Utilize o algoritmo de fatoração L D L^{\prime} para descobrir uma fatoração da forma A=L D L^{\prime} para as seguintes matrizes:
A=\left[\begin{array}{rrrr}4 & 0 & 2 & 1 \\\\ 0 & 3 & -1 & 1 \\\\ 2 & -1 & 6 & 3 \\\\ 1 & 1 & 3 & 8\end{array}\right]

Passo 1

O enunciado pede a fatoração A = L\,D\,L^{\prime} da matriz simétrica
A = \begin{pmatrix} 4 & 0 & 2 & 1\\\\ 0 & 3 & -1 & 1\\\\ 2 & -1 & 6 & 3\\\\ 1 & 1 & 3 & 8 \end{pmatrix}
usando o algoritmo LDL′. Para isso, em cada etapa k calculamos
  1. o pivô \displaystyle d_{k} = a^{(k-1)}_{kk},
  2. os multiplicadores \displaystyle \ell_{ik} = a^{(k-1)}_{ik}/d_{k} para i>k,
  3. atualizamos a submatriz remanescente \displaystyle a^{(k)}_{ij} = a^{(k-1)}_{ij} - d_{k}\,\ell_{ik}\,\ell_{jk}.

Passo 2

k =1:
d_{1} = a_{11} = 4.
Multiplicadores col ‑1:
\ell_{21} = \frac{0}{4} = 0,\quad \ell_{31} = \frac{2}{4} = \tfrac12,\quad \ell_{41} = \frac{1}{4} = \tfrac14.
Atualizando o bloco 2\le i,j\le4:
a'_{ij} = a_{ij} - 4\,\ell_{i1}\,\ell_{j1}
resulta em
A' = \begin{pmatrix} 3 & -1 & 1\\\\ -1 & 5 & \tfrac52\\\\ 1 & \tfrac52 & \tfrac{31}{4} \end{pmatrix}.

Passo 3

k =2:
d_{2} = a'_{22} = 3.
Multiplicadores col 2:
\ell_{32} = \frac{-1}{3} = -\tfrac13,\quad \ell_{42} = \frac{1}{3} = \tfrac13.
Atualizando o bloco 3\le i,j\le4:
a''_{ij} = a'_{ij} - 3\,\ell_{i2}\,\ell_{j2}
resulta em
A'' = \begin{pmatrix} \tfrac{14}{3} & \tfrac{17}{6}\\\ \tfrac{17}{6} & \tfrac{89}{12} \end{pmatrix}.

Passo 4

k =3:
d_{3} = a''_{33} = \tfrac{14}{3}, \qquad \ell_{43} = \frac{a''_{34}}{d_{3}} = \frac{\tfrac{17}{6}}{\tfrac{14}{3}} = \tfrac{17}{28}.
Finalmente,
d_{4} = a''_{44} - d_{3}\,\ell_{43}^{2} = \tfrac{89}{12} - \tfrac{14}{3}\Bigl(\tfrac{17}{28}\Bigr)^{2} = \tfrac{319}{56}.

Passo 5

Reunindo todos os \ell_{ij} e d_{k}, obtemos
L = \begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\ \tfrac12 & -\tfrac13 & 1 & 0\\\ \tfrac14 & \tfrac13 & \tfrac{17}{28} & 1 \end{pmatrix}, \qquad D = \operatorname{diag}\!\bigl(4,\;3,\;\tfrac{14}{3},\;\tfrac{319}{56}\bigr).

Resposta

A fatoração é
A = L\,D\,L^{\prime},
onde
L = \begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\ \tfrac12 & -\tfrac13 & 1 & 0\\\ \tfrac14 & \tfrac13 & \tfrac{17}{28} & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 4 & 0 & 0 & 0\\\\ 0 & 3 & 0 & 0\\\\ 0 & 0 & \tfrac{14}{3} & 0\\\\ 0 & 0 & 0 & \tfrac{319}{56} \end{pmatrix}.

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