Question

Folosiți metoda Cramer pentru a rezolva.

E posibil de rezolvat prin metoda Gauss?

Va rog frumos. Urgent

Answer 1

$\left\{\begin{array}{ccc}x+2y+3z=6\\-x+3y-2z=0\\2x+y+3z=6\end{array}\right\\\\ Metoda~Cramer:\\\\ \Delta=\left|\begin{array}{ccc}1&2&3\\-1&3&-2\\2&1&3\end{array}\right|=1 \cdot 3 \cdot 3+3 \cdot (-1) \cdot 1+2\cdot(-2)\cdot 2-3\cdot 3 \cdot 2-\\\\~~~~~~~~~~~~~~~~~~~~~~~~~-2\cdot(-1)\cdot 3-1\cdot (-2) \cdot 1=9-3-8-18+6+2=-12$

$\Delta_1=\left|\begin{array}{ccc}6&2&3\\0&3&-2\\6&1&3\end{array}\right|=6 \cdot 3\cdot 3+3 \cdot 0 \cdot 1+2 \cdot (-2) \cdot 6-3 \cdot 3 \cdot 6-2 \cdot 0 \cdot 3-\\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~-6 \cdot (-2) \cdot 1=54+0-24-54-0+12=-12$

$\Delta_2=\left|\begin{array}{ccc}1&6&3\\-1&0&-2\\2&6&3\end{array}\right|=1 \cdot 0 \cdot 3+3 \cdot (-1) \cdot 6+6 \cdot(-2)\cdot 2-3 \cdot 0 \cdot 2-\\\\~~~~~~~~~~~~~~~~~~~~-6 \cdot(-1) \cdot 3-1\cdot(-2)\cdot 6=0-18-24-0+18+12=-12$

$\Delta_3=\left|\begin{array}{ccc}1&2&6\\-1&3&0\\2&1&6\end{array}\right|=1 \cdot 3 \cdot 6+6 \cdot(-1)\cdot 1+2 \cdot 0 \cdot 2-6 \cdot 3 \cdot 2-2 \cdot(-1)\cdot6-\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1 \cdot 0 \cdot 1=18-6+0-36+12-0=-12$

$\displaystyle x=\frac{\Delta_1}{\Delta} =\frac{-12}{-12} =1;~y=\frac{\Delta_2}{\Delta}=\frac{-12}{-12}=1;~z=\frac{\Delta_3}{\Delta}=\frac{-12}{-12} =1 \\\\ S=\{(1,1,1)\}$

$Metoda~Gauss:\\\\\[ \left( {\begin{array}{ccccc} 1 & 2 & 3\\ -1 & 3 & -2 \\ 2 & 1 & 3 \\ \end{array} } \right\] \left|\begin{array}{ccc}6\\0\\6\end{array}\right) \underrightarrow{L_2-(-1)\cdot L_1} \[ \left( {\begin{array}{ccccc} 1 & 2 & 3\\ 0 & 5 & 1 \\ 2 & 1 & 3 \\ \end{array} } \right\] \left|\begin{array}{ccc}6\\6\\6\end{array}\right)\underrightarrow{L_3-2L_1}$

$\underrightarrow {L_3-2L_1} \[ \left( {\begin{array}{ccccc} 1 & 2 & 3\\ 0 & 5 & 1 \\ 0 & -3 & -3 \\ \end{array} } \right\] \left|\begin{array}{ccc}6\\6\\-6\end{array}\right)\underrightarrow{L_3-(-\frac{3}{5})\cdot L_2 } \[ \left( {\begin{array}{ccccc} 1 & 2 & 3\\ 0 & 5 & 1 \\ 0 & 0 & -\frac{12}{5} \\ \end{array} } \right\] \left|\begin{array}{ccc}6\\6\\-\frac{12}{5} \end{array}\right)$

$\displaystyle \left\{\begin{array}{ccc}x+2y+3z=6\\~~~~5y+z=6\\-\frac{12}{5}z=-\frac{12}{5} \end{array}\right\\\\-\frac{12}{5}z=-\frac{12}{5}\Rightarrow z=\frac{\cfrac{12}{5} }{\cfrac{12}{5} } \Rightarrow z=\frac{12}{5}\cdot \frac{5}{12}\Rightarrow z=1\\\\ 5y+z=6\Rightarrow 5y+1=6 \Rightarrow 5y=6-1 \Rightarrow 5y=5 \Rightarrow y=\frac{5}{5}\Rightarrow y=1\\\\x+2y+3z=6\Rightarrow x+2 \cdot 1+3 \cdot 1=6\Rightarrow x+2+3=6 \Rightarrow x+5=6 \Rightarrow \\\\ \Rightarrow x=6-5\Rightarrow x=1\\\\S=\{(1,1,1)\}$