Question

Resolva o sistema de equações lineares, utilizando método de eliminação de Gauss (escalonamento), e em seguida assinale a alternativa CORRETA que representa o conjunto solução deste sistema:
S={(3,1,2)}
S={(1,2,3)}
S={(-2,3,1)}
S={(-2,1,-3)}
S={(-2,-3,-1)}

Answer 1

$\begin{cases}\mathrm{x+y-2z=-1}\\ \mathrm{2x+y+z=0}\\ \mathrm{x+4y-6z=4}\\ \end{cases}$

$\mathrm{[A|b]^{(0)}=\left[\begin{array}{cccc}a_{11}&a_{12}&a_{13}&b_1\\a_{21}&a_{22}&a_{23}&b_2\\a_{31}&a_{32}&a_{33}&b_3\end{array}\right]=\left[\begin{array}{cccc}1&1&-2&-1\\2&1&1&0\\1&4&-6&4\end{array}\right]}\\\\\\ \mathrm{v=\dfrac{a_{21}}{a_{11}}=\dfrac{2}{1}=2\ \ \| \ \ w=\dfrac{a_{31}}{a_{11}}=\dfrac{1}{1}=1}\\\\ \mathrm{L_2^{(1)}=L_2-v.L_1=L_2-2.L_1\ \ \| \ \ L_3^{(1)}=L_3-w.L_1=L_3-L_1}\\\\ \mathrm{[A|b]^{(1)}=\left[\begin{array}{cccc}1&1&-2&-1\\0&-1&5&2\\0&3&-4&5\end{array}\right]}\\\\\\ \mathrm{k=\dfrac{a_{32}}{a_{22}}=\dfrac{3}{-1}=-3\ \ \|\ \ L_3^{(2)}=L_3^{(1)}-k.L_2^{(1)}=L_3^{(1)}+3.L_2^{(1)}}\\\\\\ \mathrm{[A|b]^{(2)}=\left[\begin{array}{cccc}1&1&-2&-1\\0&-1&5&2\\0&0&11&11\end{array}\right]=\left[\begin{array}{cccc}1&1&-2&-1\\0&-1&5&2\\0&0&1&1\end{array}\right]}$

$\begin{cases}\mathrm{x+y-2z=-1}\\ \mathrm{-y+5z=2\ \ \ \ \ \ \ \ \to}\\ \mathrm{z=1}\\ \end{cases}\ {\begin{cases}\mathrm{x=-2}\\ \mathrm{y=3\ \ \ \ \ \to}\\ \mathrm{z=1}\\ \end{cases}\ \boxed{\boxed{\mathbf{S=\{(-2,3,1)\}}}}$