Seja A=\left[\begin{array}{ccc}2 & 1 & -3 \\\\ 0 & 2 & 1 \\\\ 5 & 1 & 3\end{array}\right].
a) Encontre todos os menores de A.
b) Encontre todos os cofatores.
c) \operatorname{adj} A
d) \operatorname{det} A
e) A^{-1}

Passo 1

O enunciado dá a matriz
A = \begin{pmatrix} 2 & 1 & -3\\\\ 0 & 2 & 1\\\\ 5 & 1 & 3 \end{pmatrix}
e pede:
(a) todos os menores de A (determinantes 2\times2 obtidos ao eliminar linha i e coluna j),
(b) os cofatores C_{ij}=(-1)^{i+j}M_{ij},
(c) a matriz adjunta \operatorname{adj}A,
(d) \det A,
(e) A^{-1}.
Vamos usar definições de menor, cofator, expansão de Laplace e a fórmula
A^{-1} = \frac{1}{\det A}\,\operatorname{adj}A.

Passo 2

Calcular todos os menores M_{ij}:
M_{11}=\det\begin{pmatrix}2&1\\\ 1&3\end{pmatrix}=2\cdot3-1\cdot1=5,
M_{12}=\det\begin{pmatrix}0&1\\\ 5&3\end{pmatrix}=0\cdot3-1\cdot5=-5,
M_{13}=\det\begin{pmatrix}0&2\\\ 5&1\end{pmatrix}=0\cdot1-2\cdot5=-10,
M_{21}=\det\begin{pmatrix}1&-3\\\ 1&3\end{pmatrix}=1\cdot3-(-3)\cdot1=6,
M_{22}=\det\begin{pmatrix}2&-3\\\ 5&3\end{pmatrix}=2\cdot3-(-3)\cdot5=6+15=21,
M_{23}=\det\begin{pmatrix}2&1\\\ 5&1\end{pmatrix}=2\cdot1-1\cdot5=-3,
M_{31}=\det\begin{pmatrix}1&-3\\\ 2&1\end{pmatrix}=1\cdot1-(-3)\cdot2=7,
M_{32}=\det\begin{pmatrix}2&-3\\\ 0&1\end{pmatrix}=2\cdot1-(-3)\cdot0=2,
M_{33}=\det\begin{pmatrix}2&1\\\ 0&2\end{pmatrix}=2\cdot2-1\cdot0=4.

Passo 3

Calcular os cofatores C_{ij}=(-1)^{i+j}M_{ij}:
C_{11}=+5,\quad C_{12}=-(-5)=5,\quad C_{13}=+(-10)=-10,
C_{21}=-(6)=-6,\quad C_{22}=+21,\quad C_{23}= -(-3)=3,
C_{31}=+7,\quad C_{32}= -2,\quad C_{33}=+4.

Passo 4

Formar a matriz de cofatores e transpor para obter \operatorname{adj}A:
\text{cof}(A)= \begin{pmatrix} 5 & 5 & -10\\\\ -6 & 21 & 3\\\\ 7 & -2 & 4 \end{pmatrix}, \qquad \operatorname{adj}A = (\text{cof}(A))^{T} = \begin{pmatrix} 5 & -6 & 7\\\\ 5 & 21 & -2\\\\ -10 & 3 & 4 \end{pmatrix}.

Passo 5

Calcular \det A por expansão na primeira linha:
\det A = 2\cdot C_{11} + 1\cdot C_{12} + (-3)\cdot C_{13} =2\cdot5 + 1\cdot5 + (-3)\cdot(-10) =10+5+30=45.

Passo 6

Usar A^{-1}=\tfrac{1}{\det A}\,\operatorname{adj}A:
A^{-1} = \frac{1}{45} \begin{pmatrix} 5 & -6 & 7\\\\ 5 & 21 & -2\\\\ -10 & 3 & 4 \end{pmatrix}.

Resposta

(a) M_{11}=5,\;M_{12}=-5,\;M_{13}=-10,\;M_{21}=6,\;M_{22}=21,\;M_{23}=-3,\;M_{31}=7,\;M_{32}=2,\;M_{33}=4. (b) C_{11}=5,\;C_{12}=5,\;C_{13}=-10,\;C_{21}=-6,\;C_{22}=21,\;C_{23}=3,\;C_{31}=7,\;C_{32}=-2,\;C_{33}=4. (c) \displaystyle \operatorname{adj}A = \begin{pmatrix} 5 & -6 & 7\\\\ 5 & 21 & -2\\\\ -10 & 3 & 4 \end{pmatrix}. (d) \det A = 45. (e) \displaystyle A^{-1} = \frac{1}{45}\begin{pmatrix} 5 & -6 & 7\\\\ 5 & 21 & -2\\\\ -10 & 3 & 4 \end{pmatrix}.

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