Seja o produto \mathrm{LL}^{\top}=\mathrm{A} de uma matriz. Resolva o sistema abaixo usando a decomposição de Cholesky.
4 x_{1}+2 x_{2}+14 x_{3}=-6
2 x_{1}+17 x_{2}-5 x_{3}=9
14 x_{1}-5 x_{2}+83 x_{3}=-55
a. \mathrm{x}=\left[\begin{array}{r}2 \\\\ 0 \\\\ -1\end{array}\right] e \mathrm{L}=\left[\begin{array}{rrr}2 & 0 & 0 \\\\ 1 & 4 & 0 \\\\ -7 & -3 & 5\end{array}\right]
b. \mathrm{x}=\left[\begin{array}{r}2 \\\\ 0 \\\\ -1\end{array}\right] e \mathrm{L}=\left[\begin{array}{rrr}2 & 0 & 0 \\\\ 1 & 4 & 0 \\\\ -7 & -3 & -5\end{array}\right]
c. x=\left[\begin{array}{r}2 \\\\ 0 \\\\ -1\end{array}\right] e L=\left[\begin{array}{rrr}2 & 0 & 0 \\\\ 1 & 4 & 0 \\\\ 7 & -3 & 5\end{array}\right]
d. x=\left[\begin{array}{r}2 \\\\ 0 \\\\ -1\end{array}\right] e L=\left[\begin{array}{lll}2 & 0 & 0 \\\\ 1 & 4 & 0 \\\\ 7 & 3 & 5\end{array}\right]
e. x=\left[\begin{array}{r}2 \\\\ 0 \\\\ -1\end{array}\right] e L=\left[\begin{array}{rrr}2 & 0 & 0 \\\\ 1 & 4 & 0 \\\\ 7 & -3 & -5\end{array}\right]

Passo 1

O enunciado nos dá o sistema linear
4x_{1}+2x_{2}+14x_{3}=-6,
2x_{1}+17x_{2}-5x_{3}=9,
14x_{1}-5x_{2}+83x_{3}=-55,
cuja matriz de coeficientes
A=\begin{pmatrix} 4 & 2 & 14\\\\ 2 & 17 & -5\\\\ 14 & -5 & 83 \end{pmatrix}
é simétrica e definida positiva. Vamos usar a decomposição de Cholesky A=LL^{\top} para resolver Ax=b.

Passo 2

Para A=LL^{\top} com
L=\begin{pmatrix} l_{11} & 0 & 0\\\ l_{21} & l_{22} & 0\\\ l_{31} & l_{32} & l_{33} \end{pmatrix},
calculamos os l_{ij} da forma
l_{11}=\sqrt{a_{11}},\quad l_{21}=\frac{a_{21}}{l_{11}},\quad l_{31}=\frac{a_{31}}{l_{11}},
l_{22}=\sqrt{a_{22}-l_{21}^{2}},\quad l_{32}=\frac{a_{32}-l_{31}l_{21}}{l_{22}},\quad l_{33}=\sqrt{a_{33}-l_{31}^{2}-l_{32}^{2}}.
Substituindo os elementos de A:
l_{11}=\sqrt{4}=2,\quad l_{21}=\frac{2}{2}=1,\quad l_{31}=\frac{14}{2}=7,
l_{22}=\sqrt{17-1^{2}}=\sqrt{16}=4,\quad l_{32}=\frac{-5-7\cdot1}{4}=-3,\quad l_{33}=\sqrt{83-7^{2}-(-3)^{2}}=\sqrt{25}=5.
Logo
L=\begin{pmatrix} 2 & 0 & 0\\\\ 1 & 4 & 0\\\\ 7 & -3 & 5 \end{pmatrix}.

Passo 3

Resolvemos primeiro Ly=b com b=(-6,9,-55)^{\top}.
A equação Ly=b
y_{1}=\frac{-6}{2}=-3,
y_{2}=\frac{9 - 1\cdot(-3)}{4}=\frac{12}{4}=3,
y_{3}=\frac{-55 - 7\cdot(-3) -(-3)\cdot3}{5}=\frac{-55+21+9}{5}=-5.
Portanto y=(-3,3,-5)^{\top}.

Passo 4

Agora resolvemos L^{\top}x=y. Como
L^{\top}=\begin{pmatrix} 2 & 1 & 7\\\\ 0 & 4 & -3\\\\ 0 & 0 & 5 \end{pmatrix},
temos das linhas inferiores para cima:
5x_{3}=-5\implies x_{3}=-1,
4x_{2}-3x_{3}=3\implies 4x_{2}+3=3\implies x_{2}=0,
2x_{1}+1\cdot x_{2}+7x_{3}=-3\implies 2x_{1}-7=-3\implies x_{1}=2.
Assim x=(2,0,-1)^{\top}.

Resposta

c. \displaystyle x=\begin{pmatrix}2\\\ 0\\\ -1\end{pmatrix} e \displaystyle L=\begin{pmatrix}2&0&0\\\ 1&4&0\\\ 7&-3&5\end{pmatrix}.

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