Find the basis for the solution space of the system and describe all solutions:
$3x_1 - x_2 + x_4 = 0$
$x_1 + x_2 + x_3 + x_4 = 0$
I row reduce the matrix:
$\begin{pmatrix} 1&1&1&1\\ 0&-4&-3&-2 \end{pmatrix}$
And from here I do not know what to do.
$\endgroup$1 Answer
$\begingroup$The form of the reduced matrix tells you that everything can be expressed in terms of the free parameters $x_3$ and $x_4$. It may be helpful to take your reduction one more step and get to $$\pmatrix{4&0&1&2\cr0&4&3&2\cr}$$ Now writing $x_3=s$ and $x_4=t$ the first row says $x_1=(1/4)(-s-2t)$ and the second row says $x_2=(1/4)(-3s-2t)$. If we don't like fractions, we can instead write $x_3=-4u$, $x_4=-4v$, whence $x_1=u+2v$, $x_2=3u+2v$. So we have $$(x_1,x_2,x_3,x_4)=(u+2v,3u+2v,-4u,-4v)=u(1,3,-4,0)+v(2,2,0,-4)$$ Notice this is also equivalent to $span\{{(1,3,-4,0),(2,2,0,-4)\}}$ due to the parameters $u$ and $v$. Also notice that these two vectors are linearly independent.
Now you can read off the basis, $\{{\,(1,3,-4,0),(2,2,0,-4)\,\}}$.
This set is a basis because a) It is linearly independent, and b) because it spans the solution space
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