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 & 1 & -1 & 0 \\\\ 1 & 3 & -1 & 0 \\\\ -1 & -1 & 5 & 2 \\\\ 0 & 0 & 2 & 4\end{array}\right]

Passo 1

O enunciado pede a fatoração
A = L\,D\,L^{\prime}
para a matriz simétrica
A = \begin{pmatrix} 4 & 1 & -1 & 0\\\\ 1 & 3 & -1 & 0\\\\ -1 & -1 & 5 & 2\\\\ 0 & 0 & 2 & 4 \end{pmatrix}.
O método LDL′ consiste em, a cada etapa k, definir
D_{k} = A^{(k-1)}_{kk}, \quad \ell_{i k} = \frac{A^{(k-1)}_{ik}}{D_{k}},
e em seguida atualizar a submatriz de Schur
A^{(k)}_{ij} = A^{(k-1)}_{ij} - \ell_{i k}\,D_{k}\,\ell_{j k},
para i,j>k. Inicialmente A^{(0)}=A.

Passo 2

k =1:
D_{1} = A^{(0)}_{11} = 4.
Logo
\ell_{21} = \frac{1}{4},\quad \ell_{31} = \frac{-1}{4},\quad \ell_{41} = 0.
A submatriz 2\le i,j\le4 atualizada é
A^{(1)}_{ij} = a_{ij} - \ell_{i1}\,D_{1}\,\ell_{j1} = \begin{pmatrix} 3-\tfrac{1}{4} & -1+\tfrac{1}{4} & 0\\[6pt] -1+\tfrac{1}{4} & 5-\tfrac{1}{4} & 2\\[6pt] 0 & 2 & 4 \end{pmatrix} = \begin{pmatrix} \tfrac{11}{4} & -\tfrac{3}{4} & 0\\[3pt] -\tfrac{3}{4} & \tfrac{19}{4} & 2\\[3pt] 0 & 2 & 4 \end{pmatrix}.]

Passo 3

k =2:
D_{2} = A^{(1)}_{22} = \tfrac{11}{4}, \quad \ell_{32} = \frac{-\tfrac{3}{4}}{\tfrac{11}{4}} = -\tfrac{3}{11}, \quad \ell_{42} = 0.
Atualizando a submatriz 3\le i,j\le4:
A^{(2)}_{ij} = A^{(1)}_{ij} - \ell_{i2}\,D_{2}\,\ell_{j2} = \begin{pmatrix} \tfrac{19}{4}-\tfrac{9}{44} & 2\\[3pt] 2 & 4 \end{pmatrix} = \begin{pmatrix} \tfrac{50}{11} & 2\\[3pt] 2 & 4 \end{pmatrix}.]

Passo 4

k =3:
D_{3} = A^{(2)}_{33} = \tfrac{50}{11}, \quad \ell_{43} = \frac{2}{\tfrac{50}{11}} = \tfrac{11}{25}.
Por fim
D_{4} = A^{(2)}_{44} - \ell_{43}\,D_{3}\,\ell_{43} = 4 - \tfrac{11}{25}\cdot\tfrac{50}{11}\cdot\tfrac{11}{25} = \tfrac{78}{25}.

Passo 5

Reunindo os \ell_{ij} e os D_{k}, temos
L = \begin{pmatrix} 1 & 0 & 0 & 0\\[3pt] \tfrac14 & 1 & 0 & 0\\[3pt] -\tfrac14 & -\tfrac3{11} & 1 & 0\\[3pt] 0 & 0 & \tfrac{11}{25} & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 4 & 0 & 0 & 0\\[3pt] 0 & \tfrac{11}{4} & 0 & 0\\[3pt] 0 & 0 & \tfrac{50}{11} & 0\\[3pt] 0 & 0 & 0 & \tfrac{78}{25} \end{pmatrix}.]

Resposta

A = L\,D\,L^{\prime} \quad\text{com}\quad L = \begin{pmatrix} 1 & 0 & 0 & 0\\[3pt] \tfrac14 & 1 & 0 & 0\\[3pt] -\tfrac14 & -\tfrac3{11} & 1 & 0\\[3pt] 0 & 0 & \tfrac{11}{25} & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 4 & 0 & 0 & 0\\[3pt] 0 & \tfrac{11}{4} & 0 & 0\\[3pt] 0 & 0 & \tfrac{50}{11} & 0\\[3pt] 0 & 0 & 0 & \tfrac{78}{25} \end{pmatrix}.]

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