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Question Number 25975 by mariahameed97@gmail.com last updated on 17/Dec/17

Answered by Rasheed.Sindhi last updated on 17/Dec/17

$$\mathrm{x}+\mathrm{y}+\mathrm{z}=6$$$$\mathrm{x}-\mathrm{y}+\mathrm{z}=2$$$$2\mathrm{x}-\mathrm{y}+3\mathrm{z}=6$$$$\mathrm{D}=\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 2& -1& 3\end{array}\right|$$$$=\left|\begin{array}{ccc}1& 0& 0\\ 1& -2& 0\\ 2& -3& 1\end{array}\right|{\mathrm{C}}_2-{\mathrm{C}}_1,{\mathrm{C}}_3-{\mathrm{C}}_1$$$$=\left|\begin{array}{cc}-2& 0\\ -3& 1\end{array}\right|=-2$$$$-------$$$${\mathrm{D}}_{\mathrm{x}}=\left|\begin{array}{ccc}6& 1& 1\\ 2& -1& 1\\ 6& -1& 3\end{array}\right|$$$$=\left|\begin{array}{ccc}8& 0& 2\\ 2& -1& 1\\ 4& 0& 2\end{array}\right|{\mathrm{R}}_1+{\mathrm{R}}_2,{\mathrm{R}}_3-{\mathrm{R}}_2$$$$=\left|\begin{array}{ccc}4& 0& 0\\ 2& -1& 1\\ 4& 0& 2\end{array}\right|{\mathrm{R}}_1-{\mathrm{R}}_3$$$$=4\left|\begin{array}{cc}-1& 1\\ 0& 2\end{array}\right|=4(-2)=-8$$$$--------$$$${\mathrm{D}}_{\mathrm{y}}=\left|\begin{array}{ccc}1& 6& 1\\ 1& 2& 1\\ 2& 6& 3\end{array}\right|$$$$=\left|\begin{array}{ccc}0& 4& 0\\ 1& 2& 1\\ 1& 4& 2\end{array}\right|{\mathrm{R}}_1-{\mathrm{R}}_2,{\mathrm{R}}_3-{\mathrm{R}}_2$$$$=-4\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|=-4(2-1)=-4$$$$--------$$$${\mathrm{D}}_{\mathrm{z}}=\left|\begin{array}{ccc}1& 1& 6\\ 1& -1& 2\\ 2& -1& 6\end{array}\right|$$$$=\left|\begin{array}{ccc}1& 0& 0\\ 1& -2& -4\\ 2& -3& -6\end{array}\right|{\mathrm{C}}_2-{\mathrm{C}}_1,{\mathrm{C}}_3-{6\mathrm{C}}_1$$$$=\left|\begin{array}{cc}-2& -4\\ -3& -6\end{array}\right|=12-12=0$$$$---------$$$$\mathrm{x}=\frac{{\mathrm{D}}_{\mathrm{x}}}{\mathrm{D}},\mathrm{y}=\frac{{\mathrm{D}}_{\mathrm{y}}}{\mathrm{D}},\mathrm{z}=\frac{{\mathrm{D}}_{\mathrm{z}}}{\mathrm{D}}$$$$\mathrm{x}=\frac{-8}{-2},\mathrm{y}=\frac{-4}{-2},\mathrm{z}=\frac{0}{-2}$$$$\mathrm{x}=4,\mathrm{y}=2,\mathrm{z}=0$$

Commented by mariahameed97@gmail.com last updated on 17/Dec/17

$$\mathrm{thnxx}$$

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