Întrebare puterea unei matrice

Bua seara
Cu cine este egala:
\[A=\begin{pmatrix} 3&1\\ -1&1\\ \end{pmatrix}^{2016}\]
multumesc

Buna seara,

Incercati sa folositi Teorema Cayley - Haimilton.

Vedeti aici un exemplu: http://ssmr.ro/gazeta/didactica/2012/2/articol.pdf

(paginile 6-7 (exemplul 14.).

Buna seara
Din materialul expus de dvs.eu am extras modelul din ex.7 pag.3.
Si anume:
am considerat ca plecare matricea
\[A=\begin{pmatrix} 3&1\\ -1&1\\ \end{pmatrix} Consideram \ matricea\\= A =I_3+B=\begin{pmatrix} 3&1\\ -1&1\\ \end{pmatrix}\\

A=\begin{pmatrix} 1&0\\ 0&1\\ \end{pmatrix}+\begin{pmatrix} 2&1\\ -1&0\\ \end{pmatrix}\\
Recurenta:
B^n=\begin{pmatrix} n+1&n\\ -n&-n+1\\ \end{pmatrix}si\\ A^n=2^{n-1}\cdot(I_3+B^n)\]
pe de alta parte:
\[f(f(x))=\dfrac{3\cdot\dfrac{3x+1}{1-x}+1}{1-\dfrac{3x+1}{1-x}}=\dfrac{8x+4}{-4x}\\ A^2=\begin{pmatrix} 8&4\\ -4&0\\ \end{pmatrix}\]
\[f(f(f(x)))=\dfrac{8\cdot\dfrac{3x+1}{1-x}+4}{-4\cdot\dfrac{3x+1}{1-x}}=\dfrac{20x+12}{-12x-4}\\ A^3=\begin{pmatrix} 20&12\\ -12&-4\\ \end{pmatrix}etc\]

\[A^{2016}=2^{2015}\cdot\begin{pmatrix} 2018&2016\\ -2016&-2014\\ \end{pmatrix}\]



\[\underbrace{f\circ f\circ f\circ f\dots\circ f}_{f \ de\ 2016\ ori}(x)=\dfrac{2018x+2016}{-2016x-2014}\\ pentru \ x=4 \ ca \ in \ ipoteza \ rezulta:\\ f......(4)=\dfrac{2018\cdot 4+2016}{-2016\cdot 4-2014}=-\dfrac{5044}{5039}\\ raspuns\ valabil\ (c)\]
multumesc