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Question Number 161429 by greg_ed last updated on 17/Dec/21
$$\mathrm{help}\:\mathrm{me}\:! \\ $$$$\begin{cases}{{x}+\mathrm{3}{y}+{z}=\mathrm{2}}\\{−\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{2}{z}=\mathrm{3}}\\{−\mathrm{2}{x}+\mathrm{7}{y}+\mathrm{3}{z}=\mathrm{5}}\end{cases} \\ $$$$\boldsymbol{\mathrm{G}}\mathrm{auss}\:\mathrm{Method}... \\ $$
Commented by 1549442205PVT last updated on 18/Dec/21
$${the}\:{given}\:{system}\:{of}\:{equations}\:{is}\:{not} \\ $$$${compatible}\:{because} \\ $$$$\bigtriangleup=\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{1}}\\{−\mathrm{3}\:\:\:\mathrm{4}\:\:\:\:\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\mathrm{7}\:\:\:\mathrm{3}}\end{vmatrix}=\mathrm{12}−\mathrm{12}−\mathrm{21}+\mathrm{8}+\mathrm{27}−\mathrm{14}=\mathrm{0} \\ $$$${the}\:{system}\:{is}\:{equivalent}\:{to} \\ $$$$\begin{cases}{{x}+\mathrm{3}{y}+{z}=\mathrm{2}}\\{\mathrm{13}{y}+\mathrm{5}{z}=\mathrm{9}}\end{cases}\Rightarrow\begin{cases}{{x}=\frac{\mathrm{2}{z}−\mathrm{1}}{\mathrm{13}}}\\{{y}=\frac{\mathrm{9}−\mathrm{5}{z}}{\mathrm{13}}}\end{cases} \\ $$$${or\begin{cases}{{x}=\frac{\mathrm{2}{t}−\mathrm{1}}{\mathrm{13}}}\\{{y}=\frac{\mathrm{9}−\mathrm{5}{t}}{\mathrm{13}}}\\{{z}={t}}\end{cases}}\:\:\forall{t}\in{R} \\ $$