Question

Como se resolve isso, e qual a resolução?

2x - y - 4z + w - v = 7
-x + 3y -2z - w + 2v = 1
5x + y + 3z - 4w + v = 33
3x - 2y - 2z - 2w + 3v = 24
-4x - y - 5z + 3w - 4v = 49

Answer 1

✅ A solução do sistema linear é (x,y,z,w,v)≈(-8,5;8,2;-7,9;-30,6;-14,7)

 

✍️ Solução: Vamos utilizar o método de Gauss para a solução do SL. A matriz estendida associada é

$\large\begin{array}{lr}\rm \begin{cases}\rm 2x-y-4z+w-v=7\\\rm -x + 3y -2z - w + 2v = 1\\\rm 5x + y + 3z - 4w + v = 33\\\rm 3x - 2y - 2z - 2w + 3v = 24\\\rm -4x - y - 5z + 3w - 4v = 49\end{cases}\sim \left[\begin{array}{ccccc|c}\rm 2&\rm -1&\rm -4&\rm 1&\rm -1&\rm 7\\\rm -1&\rm 3&\rm -2&\rm -1&\rm 2&\rm 1\\\rm 5&\rm 1&\rm 3&\rm -4&\rm 1&\rm 33\\\rm 3&\rm -2&\rm -2&\rm -2&\rm 3&\rm 24\\\rm -4&\rm -1&\rm -5&\rm 3&\rm -4&\rm 49\end{array}\right]\end{array}$

Observe as operações elementares

$\large\begin{array}{lr}\rm\left[\begin{array}{ccccc|c}\rm 2&\rm -1&\rm -4&\rm 1&\rm -1&\rm 7\\\rm -1&\rm 3&\rm -2&\rm -1&\rm 2&\rm 1\\\rm 5&\rm 1&\rm 3&\rm -4&\rm 1&\rm 33\\\rm 3&\rm -2&\rm -2&\rm -2&\rm 3&\rm 24\\\rm -4&\rm -1&\rm -5&\rm 3&\rm -4&\rm 49\end{array}\right]\begin{array}{lr}\rm L_1 \leftrightarrow L_2\\\rm L_1 \to -L_1\end{array}~~\sim\left[\begin{array}{ccccc|c}\rm 1&\rm -3&\rm 2&\rm 1&\rm -2&\rm -1\\\rm 2&\rm -1&\rm -4&\rm 1&\rm -1&\rm 7\\\rm 5&\rm 1&\rm 3&\rm -4&\rm 1&\rm 33\\\rm 3&\rm -2&\rm -2&\rm -2&\rm 3&\rm 24\\\rm -4&\rm -1&\rm -5&\rm 3&\rm -4&\rm 49\end{array}\right]\end{array}$

Fazendo $\rm L_3 \to -5L_1+L_3;~L_4\to-3L_1+L_4;~L_5\to4L_1+L_5~e~L_2\to-2L_1+L_2$

$\large\sim\begin{array}{lr}\rm \left[\begin{array}{ccccc|c}\rm 1&\rm -3&\rm 2&\rm 1&\rm -2&\rm -1\\\rm 0&\rm 5&\rm -8&\rm -1&\rm 3&\rm 9\\\rm 0&\rm 16&\rm -7&\rm -9&\rm 11&\rm 38\\\rm 0&\rm 7&\rm -8&\rm -5&\rm 9&\rm 27\\\rm 0&\rm -13&\rm 3&\rm 7&\rm -12&\rm 45\end{array}\right]\end{array}$

Fazendo $\rm L_2 \to\tfrac{1}{5}L_2;~L_3\to-16L_2+L_3;~L_4\to-7L_2+L_4~e~L_5\to13L_2+L_5$

$\large\sim\begin{array}{lr}\rm \left[\begin{array}{ccccc|c}\rm 1&\rm -3&\rm 2&\rm 1&\rm -2&\rm -1\\&&&&\\\rm 0&\rm 1&\rm -\tfrac{8}{5}&\rm -\tfrac{1}{5}&\rm \tfrac{3}{5}&\rm \tfrac{9}{5}\\&&&&\\\rm 0&\rm 0&\rm \tfrac{93}{5}&\rm -\tfrac{29}{5}&\rm \tfrac{7}{5}&\rm \tfrac{46}{5}\\&&&&\\\rm 0&\rm 0&\rm \tfrac{16}{5}&\rm -\tfrac{18}{5}&\rm \tfrac{24}{5}&\rm \tfrac{72}{5}\\&&&&\\\rm 0&\rm 0&\rm -\tfrac{89}{5}&\rm \tfrac{22}{5}&\rm -\tfrac{21}{5}&\rm \tfrac{342}{5}\end{array}\right]\end{array}$

