Considere as bases B=\left\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\} e B^{\prime}=\left\{\mathbf{u}_{1}^{\prime}, \mathbf{u}_{2}^{\prime}, \mathbf{u}_{3}^{\prime}\right\} de \mathbb{R}^{3}, em que
\begin{array}{ll} \mathbf{u}_{1}=\left[\begin{array}{c} -3 \\\\ 0 \\\\ 1 \end{array}\right], & \mathbf{u}_{2}=\left[\begin{array}{l} 2 \\\\ 1 \\\\ 2 \end{array}\right], \quad \mathbf{u}_{3}=\left[\begin{array}{c} 1 \\\\ 6 \\\\ -1 \end{array}\right] \\\ \mathbf{u}_{1}^{\prime}=\left[\begin{array}{c} -6 \\\\ -6 \\\\ 0 \end{array}\right], & \mathbf{u}_{2}^{\prime}=\left[\begin{array}{c} -2 \\\\ 2 \\\\ 4 \end{array}\right], \quad \mathbf{u}_{3}^{\prime}=\left[\begin{array}{c} -2 \\\\ -3 \\\\ 7 \end{array}\right] \end{array}
(a) Encontre a matriz de transição de B para B^{\prime}.
(b) Calcule o vetor de coordenadas [\mathbf{w}]_{B}, em que
\mathbf{w}=\left[\begin{array}{c} -5 \\\\ 8 \\\\ -5 \end{array}\right]
e use (12) para calcular [\mathbf{w}]_{B^{\prime}}.
(c) Confira seu trabalho calculando [\mathbf{w}]_{B^{\prime}} diretamente.

Passo 1

O enunciado trata de mudança de base em \mathbb{R}^{3}. Temos as bases
B=\{\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3}\},\quad B'=\{\mathbf{u}_{1}',\mathbf{u}_{2}',\mathbf{u}_{3}'\},
com
\mathbf{u}_{1}=\begin{pmatrix}-3\\\ 0\\\ 1\end{pmatrix},\; \mathbf{u}_{2}=\begin{pmatrix}2\\\ 1\\\ 2\end{pmatrix},\; \mathbf{u}_{3}=\begin{pmatrix}1\\\ 6\\\ -1\end{pmatrix},\quad \mathbf{u}_{1}'=\begin{pmatrix}-6\\\ -6\\\ 0\end{pmatrix},\; \mathbf{u}_{2}'=\begin{pmatrix}-2\\\ 2\\\ 4\end{pmatrix},\; \mathbf{u}_{3}'=\begin{pmatrix}-2\\\ -3\\\ 7\end{pmatrix}.
Para achar a matriz de transição de B para B', vamos usar que, se U e U' são as matrizes cujas colunas são \mathbf{u}_{i} e \mathbf{u}_{i}' em coordenadas canônicas, então a matriz de transição
P_{B\to B'} \;=\;(U')^{-1}\,U
satisfaz [\mathbf{w}]_{B'}=P_{B\to B'}\,[\mathbf{w}]_{B}.

