Considere o seguinte sistema de equações lineares:
\left[\begin{array}{lll} 8 & 5 & 6 \\\\ 2 & 3 & 1 \\\\ 6 & 1 & 3 \end{array}\right]\left[\begin{array}{l} x_{1} \\\ x_{2} \\\ x_{3} \end{array}\right]=\left[\begin{array}{r} 19 \\\\ 6 \\\\ 10 \end{array}\right]
Caso possível, resolva o sistema dado usando o método de decomposição LU.

Passo 1

O enunciado dá um sistema linear A\mathbf{x}=\mathbf{b} com
A = \begin{pmatrix} 8 & 5 & 6\\\\ 2 & 3 & 1\\\\ 6 & 1 & 3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix}19\\\ 6\\\ 10\end{pmatrix},
e pede a solução via decomposição LU. Vamos usar o método de Doolittle, em que escrevemos A = LU com
L = \begin{pmatrix} 1 & 0 & 0\\\ \ell_{21} & 1 & 0\\\ \ell_{31} & \ell_{32} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} u_{11} & u_{12} & u_{13}\\\\ 0 & u_{22} & u_{23}\\\\ 0 & 0 & u_{33} \end{pmatrix}.

Passo 2

Calculamos os elementos de L e U a partir das fórmulas
u_{11}=a_{11}=8,\quad u_{12}=a_{12}=5,\quad u_{13}=a_{13}=6,
\ell_{21}=\frac{a_{21}}{u_{11}}=\frac{2}{8}=\frac14,\quad \ell_{31}=\frac{a_{31}}{u_{11}}=\frac{6}{8}=\frac34,
u_{22}=a_{22}-\ell_{21}u_{12} =3-\tfrac14\cdot5=\tfrac74, \quad u_{23}=a_{23}-\ell_{21}u_{13} =1-\tfrac14\cdot6=-\tfrac12,
\ell_{32}=\frac{a_{32}-\ell_{31}u_{12}}{u_{22}} =\frac{1-\tfrac34\cdot5}{\tfrac74}=-\frac{11}{7},
u_{33}=a_{33}-\ell_{31}u_{13}-\ell_{32}u_{23} =3-\tfrac34\cdot6-\Bigl(-\tfrac{11}{7}\Bigr)\Bigl(-\tfrac12\Bigr) =-\frac{16}{7}.
Portanto
L = \begin{pmatrix} 1 & 0 & 0\\\ \tfrac14 & 1 & 0\\\ \tfrac34 & -\tfrac{11}{7} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} 8 & 5 & 6\\\\ 0 & \tfrac74 & -\tfrac12\\\\ 0 & 0 & -\tfrac{16}{7} \end{pmatrix}.

Passo 3

Resolvemos primeiro L\mathbf{y}=\mathbf{b} por substituição direta:
y_1 = 19, \quad y_2 = 6 - \tfrac14\cdot19 = \tfrac54,
y_3 = 10 - \tfrac34\cdot19 -\Bigl(-\tfrac{11}{7}\Bigr)\tfrac54 = -\tfrac{16}{7}.

Passo 4

Agora resolvemos U\mathbf{x}=\mathbf{y} por retro-substituição:
-\tfrac{16}{7}\,x_3 = -\tfrac{16}{7} \quad\Longrightarrow\quad x_3 = 1,
\tfrac{7}{4}\,x_2 - \tfrac{1}{2}\cdot 1 = \tfrac{5}{4} \quad\Longrightarrow\quad x_2 = 1,
8\,x_1 + 5\cdot 1 + 6\cdot 1 = 19 \quad\Longrightarrow\quad x_1 = 1.

Resposta

x_1 = 1,\;x_2 = 1,\;x_3 = 1.

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