Question

Subpunctul c)
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Answer 1

Răspuns:

Explicație pas cu pas:

$\texttt{Sistemul are solutie unica daca det(A(a))}\neq 0\texttt{ adica }a\notin\{0,2\}\\\texttt{(/*la bacalaureat sa scrii si asta ca de nu pierzi puncte*}\backslash)\\\texttt{Cel mai simplu ar fi sa aflam solutiile prin metoda Cramer.}\\\Delta x_0=\begin{vmatrix}1&1&1\\1&a&2\\2&2&a\end{vmatrix}=a^2+2+2-2a-4-a=a^2-3a=a(a-3)\\x_0=\dfrac{\Delta x_0}{\det(A(a))}=\dfrac{a(a-3)}{a(a-2)}=\dfrac{a-3}{a-2}=1-\dfrac{1}{a-2}$

$\Delta y_0=\begin{vmatrix}1&1&1\\1&1&2\\1&2&a\end{vmatrix}=a+2+2-1-4-a=-1\\y_0=\dfrac{\Delta y_0}{\det(A(a))}=\dfrac{-1}{a(a-2)}\\\Delta z_0=\begin{vmatrix}1&1&1\\1&a&1\\1&2&2\end{vmatrix}=2a+2+1-a-2-2=a-1\\z_0=\dfrac{\Delta z_0}{\det(A(a))}=\dfrac{a-1}{a(a-2)}=\dfrac{1}{a-2}-\dfrac{1}{a(a-2)}\\x_0,y_0,z_0\in\mathbb{Z}\Leftrightarrow a-2\in D_1\texttt{ si }a\in D_1\\a-2\in D_1\Rightarrow a-2\in\{-1,1\}\\~~~~~~~~~~~~~~~~~~~~~~a\in\{1,3\}\\a\in \D_1\Rightarrow a\in \{-1,1\}$

$\texttt{Din ultimele doua relatii deducem ca }a=1 . \texttt{Observam ca pentru}\\a=1\Rightarrow~~x_0,y_0,z_0\in\mathbb{Z},~\texttt{deci e convenabil.}$