Passo 2

Construímos
U=\bigl[\mathbf{u}_{1}\;\mathbf{u}_{2}\;\mathbf{u}_{3}\bigr] =\begin{pmatrix} -3 & 2 & 1\\\\ 0 & 1 & 6\\\\ 1 & 2 & -1 \end{pmatrix},\quad U'=\bigl[\mathbf{u}_{1}'\;\mathbf{u}_{2}'\;\mathbf{u}_{3}'\bigr] =\begin{pmatrix} -6 & -2 & -2\\\\ -6 & 2 & -3\\\\ 0 & 4 & 7 \end{pmatrix}.
Para calcular (U')^{-1}U, resolvemos para cada i=1,2,3 o sistema
U'\,\mathbf{c}_{i} \;=\;\mathbf{u}_{i},
e colocamos \mathbf{c}_{i} como a i-ésima coluna de P_{B\to B'}.
(1) Para \mathbf{u}_{1}=(-3,0,1)^{T}, obtemos
\mathbf{c}_{1}=\frac{1}{48}\begin{pmatrix}17\\\ 33\\\ -12\end{pmatrix}.
(2) Para \mathbf{u}_{2}=(2,1,2)^{T}, obtemos
\mathbf{c}_{2}=\frac{1}{32}\begin{pmatrix}-13\\\ -5\\\ 12\end{pmatrix}.
(3) Para \mathbf{u}_{3}=(1,6,-1)^{T}, obtemos
\mathbf{c}_{3}=\frac{1}{48}\begin{pmatrix}-13\\\ 51\\\ -36\end{pmatrix}.
Logo
P_{B\to B'} =\begin{pmatrix} \frac{17}{48} & -\frac{13}{32} & -\frac{13}{48}\\[6pt] \frac{33}{48} & -\frac{5}{32} & \frac{51}{48}\\[6pt] -\frac{12}{48}& \frac{12}{32} & -\frac{36}{48} \end{pmatrix}.]

Passo 3

Para achar [\mathbf{w}]_{B}, resolvemos
U\begin{pmatrix}b_{1}\b_{2}\b_{3}\end{pmatrix} =\begin{pmatrix}-5\\\ 8\\\ -5\end{pmatrix}.
O sistema
\begin{cases} -3b_{1}+2b_{2}+b_{3}=-5,\\\ b_{2}+6b_{3}=8,\\\ b_{1}+2b_{2}-b_{3}=-5 \end{cases}
b_{1}=\frac{21}{25},\quad b_{2}=-\frac{52}{25},\quad b_{3}=\frac{42}{25}.
Portanto
[\mathbf{w}]_{B} =\begin{pmatrix}\tfrac{21}{25}\\[3pt]-\tfrac{52}{25}\\[3pt]\tfrac{42}{25}\end{pmatrix}.]

Passo 4

Aplicando a matriz de transição,
[\mathbf{w}]_{B'} =P_{B\to B'}\,[\mathbf{w}]_{B},
fazendo o produto obtemos
[\mathbf{w}]_{B'} =\begin{pmatrix}\tfrac{11}{16}\\[3pt]\tfrac{43}{16}\\[3pt]-\tfrac{9}{4}\end{pmatrix}.]

Passo 5

Para conferir, resolvemos diretamente
U'\,\mathbf{c}=\begin{pmatrix}-5\\\ 8\\\ -5\end{pmatrix}.
O sistema
\begin{cases} -6c_{1}-2c_{2}-2c_{3}=-5,\\\\ -6c_{1}+2c_{2}-3c_{3}=8,\\\\ 4c_{2}+7c_{3}=-5 \end{cases}
c_{1}=\frac{11}{16},\quad c_{2}=\frac{43}{16},\quad c_{3}=-\frac{9}{4},
o que coincide com [\mathbf{w}]_{B'} acima.

Resposta

(a)
P_{B\to B'} =\begin{pmatrix} \frac{17}{48} & -\frac{13}{32} & -\frac{13}{48}\\[4pt] \frac{33}{48} & -\frac{5}{32} & \frac{51}{48}\\[4pt] -\frac{12}{48}& \frac{12}{32} & -\frac{36}{48} \end{pmatrix}.
(b)
[\mathbf{w}]_{B} =\begin{pmatrix}\tfrac{21}{25}\\[2pt]-\tfrac{52}{25}\\[2pt]\tfrac{42}{25}\end{pmatrix}, \quad [\mathbf{w}]_{B'} =\begin{pmatrix}\tfrac{11}{16}\\[2pt]\tfrac{43}{16}\\[2pt]-\tfrac{9}{4}\end{pmatrix}.
(c) Conferido diretamente, obtemos o mesmo [\mathbf{w}]_{B'}.]

Exercícios relacionados