I have to find homogeneous system of linear equations whose solution space is: V = span((1,-2,4,3),(1,-1,6,4),(3,-8,8,3)).
First I found vectors were linearly dependent, so I discarded the third vector to form a new base. Next I figured system will have 2 free variables and if we would use gauss-jordan elimination on matrix of a system we would get this:
$$
\begin{bmatrix}
1 & 0 & * & *\\
0 & 1 & * & * \\
\end{bmatrix}
$$
Now I don't know how to find *'s or if I am on a right path. Hope you can help me, thanks in advance.
Actually the 3 vectors are linearly independent. For example, it is not hard to check that $$\left|\begin{array}{cccc}1 & -2 & 4 & 3 \\ 1 & -1 & 6 & 4 \\ 3 & -8 & 8 & 3 \\ 1 & 0 & 0 & 0 \end{array}\right|=-32\ne0$$ So we expect there will be just one equation. Suppose it is that the vector $(w,x,y,z)$ must satisfy $aw+bx+cy+dz=0$. Then we know that $a,b,c,d$ must satisfy:
(1) $a-2b+4c+3d=0$, (2) $a-b+6c+4d=0$, (3) $3a-8b+8c+3d=0$
Taking (2)-(1), we get $b+2c+d=0$, and taking 2(2)-(1), we get $a+8c+5d=0$. Substituting in (3) gives $d=0$. So the only possible equation is essentially $8w+2x-y=0$. It is easy to check that the three given points all satisfy it and hence all linear combinations of those points will also satisfy it.