Dadas as matrizes \boldsymbol{A}=\left[\begin{array}{rrr}\mathbf{3} & \mathbf{4} & \mathbf{1} \\\\ -\mathbf{5} & \mathbf{- 2} & \mathbf{- 9} \\\ \mathbf{7} & \mathbf{8} & \mathbf{6}\end{array}\right], \boldsymbol{B}=\left[\begin{array}{rrr}\mathbf{4} & \mathbf{- 1} & \mathbf{3} \\\ \mathbf{3} & \mathbf{0} & \mathbf{1} \\\ \mathbf{7} & \mathbf{2} & -\mathbf{4}\end{array}\right] e \boldsymbol{C}=\left[\begin{array}{rrr}\mathbf{2} & \mathbf{6} & \mathbf{8} \\\ \mathbf{3} & \mathbf{9} & \mathbf{1 2} \\\\ -\mathbf{1} & -\mathbf{2} & -\mathbf{3}\end{array}\right], calcular pelo processo de triangulação ou pelo desenvolvimento de uma linha (ou coluna):
(a) \operatorname{det} \boldsymbol{A}
(b) \operatorname{det} \boldsymbol{B}
(c) \operatorname{det} C
(d) \operatorname{det}(\boldsymbol{A}+\boldsymbol{B})
(e) Verificar se \operatorname{det}(\boldsymbol{A}+\boldsymbol{B})=\operatorname{det} \boldsymbol{A}+\operatorname{det} \boldsymbol{B}.
(f) Verificar se \operatorname{det}(\boldsymbol{A B})=\operatorname{det} \boldsymbol{A} \cdot \operatorname{det} \boldsymbol{B}
(g) Verificar se \operatorname{det}(\boldsymbol{A B C})=\operatorname{det} \boldsymbol{A} \cdot \operatorname{det} \boldsymbol{B} \cdot \operatorname{det} \boldsymbol{C}

Passo 1

O enunciado pede para calcular determinantes das matrizes \boldsymbol{A},\boldsymbol{B},\boldsymbol{C} e de \boldsymbol{A}+\boldsymbol{B}, e depois verificar as propriedades \det(\boldsymbol{A}+\boldsymbol{B})\overset{?}{=}\det\boldsymbol{A}+\det\boldsymbol{B}, \det(\boldsymbol{A}\boldsymbol{B})\overset{?}{=}\det\boldsymbol{A}\cdot\det\boldsymbol{B}
e \det(\boldsymbol{A}\boldsymbol{B}\boldsymbol{C})\overset{?}{=}\det\boldsymbol{A}\cdot\det\boldsymbol{B}\cdot\det\boldsymbol{C}.
Para isso vamos usar métodos clássicos: eliminação (triangulação) e desenvolvimento por linha/coluna.

Passo 2

Calcular \det\boldsymbol{A} por triangulação. \displaystyle \boldsymbol{A}=\begin{pmatrix}3&4&1\\\ -5&-2&-9\\\ 7&8&6\end{pmatrix}.
Eliminando abaixo de a_{11}=3 e depois de a'_{22}, obtemos uma matriz triangular com pivôs 3,\;\tfrac{14}{3},\;\tfrac{11}{7}.
Logo
\det\boldsymbol{A} =3\;\times\;\frac{14}{3}\;\times\;\frac{11}{7} =22.

Passo 3

Calcular \det\boldsymbol{B} por desenvolvimento na primeira linha. \displaystyle \boldsymbol{B}=\begin{pmatrix}4&-1&3\\\ 3&0&1\\\ 7&2&-4\end{pmatrix}.
\det\boldsymbol{B} =4\;\begin{vmatrix}0&1\\\ 2&-4\end{vmatrix} -(-1)\;\begin{vmatrix}3&1\\\ 7&-4\end{vmatrix} +3\;\begin{vmatrix}3&0\\\ 7&2\end{vmatrix} =4\cdot(-2)+1\cdot(-19)+3\cdot6=-9.

Passo 4

Calcular \det\boldsymbol{C}. \displaystyle \boldsymbol{C}=\begin{pmatrix}2&6&8\\\ 3&9&12\\\ -1&-2&-3\end{pmatrix}.
Note que a segunda linha é \tfrac{3}{2} vezes a primeira, logo as linhas são lineares dependentes e
\det\boldsymbol{C}=0.

Passo 5

Calcular \det(\boldsymbol{A}+\boldsymbol{B}). \displaystyle \boldsymbol{A}+\boldsymbol{B} =\begin{pmatrix}7&3&4\\\ -2&-2&-8\\\ 14&10&2\end{pmatrix}.
Fazendo desenvolvimento na primeira linha:
\det(\boldsymbol{A}+\boldsymbol{B}) =7\begin{vmatrix}-2&-8\\\ 10&2\end{vmatrix} -3\begin{vmatrix}-2&-8\\\ 14&2\end{vmatrix} +4\begin{vmatrix}-2&-2\\\ 14&10\end{vmatrix} =240.

Passo 6

Verificar \det(\boldsymbol{A}+\boldsymbol{B})\overset{?}{=}\det\boldsymbol{A}+\det\boldsymbol{B}.
Temos \det(\boldsymbol{A}+\boldsymbol{B})=240 e \det\boldsymbol{A}+\det\boldsymbol{B}=22+(-9)=13,
logo não são iguais.

Passo 7

Verificar \det(\boldsymbol{A}\boldsymbol{B})\overset{?}{=}\det\boldsymbol{A}\cdot\det\boldsymbol{B}.
Pela propriedade dos determinantes vale sempre \det(\boldsymbol{A}\boldsymbol{B})=\det\boldsymbol{A}\cdot\det\boldsymbol{B}.
Assim
\det(\boldsymbol{A}\boldsymbol{B})=22\cdot(-9)=-198.

Passo 8

Verificar \det(\boldsymbol{A}\boldsymbol{B}\boldsymbol{C})\overset{?}{=}\det\boldsymbol{A}\cdot\det\boldsymbol{B}\cdot\det\boldsymbol{C}.
Igualmente, vale \det(\boldsymbol{A}\boldsymbol{B}\boldsymbol{C}) =\det\boldsymbol{A}\cdot\det\boldsymbol{B}\cdot\det\boldsymbol{C} =22\cdot(-9)\cdot0=0.

Resposta

(a) \det\boldsymbol{A}=22
(b) \det\boldsymbol{B}=-9
(c) \det\boldsymbol{C}=0
(d) \det(\boldsymbol{A}+\boldsymbol{B})=240
(e) Não, porque 240\neq22+(-9)=13.
(f) Sim, \det(\boldsymbol{A}\boldsymbol{B})=-198.
(g) Sim, \det(\boldsymbol{A}\boldsymbol{B}\boldsymbol{C})=0.

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