Determine a inversa de:
A=\left[\begin{array}{lll} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\\ \frac{2}{3} & a & b \\\ \frac{2}{3} & b & a \end{array}\right]

Passo 1

Resumindo: queremos achar a inversa da matriz
A=\begin{pmatrix} \tfrac13 & \tfrac23 & \tfrac23\\\ \tfrac23 & a & b\\\ \tfrac23 & b & a \end{pmatrix}.
O conteúdo é “inversão de matriz 3\times3” via adjunta/adjugate. Vamos precisar do determinante de A e da matriz de cofatores.

Passo 2

Calculemos \det(A) expandindo pela primeira linha. Chamando
M_{11}=\det\begin{pmatrix}a&b\b&a\end{pmatrix}=a^2-b^2,
M_{12}=\det\begin{pmatrix}\tfrac23&b\\tfrac23&a\end{pmatrix} =\tfrac23\,a-\tfrac23\,b=\tfrac23(a-b), \quad M_{13}=\det\begin{pmatrix}\tfrac23&a\\tfrac23&b\end{pmatrix} =\tfrac23(b-a)=-\tfrac23(a-b).
Então
\det(A) =\tfrac13\,M_{11}-\tfrac23\,M_{12}+\tfrac23\,M_{13} =\tfrac13(a^2-b^2)-\tfrac23\cdot\tfrac23(a-b)+\tfrac23\cdot\bigl(-\tfrac23(a-b)\bigr).
Agrupando:
\det(A) =\tfrac{a^2-b^2}{3}-\tfrac{4}{9}(a-b)-\tfrac{4}{9}(a-b) =(a-b)\Bigl(\tfrac{a+b}{3}-\tfrac{8}{9}\Bigr) =\tfrac{(a-b)\bigl(3(a+b)-8\bigr)}{9}.

Passo 3

Agora obtemos a matriz de cofatores C=(C_{ij}). Os cofatores não nulos são
C_{11}=a^2-b^2, \quad C_{12}=C_{21}=-\tfrac23(a-b), \quad C_{13}=C_{31}=-\tfrac23(a-b),
C_{22}=\det\begin{pmatrix}\tfrac13&\tfrac23\\tfrac23&a\end{pmatrix} =\tfrac a3-\tfrac{4}{9}=\tfrac{3a-4}{9}, \quad C_{23}=C_{32}=-\det\begin{pmatrix}\tfrac13&\tfrac23\\tfrac23&b\end{pmatrix} =\tfrac{4-3b}{9}, \quad C_{33}=\tfrac{3a-4}{9}.

Passo 4

A adjunta é \operatorname{adj}(A)=C^{T}=C (aqui C é simétrica). Logo
A^{-1}=\frac{1}{\det(A)}\,\operatorname{adj}(A) =\frac{9}{(a-b)\bigl(3(a+b)-8\bigr)}\; \begin{pmatrix} a^2-b^2 & -\tfrac{2}{3}(a-b) & -\tfrac{2}{3}(a-b)\\[6pt] -\tfrac{2}{3}(a-b) & \tfrac{3a-4}{9} & \tfrac{4-3b}{9}\\[6pt] -\tfrac{2}{3}(a-b) & \tfrac{4-3b}{9} & \tfrac{3a-4}{9} \end{pmatrix}.]

Passo 5

Basta simplificar cada entrada. Chamando \Delta=(a-b)\bigl(3(a+b)-8\bigr), temos
A^{-1} =\frac{1}{\Delta} \begin{pmatrix} 9(a+b) & -6 & -6\\\\ -6 & 3a-4 & 4-3b\\\\ -6 & 4-3b & 3a-4 \end{pmatrix},
pois, por exemplo,
\tfrac{9}{\Delta}(a^2-b^2)=\tfrac{9(a+b)(a-b)}{\Delta}=\tfrac{9(a+b)}{3(a+b)-8}\cdot\tfrac{a-b}{a-b},
e
\tfrac{9}{\Delta}\bigl(-\tfrac23(a-b)\bigr)=-\tfrac{6}{\Delta/(a-b)}=-\tfrac{6}{3(a+b)-8}.

Resposta

A^{-1} =\frac{1}{(a-b)\bigl(3(a+b)-8\bigr)} \begin{pmatrix} 9(a+b) & -6 & -6\\\\ -6 & 3a-4 & 4-3b\\\\ -6 & 4-3b & 3a-4 \end{pmatrix},
válida desde que (a-b)\bigl(3(a+b)-8\bigr)\neq0.

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