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Solve-2x-1-1-5-2-0-2x-2-1-2-2-0-2x-3-3-1-2-0-Find-the-values-of-x-1-x-2-x-3-1-and-2-

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Question Number 208645 by Mastermind last updated on 20/Jun/24
Solve : 2x_1 − λ_1 − 5λ_2 = 0 2x_2 − λ_1 − 2λ_2 = 0 2x_3 − 3λ_1 − λ_2 = 0 Find the values of x_1 , x_2 , x_3 , λ_1 , and λ_2
$$\mathrm{Solve}\:: \\ $$$$\mathrm{2x}_{\mathrm{1}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{5}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{2}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{2}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{3}} \:−\:\mathrm{3}\lambda_{\mathrm{1}} \:−\:\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}_{\mathrm{1}} ,\:\mathrm{x}_{\mathrm{2}} ,\:\mathrm{x}_{\mathrm{3}} ,\:\lambda_{\mathrm{1}} ,\:\mathrm{and}\:\lambda_{\mathrm{2}} \\ $$
Answered by Frix last updated on 20/Jun/24
3 equations with 5 unknown, seriously? ((x_1 ),(x_2 ),(x_3 ) ) =(λ_1 /2) ((1),(1),(3) ) +(λ_2 /2) ((5),(2),(1) ) This describes a plane in R^3 . We can eliminate λ_1 , λ_2 to get 5x−14y+3z=0 Use the cross product to get the normal vector: ((1),(1),(3) ) × ((5),(2),(1) ) = (((−5)),((14)),((−3)) )
$$\mathrm{3}\:\mathrm{equations}\:\mathrm{with}\:\mathrm{5}\:\mathrm{unknown},\:\mathrm{seriously}? \\ $$$$\begin{pmatrix}{{x}_{\mathrm{1}} }\\{{x}_{\mathrm{2}} }\\{{x}_{\mathrm{3}} }\end{pmatrix}\:=\frac{\lambda_{\mathrm{1}} }{\mathrm{2}}\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:+\frac{\lambda_{\mathrm{2}} }{\mathrm{2}}\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\: \\ $$$$\mathrm{This}\:\mathrm{describes}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} .\:\mathrm{We}\:\mathrm{can}\:\mathrm{eliminate} \\ $$$$\lambda_{\mathrm{1}} ,\:\lambda_{\mathrm{2}} \:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{5}{x}−\mathrm{14}{y}+\mathrm{3}{z}=\mathrm{0} \\ $$$$\mathrm{Use}\:\mathrm{the}\:\mathrm{cross}\:\mathrm{product}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{normal}\:\mathrm{vector}: \\ $$$$\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:=\begin{pmatrix}{−\mathrm{5}}\\{\mathrm{14}}\\{−\mathrm{3}}\end{pmatrix} \\ $$
Commented by Mastermind last updated on 23/Jun/24
Thank you but kindly help me value. I appreciate
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{but}\:\mathrm{kindly}\:\mathrm{help}\:\mathrm{me}\:\mathrm{value}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$
Commented by Frix last updated on 23/Jun/24
The x_i depend on λ_1 , λ_2 ∈R, there are infinite values.
$$\mathrm{The}\:{x}_{{i}} \:\mathrm{depend}\:\mathrm{on}\:\lambda_{\mathrm{1}} ,\:\lambda_{\mathrm{2}} \:\in\mathbb{R},\:\mathrm{there}\:\mathrm{are}\:\mathrm{infinite} \\ $$$$\mathrm{values}. \\ $$

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