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2016-12-30T04:47:10-02:00
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Classificar o sistema de equações lineares:

\left\{\! \begin{array}{rcrcrcrc} \mathsf{2x}&\!\!\!+\!\!\!&\mathsf{3y}&\!\!\!+\!\!\!&\mathsf{4z}&\!\!\!=\!\!\!&\mathsf{9}&\quad\mathsf{(i)}\\ \mathsf{x}&\!\!\!-\!\!\!&\mathsf{y}&\!\!\!+\!\!\!&\mathsf{2z}&\!\!\!=\!\!\!&\mathsf{2}&\quad\mathsf{(ii)}\\ \mathsf{x}&\!\!\!+\!\!\!&\mathsf{4y}&\!\!\!+\!\!\!&\mathsf{2z}&\!\!\!=\!\!\!&\mathsf{7}&\quad\mathsf{(iii)} \end{array} \right.


Escrevendo a matriz ampliada e escalonando:

\begin{array}{cc} \begin{bmatrix}2&3&4&9\\\\1&-1&2&2\\\\1&4&2&7\end{bmatrix}\quad&\quad\begin{array}{l} L_1\leftarrow \frac{1}{2}\,L_1\\\\ \\\\ \\\\ \end{array} \end{array}\\\\\\\\ \begin{array}{cc} \begin{bmatrix}1&\frac{3}{2}&2&\frac{9}{2}\\\\1&-1&2&2\\\\1&4&2&7\end{bmatrix}\quad&\quad\begin{array}{l} \\\\ L_2\leftarrow L_2-L_1\\\\ L_3\leftarrow L_3-L_1 \end{array} \end{array}


\begin{array}{cc} \begin{bmatrix}1&\frac{3}{2}&2&\frac{9}{2}\\\\0&-\frac{5}{2}&0&-\frac{5}{2}\\\\0&\frac{5}{2}&0&\frac{5}{2}\end{bmatrix}\quad&\quad\begin{array}{l} \\\\L_2\leftarrow -\frac{2}{5}\,L_2\\\\L_3\leftarrow -\frac{2}{5}\,L_3 \end{array} \end{array}\\\\\\\\ \begin{array}{cc} \begin{bmatrix}1&\frac{3}{2}&2&\frac{9}{2}\\\\0&1&0&1\\\\0&1&0&1\end{bmatrix}\quad&\quad\begin{array}{l} \\\\L_2\leftarrow -\frac{2}{5}\,L_2\\\\L_3\leftarrow -\frac{2}{5}\,L_3 \end{array} \end{array}


\begin{array}{cc} \begin{bmatrix}1&\frac{3}{2}&2&\frac{9}{2}\\\\0&1&0&1\\\\0&1&0&1\end{bmatrix}\quad&\quad\begin{array}{l} \\\\\\\\L_3\leftarrow L_3-L_2\end{array}\\\\\\\\ \begin{bmatrix}1&\frac{3}{2}&2&\frac{9}{2}\\\\0&1&0&1\\\\0&0&0&0\end{bmatrix}\quad& \end{array}


Da matriz ampliada escalonada, podemos voltar ao sistema, obtendo assim

\left\{\! \begin{array}{rcrcrcrc} \mathsf{x}&\!\!\!+\!\!\!&\mathsf{\frac{3}{2}\,y}&\!\!\!+\!\!\!&\mathsf{2z}&\!\!\!=\!\!\!&\mathsf{\frac{9}{2}}&\quad\mathsf{(iv)}\\\\ &&\mathsf{y}&&&\!\!\!=\!\!\!&\mathsf{1}&\quad\mathsf{(v)}\\\\ &&&&\mathsf{0}&\!\!\!=\!\!\!&\mathsf{0}&\quad\mathsf{(vi)} \end{array} \right.


Da equação \mathsf{(v)} acima, tiramos diretamente que

\mathsf{y=1}\qquad\quad\checkmark


Substituindo na equação \mathsf{(iv)}, obtemos

\mathsf{x+\dfrac{3}{2}\cdot 1+2z=\dfrac{9}{2}}\\\\\\ \mathsf{x+2z=\dfrac{9}{2}-\dfrac{3}{2}}\\\\\\ \mathsf{x+2z=\dfrac{6}{2}}\\\\\\ \mathsf{x+2z=3}

\mathsf{2z=3-x}\\\\ \mathsf{z=\dfrac{1}{2}\,(3-x)\qquad\quad\checkmark}


Note que a variável \mathsf{z} fica dependendo da variável \mathsf{x}.


Fazendo \mathsf{x=\lambda\in\mathbb{R},} obtemos o conjunto solução para o sistema:

\left\{\! \begin{array}{l} \mathsf{x=\lambda}\\\\ \mathsf{y=1}\\\\ \mathsf{z=\dfrac{1}{2}\,(3-\lambda)} \end{array}\right.\qquad\qquad\textsf{com }\lambda \in\mathbb{R}


onde para cada valor que se escolha para \lambda, obtém-se uma solução para o sistema.


Conjunto solução:

\mathsf{S=\left\{(x,\,y,\,z)\in\mathbb{R}^3:~~x=\lambda,\;y=1,\;z=\dfrac{1}{2}\,(3-\lambda),~~com~~\lambda\in\mathbb{R}\right\}}.


Resposta:  alternativa b) admite infinitas soluções.


Bons estudos! :-)


Tags:  resolução classificação sistemas lineares escalonamento determinante regra de cramer matriz coeficiente solução resolver álgebra

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