Q14: Calcule o polinómio característico da matriz A abaixo:
A=\left[\begin{array}{rrrrrrrr} -2 & 1 & -1 & -2 & -3 & 2 & 3 & 6 \\\\ -4 & -6 & 5 & 6 & 1 & 7 & -7 & -4 \\\\ 0 & 0 & 1 & 1 & 0 & -2 & -7 & -5 \\\\ 0 & 0 & 20 & 0 & 0 & -6 & -4 & 1 \\\\ 0 & 0 & -6 & 7 & 5 & -2 & 5 & -4 \\\\ 0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 \\\\ 0 & 0 & 0 & 0 & 0 & 3 & -9 & -8 \end{array}\right]
Dica: explore a estrutura em blocos da matriz.
(A) p_{c}^{A}(\lambda)=(-4-\lambda)^{3}(5-\lambda)^{2}(-5-\lambda)^{2}(3-\lambda)
(B) p_{c}^{A}(\lambda)=(4-\lambda)^{3}(-2-\lambda)^{2}(-3-\lambda)^{2}(2-\lambda)
(C) p_{c}^{A}(\lambda)=(4-\lambda)^{3}(-2-\lambda)^{3}(-3-\lambda)(2-\lambda)
(D) p_{c}^{A}(\lambda)=(4-\lambda)^{2}(-2-\lambda)^{2}(-3-\lambda)^{2}(2-\lambda)^{2}
(E) p_{c}^{A}(\lambda)=(-4-\lambda)^{2}(5-\lambda)^{2}(-5-\lambda)^{2}(3-\lambda)^{2}
(F) p_{c}^{A}(\lambda)=(-4-\lambda)^{3}(5-\lambda)^{3}(-5-\lambda)(3-\lambda)
(G) p_{c}^{A}(\lambda)=(4-\lambda)^{3}(-2-\lambda)^{3}(-3-\lambda)^{2}
(H) p_{c}^{A}(\lambda)=(-4-\lambda)^{3}(5-\lambda)^{3}(-5-\lambda)^{2}
(I) Não sei

Passo 1

O enunciado pede o polinómio característico de
A=\begin{pmatrix} -2 & 1 & -1 & -2 & -3 & 2 & 3 & 6\\\\ -4 & -6 & 5 & 6 & 1 & 7 & -7 & -4\\\\ 0 & 0 & 1 & 1 & 0 & -2 & -7 & -5\\\\ 0 & 0 & 20 & 0 & 0 & -6 & -4 & 1\\\\ 0 & 0 & -6 & 7 & 5 & -2 & 5 & -4\\\\ 0 & 0 & 0 & 0 & 0 & 5 & 0 & 0\\\\ 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1\\\\ 0 & 0 & 0 & 0 & 0 & 3 & -9 & -8 \end{pmatrix}.
Como vemos zeros abaixo de certos blocos, A é triangular por blocos com submatrizes de dimensões 2,3,3. Logo
p_A(\lambda)=\det(A-\lambda I_8) =\det(A_{11}-\lambda I_2)\,\det(A_{22}-\lambda I_3)\,\det(A_{33}-\lambda I_3).

Passo 2

Identificamos os blocos:
A_{11}=\begin{pmatrix}-2&1\\\ -4&-6\end{pmatrix},\quad A_{22}=\begin{pmatrix}1&1&0\\\ 20&0&0\\\ -6&7&5\end{pmatrix},\quad A_{33}=\begin{pmatrix}5&0&0\\\ 1&-2&1\\\ 3&-9&-8\end{pmatrix}.

Passo 3

Calculamos
\det(A_{11}-\lambda I_2) =\det\begin{pmatrix}-2-\lambda&1\\\ -4&-6-\lambda\end{pmatrix} =(\lambda+2)(\lambda+6)+4 =\lambda^2+8\lambda+16 =(\lambda+4)^2.

Passo 4

Para o bloco 3\times3:
\det(A_{22}-\lambda I_3) =\det\begin{pmatrix}1-\lambda&1&0\\\ 20&-\lambda&0\\\ -6&7&5-\lambda\end{pmatrix} =(5-\lambda)\bigl[(1-\lambda)(-\lambda)-20\bigr] =(5-\lambda)(\lambda^2-\lambda-20) =(5-\lambda)(\lambda-5)(\lambda+4).

Passo 5

Para o último bloco 3\times3:
\det(A_{33}-\lambda I_3) =\det\begin{pmatrix}5-\lambda&0&0\\\ 1&-2-\lambda&1\\\ 3&-9&-8-\lambda\end{pmatrix} =(5-\lambda)\det\begin{pmatrix}-2-\lambda&1\\\ -9&-8-\lambda\end{pmatrix} =(5-\lambda)\bigl[(\lambda+2)(\lambda+8)+9\bigr] =(5-\lambda)(\lambda+5)^2.

Passo 6

Multiplicando os fatores:
p_A(\lambda) =(\lambda+4)^2\;\bigl[(5-\lambda)(\lambda-5)(\lambda+4)\bigr]\;\bigl[(5-\lambda)(\lambda+5)^2\bigr].
Agrupando raízes em \lambda+4,\lambda-5,\lambda+5 obtemos
p_A(\lambda) =(\lambda+4)^3(\lambda-5)^3(\lambda+5)^2.
Isso coincide com a fatoração de
(-4-\lambda)^3\,(5-\lambda)^3\,(-5-\lambda)^2,
ou seja, com a opção (H).

Resposta

(H)

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