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Question Number 1899 by Yozzy last updated on 22/Oct/15
Consider the system of equations 2yz+zx−5xy=2 yz−zx+2xy=1 yz−2zx+6xy=3. Show that xyz=±6 and find the possible values of x,y and z.
$${Consider}\:{the}\:{system}\:{of}\:{equations} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{yz}+{zx}−\mathrm{5}{xy}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{yz}−{zx}+\mathrm{2}{xy}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{yz}−\mathrm{2}{zx}+\mathrm{6}{xy}=\mathrm{3}. \\ $$$${Show}\:{that}\:{xyz}=\pm\mathrm{6}\: \\ $$$${and}\:{find}\:{the}\:{possible}\:{values} \\ $$$${of}\:{x},{y}\:{and}\:{z}. \\ $$
Commented by zainaltanjung last updated on 15/Oct/21
Consider the system of equations 2yz+zx−5xy=2 yz−zx+2xy=1 yz−2zx+6xy=3: Show that xyz=±6 and find the possible values of x;y and z
$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2yz}+\mathrm{zx}−\mathrm{5xy}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{yz}−\mathrm{zx}+\mathrm{2xy}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{yz}−\mathrm{2zx}+\mathrm{6xy}=\mathrm{3}: \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{xyz}=\pm\mathrm{6}\: \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{values} \\ $$$$\mathrm{of}\:\mathrm{x};\mathrm{y}\:\mathrm{and}\:\mathrm{z} \\ $$
Answered by 123456 last updated on 23/Oct/15
{ ((2yz+zx−5xy=2)),((yz−zx+2xy=1)),((yz−2zx+6xy=3)) :} a=yz,b=zx,c=xy { ((2a+b−5c=2)),((a−b+2c=1)),((a−2b+6c=3)) :} abc=yzzxxy=(xyz)^2 xyz=±(√(abc)) Δ= determinant ((2,1,(−5)),(1,(−1),2),(1,(−2),6))=−12+2+10−5−6+8=−3 Δa= determinant ((2,1,(−5)),(1,(−1),2),(3,(−2),6))=−12+6+10−15−6+8=−9 Δb= determinant ((2,2,(−5)),(1,1,2),(1,3,6))=12+4−15+5−12−12=−18 Δc= determinant ((2,1,2),(1,(−1),1),(1,(−2),3))=−6+1−4+2−3+4=−6 a=((Δa)/Δ)=((−9)/(−3))=3 b=((Δb)/Δ)=((−18)/(−3))=6 c=((Δc)/Δ)=((−6)/(−3))=2 xyz=±(√(abc))=±(√(64))=±6 { ((yz=3)),((zx=6)),((xy=2)) :} continue x=(2/y)=(6/z)⇔6y=2z y=(2/x)=(3/z)⇔3x=2z z=(3/y)=(6/x)⇔3x=6y 3x=6y=2z=t (x,y,z)=^? ((t/3),(t/6),(t/2)) { ((yz=(t^2 /(12))=3⇒t=±6)),((zx=(t^2 /6)=6⇒t=±6)),((xy=(t^2 /(18))=2⇒t=±6)) :} (x,y,z)=(±2,±1,±3)
