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Question Number 165778 by Odhiambojr last updated on 08/Feb/22

solve for the values of m,n and l  l+m+n=0  l^2 +m^2 −n^2 =0

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{m},\mathrm{n}\:\mathrm{and}\:\mathrm{l} \\ $$$$\mathrm{l}+\mathrm{m}+\mathrm{n}=\mathrm{0} \\ $$$$\mathrm{l}^{\mathrm{2}} +\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} =\mathrm{0} \\ $$

Answered by TheSupreme last updated on 08/Feb/22

l=−m−n  l^2 =m^2 +n^2 +2mn  2m^2 +2mn=0  m(2m+2n)=0  m=−n or m=0    m=−n → l=0 (l,m,n)=m(0,1,−1)    m=0  l+n=0→l(1,0,−1)

$${l}=−{m}−{n} \\ $$$${l}^{\mathrm{2}} ={m}^{\mathrm{2}} +{n}^{\mathrm{2}} +\mathrm{2}{mn} \\ $$$$\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{mn}=\mathrm{0} \\ $$$${m}\left(\mathrm{2}{m}+\mathrm{2}{n}\right)=\mathrm{0} \\ $$$${m}=−{n}\:{or}\:{m}=\mathrm{0} \\ $$$$ \\ $$$${m}=−{n}\:\rightarrow\:{l}=\mathrm{0}\:\left({l},{m},{n}\right)={m}\left(\mathrm{0},\mathrm{1},−\mathrm{1}\right) \\ $$$$ \\ $$$${m}=\mathrm{0} \\ $$$${l}+{n}=\mathrm{0}\rightarrow{l}\left(\mathrm{1},\mathrm{0},−\mathrm{1}\right) \\ $$

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