Ensino Médio ⇒ (ITA) Matriz inversa e equações matriciais Tópico resolvido

[Idel007]

Se [tex3]M=\begin{bmatrix}
1 & -1 \\
2 & 0 \\
\end{bmatrix}[/tex3] e [tex3]N=\begin{bmatrix}
2 & 1 \\
-1 & 3 \\
\end{bmatrix}[/tex3], então [tex3]MN^T-M^{-1}N[/tex3] é igual a:

a) [tex3]\begin{bmatrix}
\frac{3}{2} & -\frac{5}{2} \\
\frac{5}{2} & -\frac{3}{2} \\
\end{bmatrix}[/tex3]
b) [tex3]\begin{bmatrix}
\frac{3}{2} & -\frac{1}{2} \\
\frac{7}{2} & \frac{5}{2} \\
\end{bmatrix}[/tex3]
c) [tex3]\begin{bmatrix}
\frac{3}{2} & -\frac{11}{2} \\
\frac{13}{2} & -\frac{5}{2} \\
\end{bmatrix}[/tex3]
d) [tex3]\begin{bmatrix}
\frac{3}{2} & -\frac{5}{2} \\
\frac{13}{2} & -\frac{3}{2} \\
\end{bmatrix}[/tex3]
e) [tex3]\begin{bmatrix}
\frac{3}{2} & -\frac{1}{2} \\
\frac{13}{2} & -\frac{3}{2} \\
\end{bmatrix}[/tex3]

Resposta

---

C

[petrasMOD]

@Idel007,

A transposta de N: [tex3]N^T = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}[/tex3]

Agora, multiplicamos [tex3] MN^T = \begin{bmatrix} 1 & -1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} (1 \cdot 2) + (-1 \cdot 1) & (1 \cdot -1) + (-1 \cdot 3) \\ (2 \cdot 2) + (0 \cdot 1) & (2 \cdot -1) + (0 \cdot 3) \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 4 & -2 \end{bmatrix}[/tex3]

A inversa de M: [tex3]M^{-1} = \frac{1}{\det(M)} \text{adj}(M)[/tex3]
:Determinante de M: [tex3]\det(M) = (1 \cdot 0) - (-1 \cdot 2) = 2[/tex3]
Matriz Adjunta (para 2x2): Trocamos os elementos da diagonal principal e invertemos o sinal da secundária: [tex3] \text{adj}(M) = \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix}\\Inversa: M^{-1} = \frac{1}{2} \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1/2 \\ -1 & 1/2 \end{bmatrix}[/tex3]
Agora, multiplicamos [tex3] M^{-1}N = \begin{bmatrix} 0 & 1/2 \\ -1 & 1/2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} -1/2 & 3/2 \\ -5/2 & 1/2 \end{bmatrix}[/tex3].
[tex3]MN^T - M^{-1}N:\begin{bmatrix} 1 & -4 \\ 4 & -2 \end{bmatrix} - \begin{bmatrix} -1/2 & 3/2 \\ -5/2 & 1/2 \end{bmatrix} = \begin{bmatrix} 1 - (-1/2) & -4 - 3/2 \\ 4 - (-5/2) & -2 - 1/2 \end{bmatrix}[/tex3]
[tex3]1 + 1/2 = \mathbf{3/2}\\-4 - 3/2 = -8/2 - 3/2 = \mathbf{-11/2}\\4 + 5/2 = 8/2 + 5/2 = \mathbf{13/2}\\-2 - 1/2 = -4/2 - 1/2 = \mathbf{-5/2}[/tex3]
A matriz resultante é: [tex3] \boxed{\begin{bmatrix} 3/2 & -11/2 \\ 13/2 & -5/2 \end{bmatrix}}[/tex3]