Explicație pas cu pas:
[tex]\det(X^3)=\begin{vmatrix}5&8\\8&13\end{vmatrix}=5\cdot13-8\cdot8=65-64=1\\(\det(X))^3=1\Rightarrow \det(X)=1\\\texttt{Notam cu t urma matricii X.}\\\texttt{Conform teoremei Cayley-Hamilton vom avea:}\\X^2-tX+I_2=0\\\texttt{Adica }X^2=tX-I_2~(1)\\\texttt{Inmultim relatia cu X:}\\X^3=tX^2-X\stackrel{(1)}{=} t(tX-I_2)-X=(t^2-1)X-tI_2~(2)\\\texttt{Daca doua matrici sunt egale, inseamna ca au si urmele egale.}\\tr(X^3)=tr((t^2-1)X-tI_2)\\5+13=tr((t^2-1)X)-tr(tI_2)\\18=(t^2-1)tr(X)-t\cdot tr(I_2)[/tex]
[tex]18=(t^2-1)t-2t\\t^3-t-2t=18\\t^3-3t-18=0\\\texttt{Ecuatia se mai poate scrie:}\\(t-3)(t^2+3t+6)=0\\i) t-3=0\Rightarrow t=3\\ii)t^2+3t+6=0\\\Delta=9-4\cdot 6=9-24=-15<0,\texttt{nu convine.Deci ramane varianta}\\\texttt{cu t=3. Prin urmare }\boxed{tr(X)=3}[/tex]
[tex]\texttt{Revenim la relatia (2)}:\\X^3=(t^2-1)X-tI_2\\X^3=(3^2-1)X-3I_2\\X^3=8X-3I_2\\8X=X^3+3I_2\\8X=\begin{bmatrix}8&8\\8&16\end{bmatrix}\\\boxed{X=\begin{bmatrix}1&1\\1&2\end{bmatrix}}[/tex]
