Răspuns:
Explicație pas cu pas:
Nu e neaparat nevoie sa aplici teorema lui Cayley-Hamilton pentru x^(1/2) ... poti sa o aplici si pentru X.
In primul rand, prin trecere la determinanti, ecuatia initiala devine :
[tex]\det(X^2)=\begin{vmatrix}1&12\\-4&1 \end{vmatrix}\\\det^2X=49\Rightarrow \det X\in\{\pm 7\}[/tex]
Aplicam teorema lui Cayley-Hamilton:
[tex]X^2-trX\cdot X+\det(X)\cdot I_2=O_2\\X^2=trX\cdot X-\det(X)\cdot I_2(1)[/tex]
Prin trecere la "urma" ecuatia devine:
[tex]tr(X^2)=tr(trX\cdot X)-tr(\det(X)\cdot I_2)\\2=tr^2X-\det X\cdot tr(I_2)\\2=tr^2 X-\det X\cdot 2[/tex]
[tex]\texttt{Daca }\det X=-7,\texttt{ atunci :}\\2=tr^2 X+14\\tr^2X=-12,\texttt{ imposibil}\\\texttt{Prin urmare }\det X=7 \texttt{ si atunci:}\\2=tr^2X-14\\tr^2X=16\Rightarrow tr X\in\{\pm 4\}[/tex]
[tex]\texttt{Daca }tr(X)=4\texttt{ atunci din relatia (1) rezulta: }\\4X-7I_2=\begin{pmatrix}1&12\\-4&1\end{pmatrix}\\4X=\begin{pmatrix}1&12\\-4&1\end{pmatrix}+\begin{pmatrix}7&0\\0&7\end{pmatrix}\\4X=\begin{pmatrix}8&12\\-4&8\end{pmatrix}\\X=\begin{pmatrix}2&3\\-1&2\end{pmatrix}[/tex]
[tex]\texttt{Daca }trX=-4,\texttt{atunci tot din relatia (1) rezulta:}\\\\-4X=\begin{pmatrix}8&12\\-4&8\end{pmatrix}\\X=\begin{pmatrix}-2&-3\\1&-2\end{pmatrix}[/tex]
Raspunsul corect este ----> A