$$\begin{cases}{\mathrm{2}{yz}+{zx}−\mathrm{5}{xy}=\mathrm{2}}\\{{yz}−{zx}+\mathrm{2}{xy}=\mathrm{1}}\\{{yz}−\mathrm{2}{zx}+\mathrm{6}{xy}=\mathrm{3}}\end{cases} \\ $$$${a}={yz},{b}={zx},{c}={xy} \\ $$$$\begin{cases}{\mathrm{2}{a}+{b}−\mathrm{5}{c}=\mathrm{2}}\\{{a}−{b}+\mathrm{2}{c}=\mathrm{1}}\\{{a}−\mathrm{2}{b}+\mathrm{6}{c}=\mathrm{3}}\end{cases} \\ $$$${abc}={yzzxxy}=\left({xyz}\right)^{\mathrm{2}} \\ $$$${xyz}=\pm\sqrt{{abc}} \\ $$$$\Delta=\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{−\mathrm{5}}\\{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{6}}\end{vmatrix}=−\mathrm{12}+\mathrm{2}+\mathrm{10}−\mathrm{5}−\mathrm{6}+\mathrm{8}=−\mathrm{3} \\ $$$$\Delta{a}=\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{−\mathrm{5}}\\{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{3}}&{−\mathrm{2}}&{\mathrm{6}}\end{vmatrix}=−\mathrm{12}+\mathrm{6}+\mathrm{10}−\mathrm{15}−\mathrm{6}+\mathrm{8}=−\mathrm{9} \\ $$$$\Delta{b}=\begin{vmatrix}{\mathrm{2}}&{\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{3}}&{\mathrm{6}}\end{vmatrix}=\mathrm{12}+\mathrm{4}−\mathrm{15}+\mathrm{5}−\mathrm{12}−\mathrm{12}=−\mathrm{18} \\ $$$$\Delta{c}=\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{3}}\end{vmatrix}=−\mathrm{6}+\mathrm{1}−\mathrm{4}+\mathrm{2}−\mathrm{3}+\mathrm{4}=−\mathrm{6} \\ $$$${a}=\frac{\Delta{a}}{\Delta}=\frac{−\mathrm{9}}{−\mathrm{3}}=\mathrm{3} \\ $$$${b}=\frac{\Delta{b}}{\Delta}=\frac{−\mathrm{18}}{−\mathrm{3}}=\mathrm{6} \\ $$$${c}=\frac{\Delta{c}}{\Delta}=\frac{−\mathrm{6}}{−\mathrm{3}}=\mathrm{2} \\ $$$${xyz}=\pm\sqrt{{abc}}=\pm\sqrt{\mathrm{64}}=\pm\mathrm{6} \\ $$$$\begin{cases}{{yz}=\mathrm{3}}\\{{zx}=\mathrm{6}}\\{{xy}=\mathrm{2}}\end{cases} \\ $$$$\mathrm{continue} \\ $$$${x}=\frac{\mathrm{2}}{{y}}=\frac{\mathrm{6}}{{z}}\Leftrightarrow\mathrm{6}{y}=\mathrm{2}{z} \\ $$$${y}=\frac{\mathrm{2}}{{x}}=\frac{\mathrm{3}}{{z}}\Leftrightarrow\mathrm{3}{x}=\mathrm{2}{z} \\ $$$${z}=\frac{\mathrm{3}}{{y}}=\frac{\mathrm{6}}{{x}}\Leftrightarrow\mathrm{3}{x}=\mathrm{6}{y} \\ $$$$\mathrm{3}{x}=\mathrm{6}{y}=\mathrm{2}{z}={t} \\ $$$$\left({x},{y},{z}\right)\overset{?} {=}\left(\frac{{t}}{\mathrm{3}},\frac{{t}}{\mathrm{6}},\frac{{t}}{\mathrm{2}}\right) \\ $$$$\begin{cases}{{yz}=\frac{{t}^{\mathrm{2}} }{\mathrm{12}}=\mathrm{3}\Rightarrow{t}=\pm\mathrm{6}}\\{{zx}=\frac{{t}^{\mathrm{2}} }{\mathrm{6}}=\mathrm{6}\Rightarrow{t}=\pm\mathrm{6}}\\{{xy}=\frac{{t}^{\mathrm{2}} }{\mathrm{18}}=\mathrm{2}\Rightarrow{t}=\pm\mathrm{6}}\end{cases} \\ $$$$\left({x},{y},{z}\right)=\left(\pm\mathrm{2},\pm\mathrm{1},\pm\mathrm{3}\right) \\ $$

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