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Question Number 88778 by mathocean1 last updated on 12/Apr/20

Commented by mathocean1 last updated on 13/Apr/20

thank you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by mathmax by abdo last updated on 13/Apr/20

2i+3j=e_1 and i−2j =e_2 ⇒ { ((2i+3j =e_1 )),((i−2j =e_2 )) :} Δ = determinant (((2 3)),((1 −2)))=−7 ⇒i =(Δ_i /Δ) and j =(Δ_j /Δ) Δ_i = determinant (((e_(1 ) 3)),((e_2 −2)))=−2e_1 −3e_2 Δ_j = determinant ((( 2 e_1 )),((1 e_2 )))=2e_2 −e_1 ⇒i =−(1/7)(−2e_1 −3e_2 )=(2/7)e_1 +(3/7)e_2 j =−(1/7)(−e_1 +2e_2 ) =(1/7)e_1 −(2/7)e_2 ⇒ e_3 =4i −5j =(8/7)e_1 +((12)/7)e_2 −(5/7)e_1 +((10)/7)e_2 =(3/7)e_1 +((22)/7)e_2 ⇒ e_3 ((3/7),((22)/7)) in the base (e_1 ,e_2 )

$$\mathrm{2}{i}+\mathrm{3}{j}={e}_{\mathrm{1}} {and}\:{i}−\mathrm{2}{j}\:={e}_{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}{i}+\mathrm{3}{j}\:={e}_{\mathrm{1}} }\\{{i}−\mathrm{2}{j}\:={e}_{\mathrm{2}} }\end{cases} \\ $$$$\Delta\:=\begin{vmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{1}\:\:\:\:\:\:\:\:−\mathrm{2}}\end{vmatrix}=−\mathrm{7}\:\Rightarrow{i}\:=\frac{\Delta_{{i}} }{\Delta}\:{and}\:{j}\:=\frac{\Delta_{{j}} }{\Delta} \\ $$$$\Delta_{{i}} =\begin{vmatrix}{{e}_{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:} \:\:\mathrm{3}}\\{{e}_{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:−\mathrm{2}}\end{vmatrix}=−\mathrm{2}{e}_{\mathrm{1}} −\mathrm{3}{e}_{\mathrm{2}} \\ $$$$\Delta_{{j}} =\begin{vmatrix}{\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:{e}_{\mathrm{1}} }\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:{e}_{\mathrm{2}} }\end{vmatrix}=\mathrm{2}{e}_{\mathrm{2}} −{e}_{\mathrm{1}} \:\Rightarrow{i}\:=−\frac{\mathrm{1}}{\mathrm{7}}\left(−\mathrm{2}{e}_{\mathrm{1}} −\mathrm{3}{e}_{\mathrm{2}} \right)=\frac{\mathrm{2}}{\mathrm{7}}{e}_{\mathrm{1}} +\frac{\mathrm{3}}{\mathrm{7}}{e}_{\mathrm{2}} \\ $$$${j}\:=−\frac{\mathrm{1}}{\mathrm{7}}\left(−{e}_{\mathrm{1}} +\mathrm{2}{e}_{\mathrm{2}} \right)\:=\frac{\mathrm{1}}{\mathrm{7}}{e}_{\mathrm{1}} −\frac{\mathrm{2}}{\mathrm{7}}{e}_{\mathrm{2}} \:\Rightarrow \\ $$$${e}_{\mathrm{3}} =\mathrm{4}{i}\:−\mathrm{5}{j}\:=\frac{\mathrm{8}}{\mathrm{7}}{e}_{\mathrm{1}} +\frac{\mathrm{12}}{\mathrm{7}}{e}_{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{7}}{e}_{\mathrm{1}} +\frac{\mathrm{10}}{\mathrm{7}}{e}_{\mathrm{2}} \:=\frac{\mathrm{3}}{\mathrm{7}}{e}_{\mathrm{1}} +\frac{\mathrm{22}}{\mathrm{7}}{e}_{\mathrm{2}} \:\Rightarrow \\ $$$${e}_{\mathrm{3}} \left(\frac{\mathrm{3}}{\mathrm{7}},\frac{\mathrm{22}}{\mathrm{7}}\right)\:{in}\:{the}\:{base}\:\left({e}_{\mathrm{1}} ,{e}_{\mathrm{2}} \right) \\ $$

Commented by mathmax by abdo last updated on 13/Apr/20

you are welcome

$${you}\:{are}\:{we}\boldsymbol{{lcome}} \\ $$

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