Fazendo $\rm L_3 \to\tfrac{5}{93}L_3;~L_4\to-\tfrac{16}{5}L_3+L_4;~L_5\to\tfrac{89}{5}L_3+L_5~e~L_5\to13L_3+L_5$

$\large\sim\begin{array}{lr}\rm \left[\begin{array}{ccccc|c}\rm 1&\rm -3&\rm 2&\rm 1&\rm -2&\rm -1\\&&&&\\\rm 0&\rm 1&\rm -\tfrac{8}{5}&\rm -\tfrac{1}{5}&\rm \tfrac{3}{5}&\rm \tfrac{9}{5}\\&&&&\\\rm 0&\rm 0&\rm 1&\rm -\tfrac{29}{93}&\rm \tfrac{7}{93}&\rm \tfrac{46}{93}\\&&&&\\\rm 0&\rm 0&\rm 0&\rm -\tfrac{242}{93}&\rm \tfrac{424}{93}&\rm \tfrac{1192}{93}\\&&&&\\\rm 0&\rm 0&\rm 0&\rm -\tfrac{107}{93}&\rm -\tfrac{266}{93}&\rm \tfrac{7180}{93}\end{array}\right]\end{array}$

Fazendo $\rm L_4 \to-\tfrac{93}{242}L_4;~L_5\to\tfrac{107}{93}L_4+L_5$

$\large\sim\begin{array}{lr}\rm \left[\begin{array}{ccccc|c}\rm 1&\rm -3&\rm 2&\rm 1&\rm -2&\rm -1\\&&&&\\\rm 0&\rm 1&\rm -\tfrac{8}{5}&\rm -\tfrac{1}{5}&\rm \tfrac{3}{5}&\rm \tfrac{9}{5}\\&&&&\\\rm 0&\rm 0&\rm 1&\rm -\tfrac{29}{93}&\rm \tfrac{7}{93}&\rm \tfrac{46}{93}\\&&&&\\\rm 0&\rm 0&\rm 0&\rm 1&\rm \tfrac{-212}{121}&\rm -\tfrac{596}{121}\\&&&&\\\rm 0&\rm 0&\rm 0&\rm 0&\rm -\tfrac{-590}{121}&\rm \tfrac{8656}{121}\end{array}\right]\end{array}$

Por fim, fazendo $\rm L_5\to-\tfrac{121}{590}L_5$

$\large\sim\begin{array}{lr}\rm \left[\begin{array}{ccccc|c}\rm 1&\rm -3&\rm 2&\rm 1&\rm -2&\rm -1\\&&&&\\\rm 0&\rm 1&\rm -\tfrac{8}{5}&\rm -\tfrac{1}{5}&\rm \tfrac{3}{5}&\rm \tfrac{9}{5}\\&&&&\\\rm 0&\rm 0&\rm 1&\rm -\tfrac{29}{93}&\rm \tfrac{7}{93}&\rm \tfrac{46}{93}\\&&&&\\\rm 0&\rm 0&\rm 0&\rm 1&\rm \tfrac{-212}{121}&\rm -\tfrac{596}{121}\\&&&&\\\rm 0&\rm 0&\rm 0&\rm 0&\rm 1&\rm \tfrac{-4328}{295}\end{array}\right]\end{array}$

Com o escalonamento o sistema se reduz a esse abaixo. Observe que basta fazer substituição para trás

$\large\left.\begin{array}{lr}\rm\begin{cases}\rm x-3\rm y+2z+w-2v=-1\\\rm0x + y-\tfrac{8}{5}z-\tfrac{1}{5}w+\tfrac{3}{5}v=\tfrac{9}{5}\\\rm 0x+0y+z-\tfrac{29}{93}w+\tfrac{7}{93}v=\tfrac{46}{93} \\\rm 0x+0y+0z+w-\tfrac{212}{121}v = -\tfrac{596}{121}\\\rm 0x+0y+0z+0w+v = -\tfrac{4328}{295}\end{cases} \end{array}\right\uparrow$

✔ Daí o conjunto solução é

$\large\begin{array}{lr}\red{\underline{\boxed{\boxed{\rm\therefore\:S=\left\{(x,y,z,w,v) = \left(-\dfrac{2522}{295},~-\dfrac{2433}{295},~-\dfrac{2346}{295},~-\dfrac{9036}{295},~-\dfrac{4328}{295}\right)\right\} }}}} \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\blacksquare\!\blacksquare \end{array}$

 

❏ Seção de links para complementar o estudo sobre eliminação gaussiana, sistemas lineares, álgebra linear:

• brainly.com.br/tarefa/19598700

$\rule{7cm}{0.01mm}\\\texttt{Bons estudos! :D}\\\rule{7cm}{0.01mm}$