# how to interpret the solution of a linear system?

i have the following reduced echelon form matrix

[(1,0,1,)(0,1,0)(0,0,0)] and the solutions are (2,1,0)

EDIT This should be the system of linear equations $$\left\{ \begin{matrix} x&+&&&z&=&2\\ &&y&&&=&1\\ \end{matrix} \right.$$

now I am asked which one of the following options describes the solution of the system:

a) single point at (2,1,0) b) there are no solutions c) a line through (0,0,0) and (2,1,0) d) a line through (2,1,0) and (0,1,2)

I know that it has infinite many solutions, as as the number of unknowns#equations and it doesnt have any inconsistency , so a and b are not the right answer.

ill give z the value of t, so I have (x,y,z)=(2-t, 1, t).... and from now on, in case I have done everything well I thought that I could get the general equation of that line , and plug the values inside and see if it works...but not sure..the two equations of the line I got are: x+z-2=0 and y-1=0 and then I plug them in and I got d as a result, as it fulfills the equation....is that right???

then they ask me that suppose that z=t for all real numbers is taken as a free variable in that linear system, what statement is true?

a) x doesnt depend on t b) z does not depend on t, c)y doesnt depend on t d0 all depend on t. I think is c.

please correct me if I am wrong, and pretty newbie with that and I have an exam coming up on tuesday....:S

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I believe you are right on both counts, the set $(x,y,z)=(2-t, 1, t) = (2,1,0) + t (-1,0,1)$ describes indeed a line through $(2,1,0)$ and $(0,1,2)$. (The first point is obtained for $t = 0$, the second for $t = 2$). And $y = 1$ shows that $y$ does not depend on $t$.
For the first part, it is useful to remember that the set of solutions of a system of linear homogeneous equations will be a subspace of the appropriate vector space. Whereas if the system is non-homogeneous (as in your case) it will be an affine subspace, that is, a translate of a subspace. In your case it is a line. Since $(2,1,0)$ and $(0,1,2)$ are distinct solutions, it will be indeed the line through them.
@Maximilian1988, you have argued correctly that the set of solutions forms a line. It is easy to check that both point are on the line. But, whereas the line is correctly described by $(2,1,0)+t(-1,0,1)$, for $t \in \mathbf{R}$, it is not correctly described by your other formula. Given the two points, you construct the line for instance as $(2,1,0) + s ((2,1,0)-(0,1,2))= (2,1,0) + s (2,0,-2)$, for $s \in \mathbf{R}$, which is the same as the one just given, changing the parameter by $2 s = -t$. –  Andreas Caranti Feb 13 '13 at 16:05