Dados:
\underline{v}=\left[\begin{array}{c}2 \\\\ 1 \\\\ -3\end{array}\right] \underline{\underline{\tau}}=\left[\begin{array}{ccc}1 & 2 & -2 \\\\ 2 & 3 & -1 \\\\ -2 & -1 & 1\end{array}\right], calcular:
(b) \underline{v} \cdot \underline{\underline{\tau}}

Passo 1

O enunciado nos dá o vetor
\underline{v}=\begin{pmatrix}2\\\ 1\\\ -3\end{pmatrix}
e o tensor de segunda ordem (matriz)
\underline{\underline{\tau}}=\begin{pmatrix}1&2&-2\\\ 2&3&-1\\\ -2&-1&1\end{pmatrix}
e pede o produto contrátil \underline{v}\cdot\underline{\underline{\tau}}. Em componentes, esse produto gera o vetor cujas componentes são
(\underline{v}\cdot\underline{\underline{\tau}})_j \;=\;\sum_{i=1}^{3}v_i\,\tau_{ij}\,.

Passo 2

Escrevemos a fórmula de cada componente j:
(\underline{v}\cdot\underline{\underline{\tau}})_1 = v_1\,\tau_{11}+v_2\,\tau_{21}+v_3\,\tau_{31},
(\underline{v}\cdot\underline{\underline{\tau}})_2 = v_1\,\tau_{12}+v_2\,\tau_{22}+v_3\,\tau_{32},
(\underline{v}\cdot\underline{\underline{\tau}})_3 = v_1\,\tau_{13}+v_2\,\tau_{23}+v_3\,\tau_{33}.

Passo 3

Substituímos os valores de \underline{v} e de \underline{\underline{\tau}}:
(\underline{v}\cdot\underline{\underline{\tau}})_1 =2\cdot1 + 1\cdot2 + (-3)\cdot(-2) =2 + 2 + 6 =10,
(\underline{v}\cdot\underline{\underline{\tau}})_2 =2\cdot2 + 1\cdot3 + (-3)\cdot(-1) =4 + 3 + 3 =10,
(\underline{v}\cdot\underline{\underline{\tau}})_3 =2\cdot(-2) + 1\cdot(-1) + (-3)\cdot1 =-4 -1 -3 =-8.

Passo 4

Agrupando as componentes, obtemos o vetor resultante:
\underline{v}\cdot\underline{\underline{\tau}} =\begin{pmatrix}10\\\ 10\\\ -8\end{pmatrix}.

Resposta

\begin{pmatrix}10\\\ 10\\\ -8\end{pmatrix}

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