Question

Cum se rezolva punctul c de la exercitiul 1?

Answer 1

$A = \left(\begin{array}{cc}3&4\\2&3\end{array}\right)$

$\left(\begin{array}{c}x_{n+2}\\y_{n+2}\end{array}\right) =A\left(\begin{array}{c}x_{n+1}\\y_{n+1}\end{array}\right) = A\cdot A\left(\begin{array}{c}x_{n}\\y_{n}\end{array}\right) = A^2 \left(\begin{array}{c}x_{n}\\y_{n}\end{array}\right)$

$= \left(\begin{array}{cc}3&4\\2&3\end{array}\right)\left(\begin{array}{cc}3&4\\2&3\end{array}\right)\left(\begin{array}{c}x_{n}\\y_n\end{array}\right) =\left(\begin{array}{cc}17&24\\12&17\end{array}\right)\left(\begin{array}{c}x_{n}\\y_n\end{array}\right) = \\ \\ =\left(\begin{array}{c}17x_{n}+24y_n\\12x_n+17y_n\end{array}\right)$

$\Rightarrow \boxed{x_{n+2}=17x_n+24y_n}$

$\left(\begin{array}{c}x_{n+1}\\y_{n+1}\end{array}\right) =A\left(\begin{array}{c}x_{n}\\y_{n}\end{array}\right) = \left(\begin{array}{cc}3&4\\2&3\end{array}\right)\left(\begin{array}{c}x_{n}\\y_{n}\end{array}\right)=\left(\begin{array}{c}x_{n}\\y_{n}\end{array}\right) =\\ \\ =\left(\begin{array}{c}3x_{n}+4y_n\\2x_n+3y_n\end{array}\right)$

$\Rightarrow \boxed{x_{n+1} = 3x_n+4y_n}$

$\textbf{Astfel:}\\ x_{n+2}-6x_{n+1}+x_n = \\ =(17x_n+24y_n)-6(3x_n+4y_n)+x_n\\ = 17x_n+24y_n-18x_n-24y_n+x_n \\ =(17x_n-18x_n+x_n)+(24y_n-24y_n)\\ =0+0\\ = 0\,\,\,\,\checkmark$