<s>[INST] Determine whether the solution space of the system $Ax=0$ is a line through the origin, a plane through the origin, or the origin only Q:Determine whether the solution space of the system $Ax=0$ is a line through the origin, a plane through the origin, or the origin only. If it is a plane, find an equation for it; if it is a line, find parametric equations for it.$(i)
   \left( \begin{array}{rrr}
  1 & -2 & 7  \\
   -4 & 8 & 5  \\
   2 & -4 & 3  \\ 
   \end{array}
\right)(ii)\left( \begin{array}{rrr}
  1 & 2 & 3  \\
   2 & 5 & 3  \\
   1 & 0 & 8  \\ 
   \end{array}
\right)(iii)\left( \begin{array}{rrr}
  -1 & 1 & 1  \\
   3 & -1 & 0  \\
   2 & -4 & -5  \\ 
   \end{array}
\right)$My Approach:First of all i transforming the matrices to reduced row echelon form:$(i)
   \left( \begin{array}{rrr}
  1 & -2 & 0  \\
   0 & 0 & 1  \\
   0 & 0 & 0  \\ 
   \end{array}
\right)(ii) \left( \begin{array}{rrr}
  1 & 0 & 0  \\
   0 & 1 & 0  \\
   0 & 0 & 1  \\ 
   \end{array}
\right)(iii) \left( \begin{array}{rrr}
  1 & 0 & \frac{1}{2}  \\
   0 & 1 & \frac{3}{2}  \\
   0 & 0 & 0  \\ 
   \end{array}
\right)$But now i get stuck because i don't know how to related them geometrically.Anyone can explain it elaborately that's make my intuition on it.Any hints or solution will be appreciated.
Thanks in advance.
 [/INST] Assuming that the reduction has been done correctly, (i) has a two-dimensional image space, so its null space is one-dimensional, hence a line.
To find a parametrisation, note that $y$ can be chosen freely, $z=0$ is forced by the second equation in reduced form, and the first one forces $x=2y$, so 
$$\begin{pmatrix}2t\\ t\\0 \end{pmatrix}= t\begin{pmatrix}2\\ 1\\0 \end{pmatrix}$$ is a possible  parametrisation of the solution of $Ax=0$.
(ii) has a three-dimensional image space, so the null space is just $\{(0,0,0)\}$.
In (iii) the third variable $z$ is free, $y - \frac{3}{2}z=0$ so $y=\frac{3}{2}z$, and $x - \frac{1}{2}z=0$, so $x=\frac{1}{2}z$, so a possible parametrisation is (taking $z=2t$ to get rid of the fractions):
$$\begin{pmatrix}t\\ 3t\\2t \end{pmatrix} = t\begin{pmatrix}1\\ 3\\2 \end{pmatrix} $$
among others.
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