Determine as inversas, se existirem, usando o método da matriz adjunta.
A=\left[\begin{array}{lll}1 & 1 & 0 \\\\ 1 & 0 & 1 \\\\ 0 & 1 & 1\end{array}\right],
\mathrm{B}=\left[\begin{array}{lll}1 & 0 & 1 \\\\ 1 & 2 & 3 \\\\ 1 & 2 & 4\end{array}\right],
C=\left[\begin{array}{lll}1 & 9 & 5 \\\\ 3 & 1 & 2 \\\\ 6 & 4 & 4\end{array}\right],
\mathrm{D}=\left[\begin{array}{ccc}1 & -1 & 2 \\\\ 3 & 1 & 2 \\\\ 2 & 3 & -1\end{array}\right],
\mathrm{E}=\left[\begin{array}{lll}1 & 2 & 6 \\\\ 0 & 1 & 5 \\\\ 2 & 3 & 7\end{array}\right]
e \quad \mathrm{F}=\left[\begin{array}{ccc}2 & 3 & 0 \\\\ 1 & -2 & -1 \\\\ 2 & 0 & -1\end{array}\right].

Passo 1

O enunciado pede para determinar, pelo método da matriz adjunta, a inversa de cada uma das matrizes
A,B,C,D,E,F
quando existir. O conteúdo é “inversão de matriz 3 \times 3” usando cofactores: calculamos primeiro \det, depois a matriz de cofactores, transpondo-a para obter a adjunta, e enfim multiplicamos por 1/\det.

Passo 2

Método geral para cada matriz M\in\{A,B,C,D,E,F\}:
  1. Calcular \det(M).
  2. Se \det(M)=0, M é singular e não tem inversa.
  3. Se \det(M)\neq0, formar a matriz de cofactores \bigl(C_{ij}\bigr).
  4. Obter a adjunta \mathrm{adj}(M)=\bigl(C_{ji}\bigr).
  5. Calcular
M^{-1}=\frac{1}{\det(M)}\,\mathrm{adj}(M).

Passo 3

Para
A=\begin{pmatrix} 1 & 1 & 0\\\\ 1 & 0 & 1\\\\ 0 & 1 & 1 \end{pmatrix}
calculamos
\det(A) =1\cdot(0\cdot1-1\cdot1)\;-\;1\cdot(1\cdot1-1\cdot0) \;=\;-1-1=-2\neq0.
Os cofactores são
\mathrm{Cof}(A)= \begin{pmatrix} -1 & -1 & 1\\\\ -1 & 1 & -1\\\\ 1 & 0 & -1 \end{pmatrix},
logo
\mathrm{adj}(A)=\mathrm{Cof}(A)^{T}= \begin{pmatrix} -1 & -1 & 1\\\\ -1 & 1 & 0\\\\ 1 & -1 & -1 \end{pmatrix},
e
A^{-1}=\frac{1}{-2}\,\mathrm{adj}(A) =\begin{pmatrix} \frac12 & \frac12 & -\frac12\\\ \frac12 & -\frac12 & 0\\\\ -\frac12 & \frac12 & \frac12 \end{pmatrix}.

Passo 4

Para
B=\begin{pmatrix} 1 & 0 & 1\\\\ 1 & 2 & 3\\\\ 1 & 2 & 4 \end{pmatrix}
temos
\det(B) =1\cdot(2\cdot4-3\cdot2) -0+\;1\cdot(1\cdot2-2\cdot1) =2+0=2\neq0.
Os cofactores são
\mathrm{Cof}(B)= \begin{pmatrix} 2 & -1 & 0\\\\ 2 & 3 & -2\\\\ -2 & -2 & 2 \end{pmatrix},
de onde
\mathrm{adj}(B)= \begin{pmatrix} 2 & 2 & -2\\\\ -1 & 3 & -2\\\\ 0 & -2 & 2 \end{pmatrix},
e
B^{-1}=\frac1{2}\,\mathrm{adj}(B) =\begin{pmatrix} 1 & 1 & -1\\\\ -\tfrac12 & \tfrac32 & -1\\\\ 0 & -1 & 1 \end{pmatrix}.

Passo 5

Para
C=\begin{pmatrix} 1 & 9 & 5\\\\ 3 & 1 & 2\\\\ 6 & 4 & 4 \end{pmatrix}
calcula-se
\det\text{(C)} =1\cdot(1\cdot4-2\cdot4) -9\cdot(3\cdot4-2\cdot6) +5\cdot(3\cdot4-1\cdot6) =-4-0+30=26\neq0.
Os cofactores dão
\mathrm{Cof}\text{(C)}= \begin{pmatrix} -4 & 0 & 6\\\\ -16 & -26 & 50\\\\ 13 & 13 & -26 \end{pmatrix},
logo
\mathrm{adj}\text{(C)}= \begin{pmatrix} -4 & -16 & 13\\\\ 0 & -26 & 13\\\\ 6 & 50 & -26 \end{pmatrix},
e
C^{-1}=\tfrac1{26}\,\mathrm{adj}\text{(C)} =\begin{pmatrix} -\tfrac{2}{13} & -\tfrac{8}{13} & \tfrac12\\\\ 0 & -1 & \tfrac12\\\ \tfrac{3}{13} & \tfrac{25}{13} & -1 \end{pmatrix}.

Passo 6

Para
D=\begin{pmatrix} 1 & -1 & 2\\\\ 3 & 1 & 2\\\\ 2 & 3 & -1 \end{pmatrix}
obtemos \det(D)=0. Assim D é singular e não possui inversa.

Passo 7

Para
E=\begin{pmatrix} 1 & 2 & 6\\\\ 0 & 1 & 5\\\\ 2 & 3 & 7 \end{pmatrix}
também \det(E)=0, logo E é singular e não tem inversa.

Passo 8

Para
F=\begin{pmatrix} 2 & 3 & 0\\\\ 1 & -2 & -1\\\\ 2 & 0 & -1 \end{pmatrix}
calculamos \det(F)=1\neq0. Os cofactores são
\mathrm{Cof}(F)= \begin{pmatrix} 2 & -1 & 4\\\\ 3 & -2 & 6\\\\ -3 & 2 & -7 \end{pmatrix},
logo
\mathrm{adj}(F)= \begin{pmatrix} 2 & 3 & -3\\\\ -1 & -2 & 2\\\\ 4 & 6 & -7 \end{pmatrix},
e
F^{-1}=\mathrm{adj}(F) =\begin{pmatrix} 2 & 3 & -3\\\\ -1 & -2 & 2\\\\ 4 & 6 & -7 \end{pmatrix}.

Resposta

A inversa existe para
A^{-1}=\begin{pmatrix} \frac12 & \frac12 & -\frac12\\\ \frac12 & -\frac12 & 0\\\\ -\frac12 & \frac12 & \frac12 \end{pmatrix},\quad B^{-1}=\begin{pmatrix} 1 & 1 & -1\\\\ -\tfrac12 & \tfrac32 & -1\\\\ 0 & -1 & 1 \end{pmatrix},\quad C^{-1}=\begin{pmatrix} -\tfrac{2}{13} & -\tfrac{8}{13} & \tfrac12\\\\ 0 & -1 & \tfrac12\\\ \tfrac{3}{13} & \tfrac{25}{13} & -1 \end{pmatrix},\quad F^{-1}=\begin{pmatrix} 2 & 3 & -3\\\\ -1 & -2 & 2\\\\ 4 & 6 & -7 \end{pmatrix}.
As matrizes D e E têm determinante zero e, portanto, não são invertíveis.

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