Considere as seguintes matrizes
A=\left[\begin{array}{ccc} x & -2 & 1 \\\\ 0 & x+1 & 3 \\\\ 1 & -5 & 1 \end{array}\right], \quad B=\left[\begin{array}{ccc} x & 1 & 1 \\\\ 2 & x & 3 \\\\ -1 & 5 & x \end{array}\right] \quad \text { e } \quad C=\left[\begin{array}{cc} x^{2} & 2 x \\\\ 10 & x \end{array}\right] .
Resolva as seguintes equações: (b) \operatorname{det}(B)=\operatorname{det}\text{(C)}.

Passo 1

O enunciado pede para resolver a equação \det(B)=\det\text{(C)}, onde
B = \begin{pmatrix} x & 1 & 1\\\\ 2 & x & 3\\\\ -1 & 5 & x \end{pmatrix}, \quad C = \begin{pmatrix} x^2 & 2x\\\\ 10 & x \end{pmatrix}.
Vamos usar a fórmula do determinante de matrizes 3\times3 e 2\times2.

Passo 2

Calcular \det(B) por cofactores na primeira linha:
\det(B) = x\begin{vmatrix}x & 3\\\ 5 & x\end{vmatrix} -1\begin{vmatrix}2 & 3\\\ -1 & x\end{vmatrix} +1\begin{vmatrix}2 & x\\\ -1 & 5\end{vmatrix}.
Agora cada menor:
\begin{vmatrix}x & 3\\\ 5 & x\end{vmatrix} = x^2 - 15,\quad \begin{vmatrix}2 & 3\\\ -1 & x\end{vmatrix} = 2x + 3,\quad \begin{vmatrix}2 & x\\\ -1 & 5\end{vmatrix} = 10 + x.
Logo
\det(B)=x(x^2-15)-(2x+3)+(10+x) =x^3-15x-2x-3+10+x =x^3-16x+7.

Passo 3

Calcular \det\text{(C)} pela regra 2\times2:
\det\text{(C)} = x^2\cdot x - (2x)\cdot 10 = x^3 - 20x.

Passo 4

Igualar e resolver:
x^3-16x+7 = x^3-20x \quad\Longrightarrow\quad -16x+7=-20x \quad\Longrightarrow\quad 4x+7=0 \quad\Longrightarrow\quad x=-\frac{7}{4}.

Resposta

x = -\tfrac{7}{4}.

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