Question

x1+2x2+3x3=0
2x1+x1-2x3=0
-4x1-x2+2x3=0

quais os valores de x1,,x2,x3?

Answer 1

$\begin{cases}x_{1}+2x_{2}+3x_{3}=0\\2x_{1}+x_{2}-2x_{3}=0\\-4x_{1}-x_{2}+2x_{3}=0\end{cases}$

Escrevendo o sistema na forma matricial:

$\left[\begin{array}{cccc}~~1&~~2&~~3&~~~0\\~~2&~~1&-2&~~~0\\-4&-1&~~2&~~~0\end{array}\right]$

Vamos resolvê-lo pelo método da eliminação de Gauss:

$\left[\begin{array}{cccc}~~1&~~2&~~3&~~~0\\~~2&~~1&-2&~~~0\\-4&-1&~~2&~~~0\end{array}\right]~l_{2}\leftarrow l_{2}-2l_{1},~l_{3}\leftarrow l_{3}+4l_{1}\\\\\\\left[\begin{array}{cccc}1&~~2&~~3&~~~0\\0&-3&-8&~~~0\\0&~~7&~14&~~~0\end{array}\right]~l_{2}\leftarrow(-1)l_{2},~l_{3}\leftarrow(\frac{1}{2})l_{3}\\\\\\\left[\begin{array}{cccc}1&2&3&~~~0\\0&3&8&~~~0\\0&1&2&~~~0\end{array}\right]~l_{3}\leftarrow l_{3}-(\frac{1}{3})l_{2}$

$\left[\begin{array}{cccc}1&2&~~3&~~~0\\0&3&~~8&~~~0\\0&0&-\frac{2}{3}&~~~0\end{array}\right]~l_{3}\leftarrow(-\frac{3}{2})l_{3}\\\\\\\left[\begin{array}{cccc}1&2&3&~~~0\\0&3&8&~~~0\\0&0&1&~~~0\end{array}\right]~l_{2}\leftarrow l_{2}-8l_{3},~l_{1}\leftarrow l_{1}-3l_{2}\\\\\\\left[\begin{array}{cccc}1&2&0&~~~0\\0&3&0&~~~0\\0&0&1&~~~0\end{array}\right]~l_{2}\leftarrow(\frac{1}{3})l_{2}\\\\\\\left[\begin{array}{cccc}1&2&0&~~~0\\0&1&0&~~~0\\0&0&1&~~~0\end{array}\right]~l_{1}\leftarrow l_{1}-2l_{2}$

$\left[\begin{array}{cccc}1&0&0&~~~0\\0&1&0&~~~0\\0&0&1&~~~0\end{array}\right]$

Então, temos

$x_{1}+0x_{2}+0x_{3}=0~~~\therefore~~~\boxed{\boxed{x_{1}=0}}\\\\0x_{1}+x_{2}+0x_{3}=0~~~\therefore~~~\boxed{\boxed{x_{2}=0}}\\\\0x_{1}+0x_{2}+x_{3}=0~~~\therefore~~~\boxed{\boxed{x_{3}=0}}$

O sistema homogênio possui solução única, e essa é a solução trivial (Ou seja, os vetores (1,2,-4), (2,1,-1) e (3,-2,2) são linearmente independentes, já que a única forma de encontrar o vetor nulo a partir de combinação lineares entre esses vetores é a trivial)