Răspuns:
[tex]A=\begin{matrix}2 &1&2|3\\1&-2&1|-1\\1&2&1|3\end{matrix}=\overline{A}\\\det(A)=-4+4+1+4-4-1=0\Rightarrow rang A<2\\d_{car}=\begin{vmatrix}2&1\\1&-2\end{vmatrix}=-4-1=-5\neq 0\Rightarrow rangA=2\\d_{princ}=\begin{vmatrix}2&1&3\\1&-2&-1\\1&2&3\end{vmatrix}=-12+6-1+6+4-3=0\Rightarrow rang\overline{A}=2\\\texttt{Avem ca }rangA=rang\overline{A}=2.\texttt{deci sistemul este compatibil}\\\texttt{nedeterminat. Fie }z_0=\alpha[/tex][tex]\begin{cases}2x+y+2\alpha=3\\x-2y+\alpha =-1|\cdot 2\end{cases}\Leftrightarrow \begin{cases}2x+y+2\alpha=3\\2x-4y+2\alpha=-2\end{cases}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~----------\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~5y=5\Rightarrow \boxed{y=1}\\2x+1+2\alpha=3\\2x+2\alpha=2\\x=1-\alpha\\\texttt{Atunci }:\\x_0^2+y_0^2+z_0^2=(1-\alpha)^2+1+\alpha^2=1-2\alpha+\alpha^2+1+\alpha^2=\\=2\alpha^2-2\alpha+2=2(\alpha^2-\alpha+1)[/tex]
[tex]\texttt{Avand in vedere ca am obtinut o ecuatie de gradul 2 , vom tine }\\\texttt{cont ca valoarea minima se afla in varf. Prin urmare:}\\ \alpha= -\dfrac{b}{2a}=\dfrac{1}{2}\\\\x=1-\alpha=\dfrac{1}{2}\\z=\dfrac{1}{2}\\\texttt{Solutia este }(x_0,y_0,z_0)=\left\{\left(\dfrac{1}{2},1,\dfrac{1}{2}\right)\right\}[/tex]
