Question

ÁLGEBRA LINEAR
T:R²⇒R³ tal que T(-1 , 1)=(3, 2, 1) e T(0,1)=(1, 1, 0).


Encontre v ∈ R² tal que T(v)= (-2, 1, -3).




Gabarito: v=(3, 4)

Answer 1

Resposta:

T:R²⇒R³ tal que T(-1 , 1)=(3, 2, 1) e T(0,1)=(1, 1, 0

Seja (x,y) pertence R²

(x,y)=a*(-1,1)+b*(0,1)

(x,y)=-(-a,a)+(0,b)

-a+0=x ==>a=-x

y=a+b ==>b=y-a ==>b=y+x

(x,y)=-x*(-1,1)+(x+y)*(0,1)

T(x,y)=-x*T(-1,1)+(x+y)*T(0,1)

T(x,y)=-x*(3, 2, 1)+(x+y)*(1, 1, 0)

Sabendo que T(v)=(-2,1,-3)

(-2,1,-3)==-x*(3, 2, 1)+(x+y)*(1, 1, 0)

-2=-3x+x+y ==>-2=-2x+y ==>y=4

1=-2x+x+y ==>1=-x+y  ==>y=4

-3=-x ==>x=3

v=(3,4)

Answer 2

Resposta:

$v=(3,4)$

Explicação passo-a-passo:

$(0,0)=a(-1,1)+b(0,1)\\(0,0)=(-a,a+b)\\\left \{ {{-a=0} \atop {a+b=0}} \right=>LI\\\\(x,y)=a(-1,1)+b(0,1)\\(x,y)=(-a,a)+(0,b)\\(x,y)=(-a,a+b)\\\\x=-a=>a=-x\\y=a+b=>b=x+y\\\\(x,y)=-x(-1,1)+(x+y)(0,1)\\\\T(x,y)=-xT(-1,1)+(x+y)T(0,1)\\T(x,y)=-x(3,2,1)+(x+y)(1,1,0)\\T(x,y)=(-3x,-2x,-x)+(x+y,x+y,0)\\T(x,y)=(-2x+y,-x+y,-x)\\T(v)=(-2,1,-3)\\\\-x=-3=>x=3\\-3+y=1=>y=4\\\\T(3,4)=(-2,1,-3)=>v=(3,4)$