Question

Resolver o exercicio de algebra linear, alguem me ajuda?????

Answer 1

Explicação passo-a-passo:

$v = (1, - 2,k) \\ \\ u1 = (3,0, - 2) \\ \\u2 = (2, - 1, - 5) \\ \\ u3 = (0, 0, 1) \\ \\ (1, - 2,k) = a(3,0, - 2) + b(2, - 1, - 5) + c(0, 0, 1) \\ \\ a(3,0, - 2) + b(2, - 1, - 5) + c(0, 0, 1) = (1, - 2,k) \\ \\ (3a, 0, - 2a) + (2b, - b, - 5b) + (0, 0, c) = (1, - 2,k) \\ \\ (3a + 2b, - b, - 7ab) + (0, 0, c) = (1, - 2,k) \\ \\ (3a +2b , - b, - 7ab + c) = (1, - 2,k) \\ \\ \left\{{{ 3a +2b = 1}\atop{- b = - 2}}\atop{- 7ab + c = k}\right. \\ \\ \boxed{b = 2 }\\ \\ 3a +2b = 1 \\ \\ 3a + 2(2) = 1 \\ \\ 3a + 4 = 1 \\ \\ 3a = 1 - 4 \\ \\ 3a = - 3 \\ \\ \boxed{a = - 1 }\\ \\ - 7ab + c = k \\ \\ - 7( - 1)(2) + c = k \\ \\ 14 + c = k \\ \\\Large \boxed{\green{ \bf k = 14 + c}}$

$\Large \boxed{\underline{\bf \blue{Bons \:Estudos!}}}\\ \\ \Large \boxed{\underline{\bf26/05/2021}}$

Answer 2

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That's it!