Resolva o sistema \left[\begin{array}{rrr}2 & 1 & 0 \\\\ 1 & 6 & 4 \\\\ 0 & 4 & 11\end{array}\right] \cdot\left[\begin{array}{l}x_{1} \\\ x_{2} \\\ x_{3}\end{array}\right]=\left[\begin{array}{l}1 \\\\ 1 \\\\ 1\end{array}\right], utilizando a fatoração \mathbf{L} \cdot \mathbf{U}.

Passo 1

O enunciado pede para resolver o sistema
\begin{pmatrix} 2 & 1 & 0\\\\ 1 & 6 & 4\\\\ 0 & 4 & 11 \end{pmatrix} \begin{pmatrix} x_{1}\\\ x_{2}\\\ x_{3} \end{pmatrix} = \begin{pmatrix} 1\\\\ 1\\\\ 1 \end{pmatrix}
usando fatoração \mathbf{L}\mathbf{U}. O conteúdo é a fatoração de Doolittle (com L de diagonal unitária e U triangular superior) seguida de substituição direta (para Ly=b) e regressiva (para Ux=y).

Passo 2

Fazemos \displaystyle A=\mathbf{L}\mathbf{U} com
\mathbf{L} = \begin{pmatrix} 1 & 0 & 0\\\ \ell_{21} & 1 & 0\\\ \ell_{31} & \ell_{32} & 1 \end{pmatrix}, \quad \mathbf{U} = \begin{pmatrix} u_{11} & u_{12} & u_{13}\\\\ 0 & u_{22} & u_{23}\\\\ 0 & 0 & u_{33} \end{pmatrix}.
Calculamos passo a passo:
u_{11}=2,\quad u_{12}=1,\quad u_{13}=0,
\ell_{21}=\frac{1}{u_{11}}=\tfrac12,\quad \ell_{31}=\frac{0}{u_{11}}=0,
u_{22}=6-\ell_{21}u_{12}=6-\tfrac12\cdot1=\tfrac{11}{2},
u_{23}=4-\ell_{21}u_{13}=4-0=4,
\ell_{32}=\frac{4-\ell_{31}u_{12}}{u_{22}}=\frac{4}{11/2}=\frac{8}{11},
u_{33}=11-\ell_{31}u_{13}-\ell_{32}u_{23} =11-\frac{8}{11}\cdot4=\frac{89}{11}.

Passo 3

Agora resolvemos \mathbf{L}y=b por substituição direta:
y_{1}=1, \quad y_{2}=1-\ell_{21}y_{1}=1-\tfrac12= \tfrac12, \quad y_{3}=1-\ell_{31}y_{1}-\ell_{32}y_{2} =1-0-\frac{8}{11}\cdot\frac12 =\frac{7}{11}.

Passo 4

Em seguida resolvemos \mathbf{U}x=y por substituição regressiva:
u_{33}x_{3}=y_{3}\;\Longrightarrow\;x_{3}=\frac{7/11}{89/11}=\frac{7}{89},
u_{22}x_{2}+u_{23}x_{3}=y_{2} \;\Longrightarrow\;\frac{11}{2}x_{2}+4\cdot\frac{7}{89}=\frac12 \;\Longrightarrow\;x_{2}=\frac{3}{89},
2\,x_{1}+1\cdot x_{2}=1 \;\Longrightarrow\;2x_{1}+\frac{3}{89}=1 \;\Longrightarrow\;x_{1}=\frac{43}{89}.

Resposta

\begin{pmatrix} x_{1}\\\ x_{2}\\\ x_{3} \end{pmatrix} = \begin{pmatrix} \frac{43}{89}\\[6pt] \frac{3}{89}\\[6pt] \frac{7}{89} \end{pmatrix}.]

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