Q14: Calcule o polinômio característico da matriz A abaixo:
A=\left[\begin{array}{rrrrrrrr} 2 & 1 & -6 & 4 & 5 & -5 & 5 & -6 \\\\ -1 & 4 & 5 & -2 & 7 & 1 & 1 & -3 \\\\ 0 & 0 & 3 & 0 & 0 & 5 & 3 & 2 \\\\ 0 & 0 & 7 & -2 & 1 & 4 & 1 & -2 \\\\ 0 & 0 & 5 & -16 & 6 & -6 & 5 & -2 \\\\ 0 & 0 & 0 & 0 & 0 & -5 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 7 & -3 & 1 \\\\ 0 & 0 & 0 & 0 & 0 & 4 & -2 & -6 \end{array}\right]
Dica: explore a estrutura em blocos da matriz.
(A) p_{c}^{A}(\lambda)=(5-\lambda)^{3}(4-\lambda)^{3}(-2-\lambda)(-3-\lambda)
(B) p_{c}^{A}(\lambda)=(5-\lambda)^{3}(4-\lambda)^{2}(-2-\lambda)^{2}(-3-\lambda)
(C) p_{c}^{A}(\lambda)=(3-\lambda)^{3}(2-\lambda)^{3}(-5-\lambda)^{2}
(D) p_{c}^{A}(\lambda)=(5-\lambda)^{2}(4-\lambda)^{2}(-2-\lambda)^{2}(-3-\lambda)^{2}
(E) p_{c}^{A}(\lambda)=(3-\lambda)^{3}(2-\lambda)^{3}(-5-\lambda)(-4-\lambda)
(F) p_{c}^{A}(\lambda)=(3-\lambda)^{3}(2-\lambda)^{2}(-5-\lambda)^{2}(-4-\lambda)
(G) p_{c}^{A}(\lambda)=(5-\lambda)^{3}(4-\lambda)^{3}(-2-\lambda)^{2}
(H) p_{c}^{A}(\lambda)=(3-\lambda)^{2}(2-\lambda)^{2}(-5-\lambda)^{2}(-4-\lambda)^{2}
(I) Não sei

Passo 1

O enunciado pede o polinômio característico de A\in\mathbb{R}^{8\times8}. Observando que
A= \begin{pmatrix} B_{11} & * & * \\\\ 0 & B_{22} & * \\\\ 0 & 0 & B_{33} \end{pmatrix},
onde
B_{11}=\begin{pmatrix}2&1\\\ -1&4\end{pmatrix},\quad B_{22}=\begin{pmatrix}3&0&0\\\ 7&-2&1\\\ 5&-16&6\end{pmatrix},\quad B_{33}=\begin{pmatrix}-5&0&0\\\ 7&-3&1\\\ 4&-2&-6\end{pmatrix}.
Como A é triangular por blocos,
p_{A}(\lambda)=\det(A-\lambda I) =\det(B_{11}-\lambda I)\,\det(B_{22}-\lambda I)\,\det(B_{33}-\lambda I).

Passo 2

Calculamos \det(B_{11}-\lambda I).
B_{11}-\lambda I =\begin{pmatrix}2-\lambda&1\\\ -1&4-\lambda\end{pmatrix}, \quad \det(B_{11}-\lambda I) =(2-\lambda)(4-\lambda)-(-1)\cdot1 =\lambda^{2}-6\lambda+9 =(3-\lambda)^{2}.

Passo 3

Para B_{22}-\lambda I notamos que a primeira linha tem só o pivô não nulo:
B_{22}-\lambda I =\begin{pmatrix} 3-\lambda&0&0\\\\ 7&-2-\lambda&1\\\\ 5&-16&6-\lambda \end{pmatrix} \implies \det(B_{22}-\lambda I) =(3-\lambda)\det\begin{pmatrix}-2-\lambda&1\\\ -16&6-\lambda\end{pmatrix}.
Dentro do menor:
(-2-\lambda)(6-\lambda)-1\cdot(-16) =\lambda^{2}-4\lambda-12+16 =\lambda^{2}-4\lambda+4 =(\lambda-2)^{2}.
Logo
\det(B_{22}-\lambda I)=(3-\lambda)(\lambda-2)^{2}.

Passo 4

Para B_{33}-\lambda I idem:
B_{33}-\lambda I =\begin{pmatrix} -5-\lambda&0&0\\\\ 7&-3-\lambda&1\\\\ 4&-2&-6-\lambda \end{pmatrix} \implies \det(B_{33}-\lambda I) =(-5-\lambda)\det\begin{pmatrix}-3-\lambda&1\\\ -2&-6-\lambda\end{pmatrix}.
O menor é
(-3-\lambda)(-6-\lambda)-1\cdot(-2) =\lambda^{2}+9\lambda+18+2 =\lambda^{2}+9\lambda+20 =(\lambda+5)(\lambda+4).
Assim
\det(B_{33}-\lambda I) =(-5-\lambda)(\lambda+5)(\lambda+4) =-(\lambda+5)^{2}(\lambda+4).
Como o bloco é 3\times3, o char.\ poly é
\det(\lambda I - B_{33}) =(-1)^{3}\det(B_{33}-\lambda I) =(\lambda+5)^{2}(\lambda+4).

Passo 5

Multiplicando os três fatores, obtemos
p_{A}(\lambda) =\;(3-\lambda)^{2}\;\times\;(3-\lambda)(\lambda-2)^{2}\;\times\;(\lambda+5)^{2}(\lambda+4) =(3-\lambda)^{3}(2-\lambda)^{2}(-5-\lambda)^{2}(-4-\lambda).

Resposta

(F) p_{A}(\lambda)=(3-\lambda)^{3}(2-\lambda)^{2}(-5-\lambda)^{2}(-4-\lambda)

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