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Question Number 89534 by mathocean1 last updated on 17/Apr/20

Commented by abdomathmax last updated on 17/Apr/20
$1) we have e_1 (2,3) e_2 (1,2) and e_3 (4,5) let find x and y /e_3 =xe_1 +ye_2 ⇒ (4,5)=(2x,3x) +(y,2y)=(2x+y,3x+2y) ⇒ { ((2x+y =4)),((3x+2y=5)) :} Δ= determinant (((2 1)),((3 2)))=4−3=1≠0 ⇒ x=(Δ_x /Δ) and y=(Δ_y /Δ) Δ_x = determinant (((4 1)),((5 2)))=3 and Δ_y = determinant (((2 4)),((3 5)))=−2 ⇒ x=3 and y =−2 ⇒e_3 =3e_1 −2e_2 2) we have det(e_1 ,e_2 )= determinant (((2 1)),((3 1)))=−1≠0 ⇒ (e_1 ,e_2 )is a base of E_2 ^→$
$1) w e h a v e e_{1} (2, 3) e_{2} (1, 2) a n d e_{3} (4, 5)$ $l e t f i n d x a n d y / e_{3} = {x e}_{1} + {y e}_{2} \Rightarrow$ $(4, 5) = (2 x, 3 x) + (y, 2 y) = (2 x + y, 3 x + 2 y) \Rightarrow$ ${\begin{cases} 2 x + y = 4 \\ 3 x + 2 y = 5 \end{cases}$ $Δ = \| \begin{matrix} 2 1 \\ 3 2 \end{matrix} \| = 4 - 3 = 1 \neq 0 \Rightarrow$ $x = \frac{Δ_{x}}{Δ} a n d y = \frac{Δ_{y}}{Δ}$ $Δ_{x} = \| \begin{matrix} 4 1 \\ 5 2 \end{matrix} \| = 3 a n d Δ_{y} = \| \begin{matrix} 2 4 \\ 3 5 \end{matrix} \| = - 2 \Rightarrow$ $x = 3 a n d y = - 2 \Rightarrow e_{3} = 3 e_{1} - 2 e_{2}$ $2) w e h a v e d e t (e_{1}, e_{2}) = \| \begin{matrix} 2 1 \\ 3 1 \end{matrix} \| = - 1 \neq 0 \Rightarrow$ $(e_{1}, e_{2}) i s a b a s e o f {\vec{E}}_{2}$
Commented by mathocean1 last updated on 18/Apr/20

$thank you sir!$
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