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Question Number 15233 by tawa tawa last updated on 08/Jun/17

$$\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{homogeneous}\mathrm{system}\mathrm{of}$$$$\mathrm{equation}$$$${\mathrm{x}}_1+{\mathrm{x}}_2-{2\mathrm{x}}_3=0$$$${3\mathrm{x}}_1-{\mathrm{x}}_2-{6\mathrm{x}}_3=0$$$$-{2\mathrm{x}}_1+{3\mathrm{x}}_2+{4\mathrm{x}}_3=0$$

Answered by arnabpapu550@gmail.com last updated on 08/Jun/17

$$\mathrm{The}\mathrm{given}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{the}$$$$\mathrm{following}\mathrm{matrix},\mathrm{AX}=\mathrm{O}$$$$\mathrm{where}\mathrm{A}=\left[\begin{array}{ccc}1& 1& -2\\ 3& -1& -6\\ -2& 3& 4\end{array}\right]\mathrm{but}\det \left(\mathrm{A}\right)=2\ne 0$$$$\mathrm{Hence}\mathrm{the}\mathrm{equations}\mathrm{are}\mathrm{consistant}.$$$$\mathrm{X}=\left(\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right)\mathrm{and}\mathrm{O}=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$$$\therefore \mathrm{X}={\mathrm{A}}^{-1}.\mathrm{O}=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$$$\therefore {\mathrm{x}}_1=0,{\mathrm{x}}_2=0\mathrm{and}{\mathrm{x}}_3=0$$$$\mathrm{Is}\mathrm{it}\mathrm{right}\mathrm{Tawa}\mathrm{Tawa}?$$

Commented by tawa tawa last updated on 08/Jun/17

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{too},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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