Ajude a resolver por eliminação de gauss : x + 2y -3z +2t = -2 2x - 3y -4z + 5t = 2 -x + 2y + 4z - 4.... Pergunta de ideia deLucasMendes2712

December 2019 1 69 Report

Ajude a resolver por eliminação de gauss :

x + 2y -3z +2t = -2
2x - 3y -4z + 5t = 2
-x + 2y + 4z - 4t = 0
2x +2y + 2z - t = 1

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acidbutter $\begin{cases}x+2y-3z+2t=-2\\ 2x-3y-4z+5t=2 \\ -x+2y+4z-4t=0 \\ 2x+2y+2z-t=1 \end{cases}$

i) Escreva o sistema na forma de matriz:
$A= \left[\begin{array}{cccc}1&2&-3&2\\2&-3&-4&5\\-1&2&4&-4\\2&2&2&-1\end{array}\right] \\\\B= \left[\begin{array}{c}-2\\2\\0\\1\end{array}\right]$
onde o sistema é:
$\left[A|B\right]= \left[\begin{array}{cccc}1&2&-3&2\\2&-3&-4&5\\-1&2&4&-4\\2&2&2&-1\end{array}\right|\left.\begin{array}{c}-2\\2\\0\\1\end{array}\right]$

O método de Gauss consiste em sucessivas operações entre linhas para se obter a forma reduzida da matriz, ou seja:
$\left[\begin{array}{ccc}x_1&x_2&x_3\\0&x_5&x_6\\0&0&x_9\end{array}\right]$
onde x1, x5 e x9 são chamados pivôs ou elementos líderes:
$\left[\begin{array}{ccccc}1&2&-3&2&-2\\2&-3&-4&5&2\\-1&2&4&-4&0\\2&2&2&-1&1\end{array}\right]\to\\\\L_4:L_4-L_2=\left[\begin{array}{ccccc}1&2&-3&2&-2\\2&-3&-4&5&2\\-1&2&4&-4&0\\0&5&6&-6&-1\end{array}\right]\to\\\\L_3:L_3+L_1=\left[\begin{array}{ccccc}1&2&-3&2&-2\\2&-3&-4&5&2\\0&4&1&-2&-2\\0&5&6&-6&-1\end{array}\right]\to\\\\L_2:L_4+L_2-2L_1=\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&-2&5&-3&5\\0&4&1&-2&-2\\0&5&6&-6&-1\end{array}\right]\to\\\\L_3:L_3+2L_2=\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&-2&5&-3&5\\0&0&11&-8&8\\0&5&6&-6&-1\end{array}\right]\to$
$L_2\leftrightarrow L_5=\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&5&6&-6&-1\\0&0&11&-8&8\\0&-2&5&-3&5\end{array}\right]\to\\\\L_2:L_2+2L_5=\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&1&16&-12&9\\0&0&11&-8&8\\0&-2&5&-3&5\end{array}\right]\to\\\\L_5:L_5+2L_2=\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&1&16&-12&9\\0&0&11&-8&8\\0&0&37&-27&23\end{array}\right]\to\\\\L_5:L_5-3L_2=\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&1&16&-12&9\\0&0&11&-8&8\\0&0&4&9&-4\end{array}\right]\to\\\\L_4:L_4-3L_5=\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&1&16&-12&9\\0&0&-1&-35&20\\0&0&4&9&-4\end{array}\right]\to$
$L_5:L_5+4L_4=\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&1&16&-12&9\\0&0&-1&-35&20\\0&0&0&-131&76\end{array}\right]$
agora passamos o sistema da forma matricial para a comum:
$\left[\begin{array}{ccccc}1&2&-3&2&-2\\0&1&16&-12&9\\0&0&-1&-35&20\\0&0&0&-131&76\end{array}\right]=\begin{cases}x+ 2y -3z+ 2t= -2\\~~~~~~~y+16z-12t=9\\~~~~~~~~~~~~~-z-35t=20\\~~~~~~~~~~~~~~~~~-131t=76\end{cases}$
agora podemos resolver o sistema facilmente:
$\displaystyle \begin{cases}x+ 2y -3z+ 2t= -2\\~~~~~~~y+16z-12t=9\\~~~~~~~~~~~~~-z-35t=20\\~~~~~~~~~~~~~~~~~-131t=76\end{cases}\implies t=-\frac{76}{131}\\\\i)~~~-z-35t=20\implies -z+\frac{2660}{131}=20\implies z=\frac{40}{131}\\\\ii)~~y+16z-12t=9\implies y=-\frac{640}{131}-\frac{912}{131}\implies y=-\frac{1552}{131}\\\\iii)~~x+2y-3z+2t=-2\implies x=\frac{3104+120+156-262}{131}=\frac{3118}{131}\\\\\boxed{S=\begin{cases}x=\frac{3118}{131}\\y=-\frac{1552}{131}\\z=\frac{40}{131}\\t=-\frac{76}{131}\end{cases}}$

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Bons estudos! :)

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