Dadas as matrizes A e B, quadradas de ordem 3 :
A=\left[\begin{array}{ccc} 2 & 1 & 0 \\\\ -1 & 3 & 4 \\\\ 1 & 2 & 1 \end{array}\right] \quad \text { e } \quad B=\left[\begin{array}{ccc} 1 & 0 & 2 \\\\ 3 & -2 & 1 \\\\ 0 & 1 & 1 \end{array}\right]
Calcule os seguintes determinantes:
7. \operatorname{det}\left(B^{2} \cdot A^{T}\right)

Passo 1

O enunciado nos dá as matrizes A e B de ordem 3 e pede \det(B^2 \cdot A^T). Para resolver, vamos usar as propriedades:
\det(XY)=\det(X)\det(Y),
\det(B^2)=\det(B\cdot B)=\det(B)^2
e
\det(A^T)=\det(A).

Passo 2

Calculamos \det(A) para
A=\begin{pmatrix} 2 & 1 & 0 \\\\ -1 & 3 & 4 \\\\ 1 & 2 & 1 \end{pmatrix}.
Expandindo pela primeira linha:
\det(A) =2\det\begin{pmatrix}3 & 4 \\\\ 2 & 1\end{pmatrix} -1\det\begin{pmatrix}-1 & 4 \\\\ 1 & 1\end{pmatrix} +0\cdot(\dots).
Logo,
\det(A) =2(3\cdot1 -4\cdot2) -1((-1)\cdot1 -4\cdot1) =2(-5)-1(-5) =-10+5 =-5.

Passo 3

Calculamos \det(B) para
B=\begin{pmatrix} 1 & 0 & 2 \\\\ 3 & -2 & 1 \\\\ 0 & 1 & 1 \end{pmatrix}.
Expandindo pela primeira linha:
\det(B) =1\det\begin{pmatrix}-2 & 1 \\\\ 1 & 1\end{pmatrix} -0\cdot(\dots) +2\det\begin{pmatrix}3 & -2 \\\\ 0 & 1\end{pmatrix}.
Logo,
\det(B) =( -2\cdot1 -1\cdot1 ) +2(3\cdot1 -0\cdot(-2)) =-3+2\cdot3 =-3+6 =3.

Passo 4

Pela regra do determinante de produto e da transposta:
\det(B^2 \cdot A^T) =\det(B^2)\det(A^T) =\det(B)^2\det(A).
Substituindo \det(B)=3 e \det(A)=-5,
\det(B^2 \cdot A^T) =3^2 \cdot (-5) =9\cdot(-5) =-45.

Resposta

-45

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