Jtdylktuy

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.

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

$$begin{pmatrix}2t\ t\0 end{pmatrix}= tbegin{pmatrix}2\ 1\0 end{pmatrix}$$ 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):

$$begin{pmatrix}t\ 3t\2t end{pmatrix} = tbegin{pmatrix}1\ 3\2 end{pmatrix} $$

among others.