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Question Number 208645 by Mastermind last updated on 20/Jun/24

$$\mathrm{Solve}:$$$${2\mathrm{x}}_1-{\lambda }_1-5{\lambda }_2=0$$$${2\mathrm{x}}_2-{\lambda }_1-2{\lambda }_2=0$$$${2\mathrm{x}}_3-3{\lambda }_1-{\lambda }_2=0$$$$$$$$\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}{\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\lambda }_1,\mathrm{and}{\lambda }_2$$

Answered by Frix last updated on 20/Jun/24

$$3\mathrm{equations}\mathrm{with}5\mathrm{unknown},\mathrm{seriously}?$$$$\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\frac{{\lambda }_1}{2}\left(\begin{array}{c}1\\ 1\\ 3\end{array}\right)+\frac{{\lambda }_2}{2}\left(\begin{array}{c}5\\ 2\\ 1\end{array}\right)$$$$\mathrm{This}\mathrm{describes}\mathrm{a}\mathrm{plane}\mathrm{in}{\mathbb{R}}^3.\mathrm{We}\mathrm{can}\mathrm{eliminate}$$$${\lambda }_1,{\lambda }_2\mathrm{to}\mathrm{get}$$$$5x-14y+3z=0$$$$\mathrm{Use}\mathrm{the}\mathrm{cross}\mathrm{product}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{normal}\mathrm{vector}:$$$$\left(\begin{array}{c}1\\ 1\\ 3\end{array}\right)\times \left(\begin{array}{c}5\\ 2\\ 1\end{array}\right)=\left(\begin{array}{c}-5\\ 14\\ -3\end{array}\right)$$

Commented by Mastermind last updated on 23/Jun/24

$$\mathrm{Thank}\mathrm{you}\mathrm{but}\mathrm{kindly}\mathrm{help}\mathrm{me}\mathrm{value}.\mathrm{I}\mathrm{appreciate}$$

Commented by Frix last updated on 23/Jun/24

$$\mathrm{The}{x}_i\mathrm{depend}\mathrm{on}{\lambda }_1,{\lambda }_2\in \mathbb{R},\mathrm{there}\mathrm{are}\mathrm{infinite}$$$$\mathrm{values}.$$

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