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

Commented by mathocean1 last updated on 13/Apr/20

$thank you sir .$
Commented by mathmax by abdo last updated on 13/Apr/20
$2i+3j=e_1 and i−2j =e_2 ⇒ { ((2i+3j =e_1 )),((i−2j =e_2 )) :} Δ = determinant (((2 3)),((1 −2)))=−7 ⇒i =(Δ_i /Δ) and j =(Δ_j /Δ) Δ_i = determinant (((e_(1 ) 3)),((e_2 −2)))=−2e_1 −3e_2 Δ_j = determinant ((( 2 e_1 )),((1 e_2 )))=2e_2 −e_1 ⇒i =−(1/7)(−2e_1 −3e_2 )=(2/7)e_1 +(3/7)e_2 j =−(1/7)(−e_1 +2e_2 ) =(1/7)e_1 −(2/7)e_2 ⇒ e_3 =4i −5j =(8/7)e_1 +((12)/7)e_2 −(5/7)e_1 +((10)/7)e_2 =(3/7)e_1 +((22)/7)e_2 ⇒ e_3 ((3/7),((22)/7)) in the base (e_1 ,e_2 )$
$2 i + 3 j = e_{1} a n d i - 2 j = e_{2}$ $\Rightarrow {\begin{cases} 2 i + 3 j = e_{1} \\ i - 2 j = e_{2} \end{cases}$ $Δ = \| \begin{matrix} 2 3 \\ 1 - 2 \end{matrix} \| = - 7 \Rightarrow i = \frac{Δ_{i}}{Δ} a n d j = \frac{Δ_{j}}{Δ}$ $Δ_{i} = \| \begin{matrix} e_{1} 3 \\ e_{2} - 2 \end{matrix} \| = - 2 e_{1} - 3 e_{2}$ $Δ_{j} = \| \begin{matrix} 2 e_{1} \\ 1 e_{2} \end{matrix} \| = 2 e_{2} - e_{1} \Rightarrow i = - \frac{1}{7} (- 2 e_{1} - 3 e_{2}) = \frac{2}{7} e_{1} + \frac{3}{7} e_{2}$ $j = - \frac{1}{7} (- e_{1} + 2 e_{2}) = \frac{1}{7} e_{1} - \frac{2}{7} e_{2} \Rightarrow$ $e_{3} = 4 i - 5 j = \frac{8}{7} e_{1} + \frac{12}{7} e_{2} - \frac{5}{7} e_{1} + \frac{10}{7} e_{2} = \frac{3}{7} e_{1} + \frac{22}{7} e_{2} \Rightarrow$ $e_{3} (\frac{3}{7}, \frac{22}{7}) i n t h e b a s e (e_{1}, e_{2})$
Commented by mathmax by abdo last updated on 13/Apr/20

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