Considere as bases \mathcal{B}=\left\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right\} e \mathcal{C}=\left\{\mathbf{g}_{1}, \mathbf{g}_{2}, \mathbf{g}_{3}\right\} de \mathbb{R}^{3} relacionadas pelas seguintes equações:
\mathbf{g}_{1}=\mathbf{e}_{1}-\mathbf{e}_{3}, \quad \mathbf{g}_{2}=2 \mathbf{e}_{2}+3 \mathbf{e}_{3}, \quad \mathbf{g}_{3}=3 \mathbf{e}_{1}+\mathbf{e}_{3} .
(a) Determine as matrizes de mudança de base de \mathcal{B} para \mathcal{C}\left(P_{\mathcal{C} \leftarrow \mathcal{B}}\right) e de \mathcal{C} para \mathcal{B}\left(P_{\mathcal{B} \leftarrow \mathcal{C}}\right).

Passo 1

O enunciado nos dá duas bases de \mathbb{R}^3, \mathcal{B}=\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\} e \mathcal{C}=\{\mathbf{g}_1,\mathbf{g}_2,\mathbf{g}_3\}, relacionadas por
\mathbf{g}_1=\mathbf{e}_1-\mathbf{e}_3,\quad \mathbf{g}_2=2\,\mathbf{e}_2+3\,\mathbf{e}_3,\quad \mathbf{g}_3=3\,\mathbf{e}_1+\mathbf{e}_3.
Queremos (a) a matriz de mudança de base de \mathcal{B} para \mathcal{C},
denotada P_{\mathcal{C}\leftarrow\mathcal{B}}, e (b) a de \mathcal{C} para \mathcal{B}, P_{\mathcal{B}\leftarrow\mathcal{C}}. Para isso,
usamos que as colunas de P_{\mathcal{C}\leftarrow\mathcal{B}} são as
coordenadas de \mathbf{g}_j na base \mathcal{B}, e que P_{\mathcal{B}\leftarrow\mathcal{C}}=(P_{\mathcal{C}\leftarrow\mathcal{B}})^{-1}.

Passo 2

Em coordenadas de \mathcal{B} temos \mathbf{g}_1\leftrightarrow(1,0,-1), \mathbf{g}_2\leftrightarrow(0,2,3), \mathbf{g}_3\leftrightarrow(3,0,1). Logo
P_{\mathcal{C}\leftarrow\mathcal{B}} =\begin{pmatrix} 1 & 0 & 3\\\\ 0 & 2 & 0\\\\ -1 & 3 & 1 \end{pmatrix}.

Passo 3

Calculamos a inversa de \displaystyle P=P_{\mathcal{C}\leftarrow\mathcal{B}} =\begin{pmatrix}1&0&3\\\ 0&2&0\\\ -1&3&1\end{pmatrix}.
O determinante é \det P=8. O adjugado é
\operatorname{adj}(P) =\begin{pmatrix} 2 & 9 & -6\\\\ 0 & 4 & 0\\\\ 2 & -3 & 2 \end{pmatrix},
então
P^{-1} =\frac1{8}\begin{pmatrix} 2 & 9 & -6\\\\ 0 & 4 & 0\\\\ 2 & -3 & 2 \end{pmatrix}.

Passo 4

Portanto:
P_{\mathcal{C}\leftarrow\mathcal{B}} =\begin{pmatrix} 1 & 0 & 3\\\\ 0 & 2 & 0\\\\ -1 & 3 & 1 \end{pmatrix}, \qquad P_{\mathcal{B}\leftarrow\mathcal{C}} =\frac1{8}\begin{pmatrix} 2 & 9 & -6\\\\ 0 & 4 & 0\\\\ 2 & -3 & 2 \end{pmatrix}.

Resposta

P_{\mathcal{C}\leftarrow\mathcal{B}} =\begin{pmatrix} 1 & 0 & 3\\\\ 0 & 2 & 0\\\\ -1 & 3 & 1 \end{pmatrix} e P_{\mathcal{B}\leftarrow\mathcal{C}} =\dfrac1{8}\begin{pmatrix} 2 & 9 & -6\\\\ 0 & 4 & 0\\\\ 2 & -3 & 2 \end{pmatrix}.

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