Matematică
Viviana3333
2016-10-01 10:51:42
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
Răspunsuri la întrebare
nasuioanamaria
2016-10-01 15:31:47

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dariusflorin
2016-10-01 15:33:02

[latex]displaystyle mathtt{ left{egin{array}{ccc}mathtt{3x+y+z=1}\mathtt{x+3y+z=2}\mathtt{x+y+3z=2}end{array} ight }\ \ mathtt{Delta= left|egin{array}{ccc}mathtt3&mathtt1&mathtt1\mathtt1&mathtt3&mathtt1\mathtt1&mathtt1&mathtt3end{array} ight| =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 ot = 0}[/latex] [latex]displaystyle mathtt{Delta_x=left|egin{array}{ccc}mathtt1&mathtt1&mathtt1\mathtt2&mathtt3&mathtt1\mathtt2&mathtt1&mathtt3end{array} ight|=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}[/latex] [latex]displaystyle mathtt{Delta_y=left|egin{array}{ccc}mathtt3&mathtt1&mathtt1\mathtt1&mathtt2&mathtt1\mathtt1&mathtt2&mathtt3end{array} ight| =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}[/latex] [latex]displaystyle mathtt{Delta_z=left|egin{array}{ccc}mathtt3&mathtt1&mathtt1\mathtt1&mathtt3&mathtt2\mathtt1&mathtt1&mathtt2end{array} ight|=3 cdot 3 cdot 2+1 cdot 1 cdot 1+1cdot 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}[/latex] [latex]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} }[/latex]

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