Q14: Calcule o polinônio característico da matriz A abaixo:
A=\left[\begin{array}{rrrrrrrr} 5 & 0 & 0 & -7 & -6 & 7 & -2 & -7 \\\\ 6 & -3 & 1 & 2 & 1 & 2 & -1 & -1 \\\\ 5 & -64 & 13 & -5 & 6 & -1 & -1 & 5 \\\\ 0 & 0 & 0 & -2 & 1 & 0 & -5 & -1 \\\\ 0 & 0 & 0 & -9 & -8 & 0 & 6 & -2 \\\\ 0 & 0 & 0 & 5 & 4 & -5 & -4 & -2 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & -2 & 1 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & -16 & 6 \end{array}\right]
Dica: explore a estrutura em blocos da matriz.
(A) p_{c}^{A}(\lambda)=(5-\lambda)^{3}(-5-\lambda)^{3}(2-\lambda)(-2-\lambda)
(B) p_{c}^{A}(\lambda)=(-3-\lambda)^{3}(3-\lambda)^{2}(4-\lambda)^{2}(-4-\lambda)
(C) p_{c}^{A}(\lambda)=(-3-\lambda)^{2}(3-\lambda)^{2}(4-\lambda)^{2}(-4-\lambda)^{2}
(D) p_{c}^{A}(\lambda)=(5-\lambda)^{2}(-5-\lambda)^{2}(2-\lambda)^{2}(-2-\lambda)^{2}
(E) p_{c}^{A}(\lambda)=(5-\lambda)^{3}(-5-\lambda)^{3}(2-\lambda)^{2}
(F) p_{c}^{A}(\lambda)=(-3-\lambda)^{3}(3-\lambda)^{3}(4-\lambda)^{2}
(G) p_{c}^{A}(\lambda)=(5-\lambda)^{3}(-5-\lambda)^{2}(2-\lambda)^{2}(-2-\lambda)
(H) p_{c}^{A}(\lambda)=(-3-\lambda)^{3}(3-\lambda)^{3}(4-\lambda)(-4-\lambda)
(I) Não sei

Passo 1

O enunciado pede o polinômio característico de
A=\begin{pmatrix} 5 & 0 & 0 & -7 & -6 & 7 & -2 & -7 \\\\ 6 & -3 & 1 & 2 & 1 & 2 & -1 & -1 \\\\ 5 & -64 & 13 & -5 & 6 & -1 & -1 & 5 \\\\ 0 & 0 & 0 & -2 & 1 & 0 & -5 & -1 \\\\ 0 & 0 & 0 & -9 & -8 & 0 & 6 & -2 \\\\ 0 & 0 & 0 & 5 & 4 & -5 & -4 & -2 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & -2 & 1 \\\\ 0 & 0 & 0 & 0 & 0 & 0 & -16 & 6 \end{pmatrix}
Para resolver vamos usar que, se uma matriz é triangular por blocos
A=\begin{pmatrix} A_{11}&* &*\\\\ 0&A_{22}&*\\\\ 0&0&A_{33} \end{pmatrix},
então
\det(A-\lambda I)=\det(A_{11}-\lambda I)\,\det(A_{22}-\lambda I)\,\det(A_{33}-\lambda I).

Passo 2

Separemos os blocos diagonais de A:
A_{11}=\begin{pmatrix} 5 & 0 &0\\\\ 6 & -3 &1\\\\ 5 & -64 &13 \end{pmatrix},\quad A_{22}=\begin{pmatrix} -2 &1&0\\\\ -9&-8&0\\\\ 5&4&-5 \end{pmatrix},\quad A_{33}=\begin{pmatrix} -2&1\\\\ -16&6 \end{pmatrix}.
Logo
p_A(\lambda)=\det(A-\lambda I) =\det(A_{11}-\lambda I_3)\,\det(A_{22}-\lambda I_3)\,\det(A_{33}-\lambda I_2).

Passo 3

Calculemos \det(A_{11}-\lambda I_3). Como a primeira linha é \bigl(5-\lambda,0,0\bigr),
\det(A_{11}-\lambda I) =(5-\lambda)\det\begin{pmatrix}-3-\lambda&1\\\ -64&13-\lambda\end{pmatrix}.
E
\det\begin{pmatrix}-3-\lambda&1\\\ -64&13-\lambda\end{pmatrix} =(-3-\lambda)(13-\lambda)+64 =\lambda^2-10\lambda+25=(\lambda-5)^2.
Assim
\det(A_{11}-\lambda I)=(5-\lambda)(\lambda-5)^2=-(\lambda-5)^3,
isto é, gera o fator (5-\lambda)^3.

Passo 4

Agora \det(A_{22}-\lambda I_3). Expandindo pela terceira coluna (só o elemento -5-\lambda de baixo é não-nulo),
\det(A_{22}-\lambda I) =(-5-\lambda)\det\begin{pmatrix}-2-\lambda&1\\\ -9&-8-\lambda\end{pmatrix}.
E
\det\begin{pmatrix}-2-\lambda&1\\\ -9&-8-\lambda\end{pmatrix} =(\lambda+2)(\lambda+8)+9 =\lambda^2+10\lambda+25=(\lambda+5)^2.
Logo
\det(A_{22}-\lambda I)=(-5-\lambda)(\lambda+5)^2=-(\lambda+5)^3,
produzindo o fator (-5-\lambda)^3.

Passo 5

Por fim \det(A_{33}-\lambda I_2).
A_{33}-\lambda I =\begin{pmatrix}-2-\lambda&1\\\ -16&6-\lambda\end{pmatrix},
\det(A_{33}-\lambda I) =(-2-\lambda)(6-\lambda)+16 =\lambda^2-4\lambda+4=(\lambda-2)^2,
isto é, fator (2-\lambda)^2.

Passo 6

Multiplicando os três blocos:
p_A(\lambda) =(5-\lambda)^3\,(-5-\lambda)^3\,(2-\lambda)^2.
Entre as alternativas, corresponde à letra (E).

Resposta

(E) p_{c}^{A}(\lambda)=(5-\lambda)^{3}(-5-\lambda)^{3}(2-\lambda)^{2}

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