Seja o produto \mathrm{LL}^{\top}=\mathrm{A} de uma matriz. Aplique o método da decomposição de Cholesky à matriz A abaixo e marque a alternativa que representa a matriz L obtida.
A=\left[\begin{array}{rrrr} 1 & 2 & 3 & 0 \\\\ 2 & 8 & 10 & -8 \\\\ 3 & 10 & 14 & -5 \\\\ 0 & -8 & -5 & 29 \end{array}\right]
a. \mathrm{L}=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\\\ 2 & 2 & 0 & 0 \\\\ 3 & 2 & 1 & 0 \\\\ 0 & -4 & 3 & 2\end{array}\right]
b. \mathrm{L}=\left[\begin{array}{llll}1 & 0 & 0 & 0 \\\\ 2 & 2 & 0 & 0 \\\\ 3 & 2 & 1 & 0 \\\\ 0 & 4 & 3 & 2\end{array}\right]
c. \mathrm{L}=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\\\ 2 & 2 & 0 & 0 \\\\ -3 & 2 & 1 & 0 \\\\ 0 & -4 & 3 & 2\end{array}\right]
d. \mathrm{L}=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\\\ 2 & 2 & 0 & 0 \\\\ 3 & 2 & 1 & 0 \\\\ 0 & -4 & -3 & 2\end{array}\right]
e. \mathrm{L}=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\\\ 2 & 2 & 0 & 0 \\\\ 3 & 2 & 1 & 0 \\\\ 0 & -4 & 3 & -2\end{array}\right]

Passo 1

O enunciado dá uma matriz simétrica e definida-positiva
A = \begin{pmatrix} 1 & 2 & 3 & 0\\\\ 2 & 8 & 10 & -8\\\\ 3 & 10 & 14 & -5\\\\ 0 & -8 & -5 & 29 \end{pmatrix}
e pede a decomposição de Cholesky, ou seja, encontrar a matriz triangular inferior L tal que
L\,L^\top = A.
Para isso, vamos calcular L coluna a coluna, usando as fórmulas
L_{ii} = \sqrt{A_{ii} - \sum_{k=1}^{i-1}L_{ik}^2}, \quad L_{ji} = \frac{1}{L_{ii}}\Bigl(A_{ji} - \sum_{k=1}^{i-1}L_{jk}L_{ik}\Bigr) \quad(j>i).

Passo 2

Primeira coluna i=1:
L_{11} = \sqrt{A_{11}} = \sqrt{1} = 1,
L_{21} = \frac{A_{21}}{L_{11}} = \frac{2}{1} = 2,\quad L_{31} = \frac{A_{31}}{L_{11}} = \frac{3}{1} = 3,\quad L_{41} = \frac{A_{41}}{L_{11}} = \frac{0}{1} = 0.

Passo 3

Segunda coluna i=2:
L_{22} = \sqrt{A_{22} - L_{21}^2} = \sqrt{8 - 2^2} = \sqrt{4} = 2,
L_{32} = \frac{A_{32} - L_{31}L_{21}}{L_{22}} = \frac{10 - 3\cdot2}{2} = \frac{4}{2} = 2,
L_{42} = \frac{A_{42} - L_{41}L_{21}}{L_{22}} = \frac{-8 - 0\cdot2}{2} = -4.

Passo 4

Terceira coluna i=3:
L_{33} = \sqrt{A_{33} - (L_{31}^2 + L_{32}^2)} = \sqrt{14 - (3^2 + 2^2)} = \sqrt{14 - 13} = 1,
L_{43} = \frac{A_{43} - (L_{41}L_{31} + L_{42}L_{32})}{L_{33}} = \frac{-5 - (0\cdot3 + (-4)\cdot2)}{1} = -5 - (-8) = 3.

Passo 5

Quarta coluna i=4:
L_{44} = \sqrt{A_{44} - (L_{41}^2 + L_{42}^2 + L_{43}^2)} = \sqrt{29 - (0^2 + (-4)^2 + 3^2)} = \sqrt{29 - (16 + 9)} = \sqrt{4} = 2.

Resposta

A matriz L obtida é
L = \begin{pmatrix} 1 & 0 & 0 & 0\\\\ 2 & 2 & 0 & 0\\\\ 3 & 2 & 1 & 0\\\\ 0 & -4 & 3 & 2 \end{pmatrix},
que corresponde à alternativa (a).

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