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Question Number 67189 by mathmax by abdo last updated on 23/Aug/19

$$solveinside{R}^{3}thesystem\{\begin{array}{l}2x+y+z=1\\ x+2y+z=2\end{array}$$$$\{x+y+2z=3$$

Answered by MJS last updated on 23/Aug/19

$$D=\left|\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right|=4$$$$D_x=\left|\begin{array}{ccc}1& 1& 1\\ 2& 2& 1\\ 3& 1& 2\end{array}\right|=-2x=\frac{D_x}{D}=-\frac{1}{2}$$$$D_y=\left|\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 3& 2\end{array}\right|=2y=\frac{D_y}{D}=\frac{1}{2}$$$$D_z=\left|\begin{array}{ccc}2& 1& 1\\ 1& 2& 2\\ 1& 1& 3\end{array}\right|=6z=\frac{D_z}{D}=\frac{3}{2}$$

Commented by turbo msup by abdo last updated on 24/Aug/19

$$thankyousirmjs.$$

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