Utilizando Cholesky, encontre a inversa da seguinte matriz:
A=\left(\begin{array}{ccc} 4 & 2 & -4 \\ 2 & 10 & 4 \\ -4 & 4 & 9 \end{array}\right)

Passo 1

O enunciado pede a inversa da matriz
A=\begin{pmatrix} 4 & 2 & -4\\\\ 2 & 10 & 4\\\\ -4 & 4 & 9 \end{pmatrix}
usando a decomposição de Cholesky. Como A é simétrica e definida positiva, podemos escrever
A = L\,L^T
onde L é triangular inferior. Depois, invertendo L e usando
A^{-1} = (L^{-1})^T\,L^{-1},
obtemos A^{-1}.

Passo 2

Vamos achar L=(\ell_{ij}) tal que A=L\,L^T. Escrevendo
L=\begin{pmatrix} \ell_{11}&0&0\\\ \ell_{21}&\ell_{22}&0\\\ \ell_{31}&\ell_{32}&\ell_{33} \end{pmatrix},
temos:
\ell_{11}=\sqrt{4}=2,\quad \ell_{21}=\frac{2}{2}=1,\quad \ell_{31}=\frac{-4}{2}=-2.
Em seguida,
\ell_{22}=\sqrt{10-\ell_{21}^2}=\sqrt{10-1}=3,
\ell_{32}=\frac{4-\ell_{31}\,\ell_{21}}{\ell_{22}} =\frac{4-(-2)\cdot1}{3}=\frac{6}{3}=2,
\ell_{33}=\sqrt{9-\ell_{31}^2-\ell_{32}^2} =\sqrt{9-4-4}=1.
Logo,
L=\begin{pmatrix} 2&0&0\\\\ 1&3&0\\\\ -2&2&1 \end{pmatrix}.

Passo 3

Agora invertemos L por substituição direta em L\,X=I. Chamando M=L^{-1}, resolvemos coluna a coluna:
coluna 1 ( Lx=e_1) dá
M(:,1)=\begin{pmatrix}1/2\\\ -1/6\\\ 4/3\end{pmatrix},
coluna 2 ( Lx=e_2) dá
M(:,2)=\begin{pmatrix}0\\\ 1/3\\\ -2/3\end{pmatrix},
coluna 3 ( Lx=e_3) dá
M(:,3)=\begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix}.
Assim,
L^{-1}=M =\begin{pmatrix} \tfrac12&0&0\\\\ -\tfrac16&\tfrac13&0\\\ \tfrac43&-\tfrac23&1 \end{pmatrix}.

Passo 4

Finalmente,
A^{-1}=(L^{-1})^T\,L^{-1} =M^T\,M.
Calculando elemento a elemento,
A^{-1} =\begin{pmatrix} \displaystyle\sum_kM_{k1}^2 & \displaystyle\sum_kM_{k1}M_{k2} & \displaystyle\sum_kM_{k1}M_{k3}\\[1em] \displaystyle\sum_kM_{k2}M_{k1} & \displaystyle\sum_kM_{k2}^2 & \displaystyle\sum_kM_{k2}M_{k3}\\[1em] \displaystyle\sum_kM_{k3}M_{k1} & \displaystyle\sum_kM_{k3}M_{k2} & \displaystyle\sum_kM_{k3}^2 \end{pmatrix} =\begin{pmatrix} \tfrac{37}{18}&-\tfrac{17}{18}&\tfrac{4}{3}\\\\ -\tfrac{17}{18}&\tfrac{5}{9}&-\tfrac{2}{3}\\\ \tfrac{4}{3}&-\tfrac{2}{3}&1 \end{pmatrix}.]

Resposta

A^{-1} =\begin{pmatrix} \tfrac{37}{18}&-\tfrac{17}{18}&\tfrac{4}{3}\\\\ -\tfrac{17}{18}&\tfrac{5}{9}&-\tfrac{2}{3}\\\ \tfrac{4}{3}&-\tfrac{2}{3}&1 \end{pmatrix}.

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