Now I know that x3 is the free variable in the below case:
$$ \begin{pmatrix} 1 & 0 & -5 & 0 \\ 0 & 1 & -1 & 0\\ 0& 0& 0&0 \end{pmatrix} $$ Note that the last column of the matrix column is the augmented column
A theorem in linear algebra state: "If there are more variables than equation, then there is a non trivial solution(in fact infinitely many)."
But, I'm unable to determine in the following case:
$$ \begin{pmatrix} 1 & 0 & -1 & 0\\ 0& 1& 2&0 \end{pmatrix} $$
For a free variable it has to be the case that: x3 = 0, isn't it? but in the above case, the equations become:
x1-x3 = 0
x2+x3 = 0
Is there a free variable here? IF so how and which?
$x_3$ is a free variable.
Let $x_3=t$, can you express $x_1=t$ and $x_2=-2t$. You get to freely choose $x_3$ and it determine $x_1$ and $x_2$.
Remark:
$x_2+x_3=0$ should be $x_2+2x_3=0$.