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solve-inside-R-3-the-system-2x-y-z-1-x-2y-z-2-x-y-2z-3-




Question Number 67189 by mathmax by abdo last updated on 23/Aug/19
solve inside R^3  the system  { ((2x+y+z =1)),((x+2y+z =2)) :}                                                               {x+y+2z =3
$${solve}\:{inside}\:{R}^{\mathrm{3}} \:{the}\:{system}\:\begin{cases}{\mathrm{2}{x}+{y}+{z}\:=\mathrm{1}}\\{{x}+\mathrm{2}{y}+{z}\:=\mathrm{2}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{x}+{y}+\mathrm{2}{z}\:=\mathrm{3}\right. \\ $$
Answered by MJS last updated on 23/Aug/19
D= determinant ((2,1,1),(1,2,1),(1,1,2))=4  D_x = determinant ((1,1,1),(2,2,1),(3,1,2))=−2     x=(D_x /D)=−(1/2)  D_y = determinant ((2,1,1),(1,2,1),(1,3,2))=2     y=(D_y /D)=(1/2)  D_z = determinant ((2,1,1),(1,2,2),(1,1,3))=6     z=(D_z /D)=(3/2)
$${D}=\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{2}}\end{vmatrix}=\mathrm{4} \\ $$$${D}_{{x}} =\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{2}}\end{vmatrix}=−\mathrm{2}\:\:\:\:\:{x}=\frac{{D}_{{x}} }{{D}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${D}_{{y}} =\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{3}}&{\mathrm{2}}\end{vmatrix}=\mathrm{2}\:\:\:\:\:{y}=\frac{{D}_{{y}} }{{D}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${D}_{{z}} =\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}\end{vmatrix}=\mathrm{6}\:\:\:\:\:{z}=\frac{{D}_{{z}} }{{D}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by turbo msup by abdo last updated on 24/Aug/19
thank you sir mjs.
$${thank}\:{you}\:{sir}\:{mjs}. \\ $$

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