Răspuns:
[tex](a) \boldsymbol{ \red{S = \{-1;1\}}}; \ (b) \boldsymbol{ \red{A_x^2 = 2xA_x + (1 - x^2) I_2}}; \ (c) \boldsymbol{ \red{0}}[/tex]
Rezolvare:
[tex]a) \ detA_x = \begin{vmatrix} x&1\\1&x\end{vmatrix} = x^2 - 1 \Rightarrow x^2 - 1 = 0 \Rightarrow (x-1)(x+1) = 0 \\[/tex]
[tex]\Rightarrow x = -1; x = 1[/tex]
b) Ecuația Cayley-Hamilton:
[tex]\boldsymbol {X^2 - Tr(A) \cdot X +det (A) \cdot I_2 = O_2}, \ \ \forall A \in \mathcal{M}_2 \in \Bbb{C}[/tex]
[tex]\Rightarrow \boldsymbol {\red{A_x^2 - Tr(A_x) \cdot A_x + det (A_x) \cdot I_2 = O_2}}[/tex]
Verificăm această relație:
[tex]Tr(A_x) = x + x = 2x; \ \ \ det(A_x) = x^2 - 1\\[/tex]
[tex]A_x^2 = \begin{pmatrix} x&1\\1&x\end{pmatrix}^2 = \begin{pmatrix} x&1\\1&x\end{pmatrix} \begin{pmatrix} x&1\\1&x\end{pmatrix} = \begin{pmatrix} x^2+1&2x\\2x&x^2+1\end{pmatrix}\\[/tex]
[tex]A_x^2 - Tr(A_x) \cdot A_x +det (A_x) \cdot I_2 = \boldsymbol {A_x^2 - 2x \cdot A_x + (x^2 - 1) \cdot I_2} =\\[/tex]
[tex]= \begin{pmatrix} x^2+1&2x\\2x&x^2+1\end{pmatrix} - 2x\begin{pmatrix} x&1\\1&x\end{pmatrix} + (x^2 - 1)\begin{pmatrix} 1&0\\0&1\end{pmatrix}\\[/tex]
[tex]= \begin{pmatrix} x^2+1&2x\\2x&x^2+1\end{pmatrix} - \begin{pmatrix} 2x^2&2x\\2x&2x^2\end{pmatrix} + \begin{pmatrix} x^2 - 1&0\\0&x^2 - 1\end{pmatrix}\\[/tex]
[tex]= \begin{pmatrix} x^2+1-2x^2+x^2-1&2x-2x\\2x-2x&x^2+1-2x^2+x^2-1\end{pmatrix} =\begin{pmatrix} 0&0\\0&0\end{pmatrix} = O_2\\[/tex]
Confirmăm că:
[tex]\Rightarrow \boldsymbol { A_x^2 = 2xA_x + (1 - x^2) I_2}[/tex]
[tex]c) \ A_x^2 = I_2 \Rightarrow \begin{pmatrix} x^2+1&2x\\2x&x^2+1\end{pmatrix} = \begin{pmatrix} 1&0\\0&1\end{pmatrix}[/tex]
[tex]\begin{cases} x^2+1=1 \\2x = 0\\2x = 0\\x^2+1=1\end{cases} \Rightarrow \begin{cases} x^2=0 \\x = 0\\x = 0\\x^2=0\end{cases} \Rightarrow x = 0[/tex]
