Question

Encontre a lei que define a transformação: T: R² -> R², tal que T(1,1) = (1,0) e T(-1,1) = (1,2)

Answer 1

$\Large\boxed{\begin{array}{l}\tt dados\begin{cases}\sf T(1,1)=(1,0)\\\sf T(-1,1)=(1,2)\\\sf T(x,y)=?\end{cases}\\\sf (x,y)=a\cdot(1,1)+b\cdot(-1,1)\\\sf (x,y)=(a,a)+(-b,b)\\\sf(x,y)=(a-b,a+b)\\+\underline{\begin{cases}\sf a-\diagup\!\!\!b=x\\\sf a+\diagup\!\!\!b=y\end{cases}}\\\sf 2a=x+y\\\sf a=\dfrac{x+y}{2}\\\sf a+b=y\\\sf \dfrac{x+y}{2}+b=y\cdot(2)\\\sf x+y+2b=2y\\\sf 2b=2y-y-x\\\sf b=\dfrac{y-x}{2}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf(x,y)=\dfrac{x+y}{2}\cdot(1,1)+\dfrac{y-x}{2}\cdot(-1,1)\\\\\sf T(x,y)=\dfrac{x+y}{2}\cdot T(1,1)+\dfrac{y-x}{2}\cdot T(-1,1)\\\\\sf T(x,y)=\dfrac{x+y}{2}\cdot(1,0)+\dfrac{y-x}{2}\cdot(1,2)\\\\\sf T(x,y)=\bigg(\dfrac{x+y}{2},0\bigg)+\bigg(\dfrac{y-x}{2},y-x\bigg)\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf T:\mathbb{R}^2\longrightarrow \mathbb{R}^2 :T(x,y)=(y,y-x)}}}}\end{array}}$