Considere o sistema linear \underset{\sim}{A} \cdot \underset{\sim}{x}=\underset{\sim}{b}, onde \underset{\sim}{A}=\left(\begin{array}{cc}4 & 1 \\\\ 2 & -1\end{array}\right), \underset{\sim}{x}=\binom{x}{y} e \underset{\sim}{b}=\binom{7}{5}. Resolva o sistema encontrando a matriz inversa {\underset{\sim}{A}}^{-1} e calculando \underset{\sim}{x}={\underset{\sim}{A}}^{-1} \cdot \underset{\sim}{b}.

Passo 1

O enunciado dá o sistema linear \underset{\sim}{A}\,\underset{\sim}{x}=\underset{\sim}{b} com
\underset{\sim}{A} = \begin{pmatrix}4 & 1\\[6pt]2 & -1\end{pmatrix}, \quad \underset{\sim}{x} = \binom{x}{y}, \quad \underset{\sim}{b} = \binom{7}{5}.
Queremos, primeiro, encontrar a matriz inversa \underset{\sim}{A}^{-1} (conteúdo: inversa de matriz 2\times2) e depois calcular \underset{\sim}{x} = \underset{\sim}{A}^{-1}\,\underset{\sim}{b}.]

Passo 2

Calculamos o determinante de \underset{\sim}{A}. Para
\underset{\sim}{A} = \begin{pmatrix}a & b\\[3pt]c & d\end{pmatrix}
temos \det(\underset{\sim}{A}) = ad - bc. Aqui
a = 4,\; b = 1,\; c = 2,\; d = -1
logo
\det(\underset{\sim}{A}) = 4 \cdot (-1) - 1 \cdot 2 = -4 - 2 = -6.]

Passo 3

Como \det(\underset{\sim}{A}) \neq 0, a inversa existe e, para matriz 2\times2, vale
\underset{\sim}{A}^{-1} = \frac{1}{\det(\underset{\sim}{A})} \begin{pmatrix} d & -b\\[3pt] -c & a \end{pmatrix} = \frac{1}{-6} \begin{pmatrix} -1 & -1\\[3pt] -2 & 4 \end{pmatrix}.
Fazendo a multiplicação escalar, obtemos
\underset{\sim}{A}^{-1} = \begin{pmatrix} \frac{1}{6} & \frac{1}{6}\\[4pt] \frac{1}{3} & -\frac{2}{3} \end{pmatrix}.]

Passo 4

Agora multiplicamos \underset{\sim}{A}^{-1} por \underset{\sim}{b}:
\underset{\sim}{x} = \underset{\sim}{A}^{-1}\,\underset{\sim}{b} = \begin{pmatrix} \frac{1}{6} & \frac{1}{6}\\[4pt] \frac{1}{3} & -\frac{2}{3} \end{pmatrix} \binom{7}{5} = \binom{\frac{1}{6}\cdot7 + \frac{1}{6}\cdot5}{\frac{1}{3}\cdot7 - \frac{2}{3}\cdot5} = \binom{\frac{12}{6}}{\frac{7-10}{3}} = \binom{2}{-1}.]

Resposta

\underset{\sim}{A}^{-1} = \begin{pmatrix} \frac{1}{6} & \frac{1}{6}\\[4pt] \frac{1}{3} & -\frac{2}{3} \end{pmatrix}, \quad \underset{\sim}{x} = \binom{x}{y} = \binom{2}{-1}.]

Exercícios relacionados