Sejam A=\left[\begin{array}{ccc}\frac{1}{3} & 0 & 0 \\\\ 0 & \frac{1}{3} & 0 \\\\ 0 & 0 & \frac{1}{3}\end{array}\right], \mathbf{u}=\left[\begin{array}{r}3 \\\\ 6 \\\\ -9\end{array}\right]. e, \mathbf{v}=\left[\begin{array}{c}a \\\ b \\\ c\end{array}\right]. Defina T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} por T(\mathbf{x})=A \mathbf{x}. Calcule T(\mathbf{u}) e T(\mathbf{v}).

Passo 1

O enunciado nos dá a transformação linear T:\mathbb{R}^3\to\mathbb{R}^3 definida por
T(\mathbf{x}) = A\,\mathbf{x},
onde
A = \begin{pmatrix}\tfrac{1}{3}&0&0\\[6pt]0&\tfrac{1}{3}&0\\[6pt]0&0&\tfrac{1}{3}\end{pmatrix}, \quad \mathbf{u} = \begin{pmatrix}3\\[3pt]6\\[3pt]-9\end{pmatrix}, \quad \mathbf{v} = \begin{pmatrix}a\\[3pt]b\\[3pt]c\end{pmatrix}.
O conteúdo aqui é multiplicação de matriz diagonal por vetor. Vamos aplicar a definição de T para obter T(\mathbf{u}) e T(\mathbf{v}).]

Passo 2

Para calcular T(\mathbf{u}) fazemos
T(\mathbf{u}) = A\,\mathbf{u} = \begin{pmatrix}\tfrac{1}{3}&0&0\\[3pt]0&\tfrac{1}{3}&0\\[3pt]0&0&\tfrac{1}{3}\end{pmatrix} \begin{pmatrix}3\\[3pt]6\\[3pt]-9\end{pmatrix} = \begin{pmatrix}\tfrac{1}{3}\cdot 3 + 0\cdot6 + 0\cdot(-9)\\[6pt]0\cdot3 + \tfrac{1}{3}\cdot6 + 0\cdot(-9)\\[6pt]0\cdot3 + 0\cdot6 + \tfrac{1}{3}\cdot(-9)\end{pmatrix} = \begin{pmatrix}1\\[3pt]2\\[3pt]-3\end{pmatrix}.]

Passo 3

Para calcular T(\mathbf{v}) aplicamos a mesma ideia:
T(\mathbf{v}) = A\,\mathbf{v} = \begin{pmatrix}\tfrac{1}{3}&0&0\\[3pt]0&\tfrac{1}{3}&0\\[3pt]0&0&\tfrac{1}{3}\end{pmatrix} \begin{pmatrix}a\\[3pt]b\\[3pt]c\end{pmatrix} = \begin{pmatrix}\tfrac{1}{3}a\\[3pt]\tfrac{1}{3}b\\[3pt]\tfrac{1}{3}c\end{pmatrix}.]

Resposta

T(\mathbf{u}) = \begin{pmatrix}1\\[3pt]2\\[3pt]-3\end{pmatrix}, \qquad T(\mathbf{v}) = \begin{pmatrix}\tfrac{a}{3}\\[3pt]\tfrac{b}{3}\\[3pt]\tfrac{c}{3}\end{pmatrix}.]

Exercícios relacionados