Seja A=\left[\begin{array}{lll}1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 2 & 1 & 3 .\end{array}\right].
(a) Calcule adj(A).
(b) Calcule \operatorname{det}(A).
(c) Calcule A^{-1}.

Passo 1

O enunciado dá a matriz
A = \begin{pmatrix} 1 & 1 & 0\\\\ 0 & 1 & 1\\\\ 2 & 1 & 3 \end{pmatrix}
e pede (a) a matriz adjunta \mathrm{adj}(A), (b) o determinante \det(A) e (c) a inversa A^{-1}. Para isso vamos usar:
  1. Cálculo de \det(A) pelo Regra de Sarrus.
  2. Cálculo dos cofatores C_{ij}=(-1)^{i+j}M_{ij}, onde M_{ij} é o menor de ordem 2 obtido eliminando linha i e coluna j.
  3. \mathrm{adj}(A) = (C_{ij})^{T}.
  4. A^{-1} = \frac{1}{\det(A)}\,\mathrm{adj}(A).

Passo 2

Aplicando Sarrus em
A = \begin{pmatrix} 1 & 1 & 0\\\\ 0 & 1 & 1\\\\ 2 & 1 & 3 \end{pmatrix}
temos
\det(A) =1\cdot1\cdot3 + 1\cdot1\cdot2 + 0\cdot0\cdot1 \;-\;\bigl(0\cdot1\cdot0 + 1\cdot1\cdot2 + 3\cdot1\cdot1\bigr) =3+2+0 - (0+2+3) =5-5=4.

Passo 3

Calculamos os cofatores C_{ij}=(-1)^{i+j}M_{ij}: M_{11}=\det\begin{pmatrix}1&1\\\ 1&3\end{pmatrix}=2, C_{11}=+2. M_{12}=\det\begin{pmatrix}0&1\\\ 2&3\end{pmatrix}=-2, C_{12}=-(-2)=+2. M_{13}=\det\begin{pmatrix}0&1\\\ 2&1\end{pmatrix}=-2, C_{13}=+(-2)=-2. M_{21}=\det\begin{pmatrix}1&0\\\ 1&3\end{pmatrix}=3, C_{21}=-3. M_{22}=\det\begin{pmatrix}1&0\\\ 2&3\end{pmatrix}=3, C_{22}=+3. M_{23}=\det\begin{pmatrix}1&1\\\ 2&1\end{pmatrix}=-1, C_{23}=-(-1)=+1. M_{31}=\det\begin{pmatrix}1&0\\\ 1&1\end{pmatrix}=1, C_{31}=+1. M_{32}=\det\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}=1, C_{32}=-1. M_{33}=\det\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}=1, C_{33}=+1.
Assim a matriz de cofatores é
(C_{ij}) = \begin{pmatrix} 2 & 2 & -2\\\\ -3 & 3 & 1\\\\ 1 & -1 & 1 \end{pmatrix}.

Passo 4

A adjunta \mathrm{adj}(A) é a transposta de (C_{ij}), ou seja
\mathrm{adj}(A) =\begin{pmatrix} 2 & -3 & 1\\\\ 2 & 3 & -1\\\\ -2 & 1 & 1 \end{pmatrix}.

Passo 5

Finalmente, como \det(A)=4, a inversa é
A^{-1} = \frac{1}{4}\,\mathrm{adj}(A) = \frac{1}{4} \begin{pmatrix} 2 & -3 & 1\\\\ 2 & 3 & -1\\\\ -2 & 1 & 1 \end{pmatrix}.

Resposta

(a)
\mathrm{adj}(A) =\begin{pmatrix} 2 & -3 & 1\\\\ 2 & 3 & -1\\\\ -2 & 1 & 1 \end{pmatrix}.
(b) \det(A)=4.
(c)
A^{-1} = \frac{1}{4} \begin{pmatrix} 2 & -3 & 1\\\\ 2 & 3 & -1\\\\ -2 & 1 & 1 \end{pmatrix}.

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