I am self-studying Linear Algebra, mainly through Khan Academy and some YouTube videos. I have encountered the following problem.
Given
$$A = \begin{bmatrix} 2 & -1 & -2\\ -4 & 2 & 4\\ -8 & 4 & 8 \end{bmatrix}$$
how can I find the solutions of $Ax = 0$? And how could these solutions be a linear combination of two vectors in $\mathbb{R}^{3}$?
As I tried reducing the matrix by the Gauss method and got $0$'s. I am quite confused as to what to do. Are there any videos that could explain this? Thank you for any help. When I reduced it with the Gauss method, I got the following:
Once you reduced the matrix, you can do an extra step and divide the first row by $2$. Then every row of zeros represents a free variable. So let $x_2=s$ and $x_3=t$, with $s,t \in \mathbb{R}$.
Now go back to a system of equations and you get $x_1-\frac{1}{2}x_2-x_3=0$, i.e. $x_1=\frac{1}{2}x_2+x_3=\frac{1}{2}s+t$. Therefore the solutions of the system are of the form $$ \pmatrix{\frac{1}{2}s+t \\ s \\ t } $$ or $$ \pmatrix{\frac{1}{2}s \\ s \\ 0 }+\pmatrix{t \\ 0 \\ t }=s\pmatrix{\frac{1}{2}\\ 1 \\ 0 }+t\pmatrix{1 \\ 0 \\ 1 } $$