I'm working through an introductory linear algebra textbook and one exercise gives the system
$2x+3y+5z+2w=0$
$-5x+6y-17z-3w=0$
$7x-4y+3z+13w=0$
And asks why, without doing any calculations, it has infinitely many solutions. Now, a previous exercise gives the same system without the fourth column and asks why, without any calculation, you can tell it's consistent, and I realized that it's because it has the trivial solution (0,0,0). But I'm struggling to see how that implies that this new system has infinitely many solutions.
I did some research and found that if an underdetermined linear system has a solution then it has infinitely many, but the explanations of this seem to talk about rank and other things that I'm not familiar with.
So if someone could please explain why you can just tell without doing any calculation why this system has infinitely many solutions (I'm guessing it has something to do with the previous problem that's the same just without that fourth column of variables) from the authors perspective (i.e. they're only assuming we have algebra 2 at this early point in the book) it would be much appreciated.
Best Answer
Note that the system without the fourth column is not only consistent but also determined ( the rows are linearly independent), this means that also the system: $$ \begin{cases} 2x+3y+5z=-2\\ -5x+6y-17z=3\\ 7x-4y+3z=-13 \end{cases} $$ is detrmined, i.e has one solution $(x,y,z)=(a,b,c)$.
Now, your system is
$$ \begin{cases} 2x+3y+5z=-2w\\ -5x+6y-17z=3w\\ 7x-4y+3z=-13w \end{cases} $$ so, by linearity, has the infinitely many solutions $(x,y,z)=(aw,bw,cw) \quad \forall w \in \mathbb{R}$