IME / ITA ⇒ (Escola Naval) Matrizes Tópico resolvido

[ALDRINMOD]

Se [tex3]\begin{bmatrix}2 & -1\\\frac{-5}{2} & \frac{3}{2}\end{bmatrix}[/tex3] é a matriz inversa de [tex3]\begin{bmatrix}x & log_y 4\\ 5 & y^2\end{bmatrix}[/tex3], então [tex3]x^y[/tex3] é igual a:

(A) [tex3]-8[/tex3]
(B) [tex3]\frac{1}{9}[/tex3]
(C) [tex3]6[/tex3]
(D) [tex3]8[/tex3]
(E) [tex3]9[/tex3]

[LucasDN684]

[tex3]
\begin{bmatrix}
2& -1\\
-\frac{5}{2} & \frac{3}{2}
\end{bmatrix}.\begin{bmatrix}
x & \log_{y}4 \\
5 & y^{2}
\end{bmatrix}=\begin{bmatrix}
2x-5 & 4\log_{y}2-y^{2} \\
-\frac{5}{2}x +\frac{15}{2} & -5\log_{y}2+\frac{3}{2}y^{2}
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix},\, 0< y\neq 1\, \, \, (I)
[/tex3]

Então segue:

[tex3](I)\Rightarrow \left\{\begin{matrix}
2x-5=1\\
4\log_{y}2-y^{2}=0\\
-5\log_{y}2+\frac{3}{2}y^{2}=1
\end{matrix}\right.\overset{\log_{y}2 \text{ em função de y}} \Leftrightarrow \, \, \left\{\begin{matrix}
x=3\\
y=2
\end{matrix}\right.\therefore x^{y}=3^{2}=9[/tex3]

[petrasMOD]

ALDRIN,
Outro modo:
Calculando a inversa:

[tex3]\begin{vmatrix}
\underbrace{\dfrac{y^2}{xy^2-10\log _y2}}_k& \dfrac{-5}{k} \\
\dfrac{-2\log _y2}{k}& \dfrac{x}{k} \\
\end{vmatrix}\\
Igualando: -\frac{5}{k} = -\frac{5}2 \implies k =2 \\
\therefore \frac{x}{k}=\frac{3}{2} \implies x = 3\\
-\frac{2\log _y2}{k}=-1 \implies y = 2\\
\therefore \boxed{x^y=3^2=9} [/tex3]