Considere a matriz \mathrm{A}=\left(\begin{array}{ccc}1 & 0 & 1 \\\\ 0 & 1 & 5 \\\\ 3 & 0 & 1\end{array}\right) e a matriz \mathrm{B}=\left(\begin{array}{lll}2 & 2 & 2 \\\\ 0 & 2 & 2 \\\\ 1 & 1 & 2\end{array}\right). Calcule:
\operatorname{det}\left(A^{3} B^{2}\right);

Passo 1

O enunciado dá as matrizes
A=\begin{pmatrix}1 & 0 & 1 \\[4pt] 0 & 1 & 5 \\[4pt] 3 & 0 & 1\end{pmatrix} \quad\text{e}\quad B=\begin{pmatrix}2 & 2 & 2 \\[4pt] 0 & 2 & 2 \\[4pt] 1 & 1 & 2\end{pmatrix}
e pede \det(A^3B^2). Para isso vamos usar duas propriedades fundamentais do determinante:
  1. \det(XY)=\det(X)\,\det(Y).
  2. \det(X^k)=(\det(X))^k.]

Passo 2

Calcular \det(A) expandindo, por exemplo, pela primeira linha.
\det(A) =1\cdot\det\begin{pmatrix}1 & 5\\[2pt]0 & 1\end{pmatrix} -0\cdot\bigl(\cdots\bigr) +1\cdot\det\begin{pmatrix}0 & 1\\[2pt]3 & 0\end{pmatrix} =1\cdot(1\cdot1-5\cdot0)+1\cdot(0\cdot0-1\cdot3) =1-3=-2.]

Passo 3

Calcular \det(B) expandindo pela primeira coluna:
\det(B) =2\cdot\det\begin{pmatrix}2 & 2\\[2pt]1 & 2\end{pmatrix} -0\cdot\bigl(\cdots\bigr) +1\cdot\det\begin{pmatrix}2 & 2\\[2pt]0 & 2\end{pmatrix} =2\cdot(2\cdot2-2\cdot1)+1\cdot(2\cdot2-2\cdot0) =2\cdot2+1\cdot4 =4+4=8.]

Passo 4

Agora aplicamos
\det(A^3B^2)=\det(A^3)\,\det(B^2) =(\det A)^3\;(\det B)^2 =(-2)^3\cdot8^2 =-8\cdot64 =-512.

Resposta

-512

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