Encontre a inversa da matriz (se existir).
Exercício 31.
\left(\begin{array}{ccc} 1 & -2 & -1 \\\\ 2 & -3 & 1 \\\\ 3 & -4 & 4 \end{array}\right)

Passo 1

O enunciado pede a inversa da matriz
A = \begin{pmatrix} 1 & -2 & -1\\\\ 2 & -3 & 1\\\\ 3 & -4 & 4 \end{pmatrix}.
O conteúdo é inversão de matrizes 3×3 via determinante e matriz dos cofatores.

Passo 2

Calcular o determinante de A expandindo pela primeira linha:
\det(A) =1\cdot\det\begin{pmatrix}-3 & 1\\\ -4 & 4\end{pmatrix} -(-2)\cdot\det\begin{pmatrix}2 & 1\\\ 3 & 4\end{pmatrix} +(-1)\cdot\det\begin{pmatrix}2 & -3\\\ 3 & -4\end{pmatrix}.
Os menores são
\det\begin{pmatrix}-3 & 1\\\ -4 & 4\end{pmatrix} =(-3)\cdot4-1\cdot(-4)=-12+4=-8,
\det\begin{pmatrix}2 & 1\\\ 3 & 4\end{pmatrix} =2\cdot4-1\cdot3=8-3=5,
\det\begin{pmatrix}2 & -3\\\ 3 & -4\end{pmatrix} =2\cdot(-4)-(-3)\cdot3=-8+9=1.
Logo,
\det(A) = 1\cdot(-8) +2\cdot5 +(-1)\cdot1 = -8+10-1 = 1\neq0,
portanto A é invertível.

Passo 3

Construir a matriz dos cofatores C. Para cada elemento a_{ij} calculamos o menor M_{ij} e aplicamos o sinal (-1)^{i+j}:
i=1:
C_{11}=+\det\begin{pmatrix}-3&1\\\ -4&4\end{pmatrix}=-8,\quad C_{12}=-\det\begin{pmatrix}2&1\\\ 3&4\end{pmatrix}=-5,\quad C_{13}=+\det\begin{pmatrix}2&-3\\\ 3&-4\end{pmatrix}=1.
i=2:
C_{21}=-\det\begin{pmatrix}-2&-1\\\ -4&4\end{pmatrix} =-\bigl((-2)\cdot4-(-1)\cdot(-4)\bigr)=-(-8-4)=12,
C_{22}=+\det\begin{pmatrix}1&-1\\\ 3&4\end{pmatrix} =1\cdot4-(-1)\cdot3=7,\quad C_{23}=-\det\begin{pmatrix}1&-2\\\ 3&-4\end{pmatrix} =-\bigl(1\cdot(-4)-(-2)\cdot3\bigr)=-(-4+6)=-2.
i=3:
C_{31}=+\det\begin{pmatrix}-2&-1\\\ -3&1\end{pmatrix} =-2\cdot1-(-1)\cdot(-3)=-2-3=-5,
C_{32}=-\det\begin{pmatrix}1&-1\\\ 2&1\end{pmatrix} =-\bigl(1\cdot1-(-1)\cdot2\bigr)=-(1+2)=-3,
C_{33}=+\det\begin{pmatrix}1&-2\\\ 2&-3\end{pmatrix} =1\cdot(-3)-(-2)\cdot2=-3+4=1.
Assim
C=\begin{pmatrix} -8 & -5 & 1\\\\ 12 & 7 & -2\\\\ -5 & -3 & 1 \end{pmatrix}.

Passo 4

A adjunta de A é a transposta de C:
\operatorname{adj}(A)=C^T =\begin{pmatrix} -8 & 12 & -5\\\\ -5 & 7 & -3\\\\ 1 & -2 & 1 \end{pmatrix}.

Passo 5

Como \det(A)=1, a inversa é
A^{-1}=\frac{1}{\det(A)}\,\operatorname{adj}(A) =\operatorname{adj}(A).

Resposta

A^{-1} = \begin{pmatrix} -8 & 12 & -5\\\\ -5 & 7 & -3\\\\ 1 & -2 & 1 \end{pmatrix}.

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