array}\right] = 2*A$

$A^{3} = A^2*A = (2*A)*A=2*(A*A) = 2*A^2 = 2*(2*A) = (2*2)*A=2^2*A$

$A^{4} = A^3*A = (2^2*A)*A=2^2*(A*A) = 2^2*A^2 = 2^2*(2*A) = (2^2*2)*A = 2^3*A$

Prin inductie se demonstreaza ca :

$A^n} = 2^{(n-1)}*A$ , ∀ n ∈ N, n>0

Intradevar:

$A ^1= 1*A =2^0*A= 2^{1-1}*A$

$A^2 = 2*A = 2^{2-1}*A$

Daca $A^n} = 2^{(n-1)}*A$, atunci aratam ca $A^{n+1} = 2^{[(n+1)-1]}*A$

$A^{n+1} = A^n*A = [2^{(n-1)}*A]*A = 2^{(n-1)}*[A*A] = 2^{(n-1)}*A^2 =$

$=2^{(n-1)}*(2*A) = [2^{(n-1)}*2]*A) = 2^{[(n-1)+1]}*A = 2^{[(n+1)-1]}*A$

Asadar:

$A^n} = 2^{(n-1)}*A = 64 A$

$2^{(n-1)} = 64 = 2^6$

$n-1 = 6\\n = 7$