Se \alpha=\left\{u_{1}, u_{2}, u_{3}\right\} e \beta=\left\{v_{1}, v_{2}, v_{3}\right\} são bases de \mathbb{R}^{3} cujos vetores relacionam-se pelas equações v_{1}=u_{1}+u_{2}, v_{2}=2 u_{1}+u_{2}+u_{3} e v_{3}= u_{1}+2 u_{2}+u_{3}, determine as matrizes [I]_{\beta}^{\alpha} \mathrm{e}[I]_{\alpha}^{\beta}.

Passo 1

O enunciado fornece duas bases \alpha=\{u_{1},u_{2},u_{3}\} e \beta=\{v_{1},v_{2},v_{3}\} de \mathbb{R}^3 relacionadas por
v_{1}=u_{1}+u_{2},\quad v_{2}=2u_{1}+u_{2}+u_{3},\quad v_{3}=u_{1}+2u_{2}+u_{3}.
Pede-se as matrizes de mudança de coordenadas [I]_{\beta}^{\alpha} e [I]_{\alpha}^{\beta}. Para isso, usamos que [I]_{\beta}^{\alpha} tem colunas dadas pelas coordenadas de cada v_{j} na base \alpha, e que [I]_{\alpha}^{\beta} é a inversa de [I]_{\beta}^{\alpha}.

Passo 2

Cada vetor v_j está expresso em função de \{u_1,u_2,u_3\}. Logo, suas coordenadas em \alpha são
[v_1]_{\alpha}=\begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix},\quad [v_2]_{\alpha}=\begin{pmatrix}2\\\ 1\\\ 1\end{pmatrix},\quad [v_3]_{\alpha}=\begin{pmatrix}1\\\ 2\\\ 1\end{pmatrix}.
As colunas de [I]_{\beta}^{\alpha} são exatamente esses vetores, ou seja,
[I]_{\beta}^{\alpha} =\begin{pmatrix} 1 & 2 & 1\\\\ 1 & 1 & 2\\\\ 0 & 1 & 1 \end{pmatrix}.

Passo 3

Para achar [I]_{\alpha}^{\beta}, notamos que
[I]_{\alpha}^{\beta}=\bigl([I]_{\beta}^{\alpha}\bigr)^{-1}.
Chamando
P=\begin{pmatrix} 1 & 2 & 1\\\\ 1 & 1 & 2\\\\ 0 & 1 & 1 \end{pmatrix},
vamos calcular P^{-1}.

Passo 4

O determinante de P é
\det(P) =1\cdot(1\cdot1-2\cdot1) -2\cdot(1\cdot1-2\cdot0) +1\cdot(1\cdot1-1\cdot0) =-2.
Calculando a matriz adjunta e dividindo por \det(P), obtemos
P^{-1} =\begin{pmatrix} \tfrac12 & \tfrac12 & -\tfrac32\\\ \tfrac12 & -\tfrac12 & \tfrac12\\\\ -\tfrac12 & \tfrac12 & \tfrac12 \end{pmatrix}.
Portanto [I]_{\alpha}^{\beta}=P^{-1}.

Resposta

[I]_{\beta}^{\alpha}=\begin{pmatrix} 1 & 2 & 1\\\\ 1 & 1 & 2\\\\ 0 & 1 & 1 \end{pmatrix}
[I]_{\alpha}^{\beta}=\begin{pmatrix} \tfrac12 & \tfrac12 & -\tfrac32\\\ \tfrac12 & -\tfrac12 & \tfrac12\\\\ -\tfrac12 & \tfrac12 & \tfrac12 \end{pmatrix}

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