How to solve this homogeneous system, with a missing column?

Question

Find the solution set of triplets $(x,y,z)$ that fulfil this system using Gauss-Jordan:

$$\begin {cases} -x + 2z = 0\\ 3x - 6z = 0\\2x - 4z = 0\end {cases}$$

First of all, I don't see any $y$ variable there. I suppose it doesn't matter and I proceed normally:

\begin{bmatrix} -1 & 2 & 0\\ 3 & -6 & 0\\2 & -4 & 0 \end{bmatrix}

$$-f_1$$

\begin{bmatrix} 1 & -2 & 0\\ 3 & -6 & 0\\2 & -4 & 0 \end{bmatrix}

$$-3f_1+f_2$$

\begin{bmatrix} 1 & -2 & 0\\ 0 & 0 & 0\\2 & -4 & 0 \end{bmatrix}

$$-2f_1+f_3$$

\begin{bmatrix} 1 & -2 & 0\\ 0 & 0 & 0\\0 & 0 & 0 \end{bmatrix}

So, this is the staggered reduced form.

This is an homogeneous system (because of the null column), thus one solution is $(0,0,0)$.

Other than that, I have to check out the range of the system. The range is $1$, which is less than the number of columns... what is the number of columns? Is it $2$ because of the matrix I was working on, or is it $3$ because there was a hidden column for $y$? (Hidden because $y = 0$).

Anyway. The reason I want to know the "number of columns" is to calculate $$columns - range$$

Which yields the amount of parameters this system's solution depends on.

Answer 1

You are almost at the end.
The number of parameters is $3-1=2$, so we have two parameter say $y=t, z=s$ and hence $x=2s$ so the set of solutions are of the form $$(x,y,z)=(2s,t,s)=(0,0,0)+s(2,0,1)+t(0,1,0)$$ which is a plane or 2 dimensional subspace

Answer 2

You have two variables. and three equations. Solve for $x$ and $z$ using any two, get $0,0$. Confirm that it is consistent with your third.

Now deduce that all triples $(x,y,z)=(0,a,0)$ where $a\in\Bbb R$ satisfy this, because your equations don't depend on $y$.

Answer 3

Your solution is $x-2y=0$ what implies:

$y=1/2 x$

and because of $ 0 z=0$, $z$ is a free parameter.
and you have a vector solution:

$f=[x;y;z]=[x;1/2x; z]=x[1;1/2;0]+z[0;0;1]$

and now you have a whole space of solutions described with this vector, for particular solution just put an value of arbitrary $x$ and $z$ in the solution.

so the number of parameters is 2 ($x$ and $z$).

Answer 4

The column of $y$ is not hidden, it's just (apparently) omitted by you which I consider to be an error on your part. In matrix form your system of equations can be written as $$ \begin{bmatrix} -1 & 0 & 2 \\ 3 & 0 & -6 \\2 & 0 &-4 \end{bmatrix} \begin{bmatrix} x \\y\\z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$ which is also equivalent to $$ \begin{bmatrix} -1 & 2 & 0 \\ 3 & -6 & 0 \\ 2 & -4 & 0 \end{bmatrix} \begin{bmatrix} x \\ z \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$ and can be written in augmented form as $$\left[ \begin{array}{ccc|c} -1 & 2 & 0 & 0 \\ 3& -6 & 0 &0 \\2 & -4& 0 & 0 \end{array}\right] $$. From here row reduction will produce an almost identical system to the one you found, but you have to account for the $y$'s.

Answer 5

you can see your equation as$$-x+0y+2z=0$$ in fact the second and third equation are the same thing. so $x=2z$, then your system $(x,y,z)$ is of order $2$. You can represent it as $(2z,y,z)$ or $(x,y,x/2)$.

explanation of order 2: $2=3-1$,here $3$ is the number of dimension of your workspace, that is $(x,y,z)$(I think that is also your number of column); $1$ is dimension of nullity of your matrix. This equality is the "Rank–nullity theorem".

hope this explains your column and range.