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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.

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up vote 2 down vote accepted

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 $$(x_1,x_2,x_3,x_4)=(u+2v,3u+2v,-4u,-4v)=u(1,3,-4,0)+v(2,2,0,-4)$$ Now you can read off the basis, $\{{\,(1,3,-4,0),(2,2,0,-4)\,\}}$.

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What happened to the Tex? Some of your letters overlap and I cannot read it that well. – CodeKingPlusPlus Oct 4 '12 at 3:53
Nvm. It is rendering fine now. But, where does the $u$ and $v$ come from exactly? – CodeKingPlusPlus Oct 4 '12 at 3:59
Could you provide a short explanation? – CodeKingPlusPlus Oct 4 '12 at 4:05
If you don't like $u$ and $v$, you can use $s$ and $t$, instead. You'll get an answer with fractions in it. There's nothing wrong with that, but I prefer to avoid fractions, so I multiplied everything by 4. If $a,b$ is a basis for a vector space, so is $4a,4b$. – Gerry Myerson Oct 4 '12 at 4:06

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