Solving the homogeneous system associated with a matrix

Question

So i have this matrix

$$ \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & -6 \\ 0 & 0 & 0 \\ \end{pmatrix} $$

and I want to solve the homogeneous system associated with it

I re-write the matrix as

$$ \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & -6 \\ 0 & 0 & 0 \\ \end{pmatrix} · (x,y,z)= $$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ $$

However, I do not know how to proceed from here as I've started studying matrices as a solo student from a couple of days.

The solutions are the linear combinations of (1,1,1) and (-1,1,0).

I am confused on how to get to these solutions which are the linear combinations of (1,1,1) and (-1,1,0) and if anybody could show me the steps the simplest way possible it would be great ! Thanks!

Answer 1

The homogeneous system of linear equations whose coefficients matrix is your matrix is equivalent with the homogeneous system whose coefficients matrix is

$$\begin{pmatrix} 0&0&0\\0&0&0\\ 1&1&-2\end{pmatrix}\implies x+y-2z=0\implies \text{Span}\,\left\{\;\begin{pmatrix}1\\-1\\0\end{pmatrix}\;,\;\;\begin{pmatrix}2\\0\\1\end{pmatrix}\;\right\}$$

Observe that both vectors you mention at the end of your question belong to the solution space (the span) above, and they both are linearly independents, so you can choose those two as basis or else you can take the basis I wrote above.

Answer 2

It's quite simple: your system is equivalent to the single equation $$x+y-2z=0\iff z=\tfrac12(x+y)$$ so the subspace $S$ of solutions is isomorphic to $K^2$ ($K$ is the base field) via the isomorphism: \begin{align} K^2&\longrightarrow S\subset K^3\\ (x,y)&\longmapsto (x,y,\tfrac12(x+y)) \end{align} Now an isomorphism maps a basis onto a basis, so determine a basis of $K^2$ that maps onto the indicated basis.