Question

Utilizand regula lui Cromer sa se rezolve sistemul de exuatii
3x+y+z=1
{ x+3y+z=2
x+y+3z=2
in acualada sunt incluse toate 3 rinduri,in 1 intreaga
Am nevoie repede, dau 50 baluri

Answer 1

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Answer 2

$\displaystyle \mathtt{ \left\{\begin{array}{ccc}\mathtt{3x+y+z=1}\\\mathtt{x+3y+z=2}\\\mathtt{x+y+3z=2}\end{array}\right }\\ \\ \mathtt{\Delta= \left|\begin{array}{ccc}\mathtt3&\mathtt1&\mathtt1\\\mathtt1&\mathtt3&\mathtt1\\\mathtt1&\mathtt1&\mathtt3\end{array}\right| =3 \cdot 3 \cdot 3+1 \cdot 1 \cdot 1+1 \cdot 1 \cdot 1-1 \cdot 3 \cdot 1-1 \cdot 1 \cdot 3-}\\ \\ \mathtt{-3 \cdot 1 \cdot 1=27+1+1-3-3-3=20}\\ \\ \mathtt{\Delta=20 \not = 0}$

$\displaystyle \mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt1\\\mathtt2&\mathtt3&\mathtt1\\\mathtt2&\mathtt1&\mathtt3\end{array}\right|=1 \cdot 3 \cdot 3+1 \cdot 2 \cdot 1+1 \cdot 1 \cdot 2-1 \cdot 3 \cdot 2-1 \cdot 2 \cdot 3- } \\ \\ \mathtt{-1 \cdot 1 \cdot 1=9+2+2-6-6-1=0}\\ \\ \mathtt{\Delta_x=0}$

$\displaystyle \mathtt{\Delta_y=\left|\begin{array}{ccc}\mathtt3&\mathtt1&\mathtt1\\\mathtt1&\mathtt2&\mathtt1\\\mathtt1&\mathtt2&\mathtt3\end{array}\right| =3 \cdot 2 \cdot 3+1 \cdot 1 \cdot 2+1 \cdot 1 \cdot 1-1 \cdot 2 \cdot 1-1 \cdot 1 \cdot 3-}\\ \\ \mathtt{-3 \cdot 1 \cdot 2=18+2+1-2-3-6=10}\\ \\ \mathtt{\Delta_y=10}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt3&\mathtt1&\mathtt1\\\mathtt1&\mathtt3&\mathtt2\\\mathtt1&\mathtt1&\mathtt2\end{array}\right|=3 \cdot 3 \cdot 2+1 \cdot 1 \cdot 1+1\cdot 2 \cdot 1-1 \cdot 3 \cdot 1-1 \cdot 1 \cdot 2-}\\ \\ \mathtt{-3 \cdot 2 \cdot 1=18+1+2-3-2-6=10}\\ \\ \mathtt{\Delta_z=10}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta} = \frac{0}{20} =0}\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta}= \frac{10}{20} = \frac{1}{2} }\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta} = \frac{10}{20} = \frac{1}{2} }\\ \\ \mathtt{x=0;~y= \frac{1}{2} ;~z= \frac{1}{2} }$