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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 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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up vote 2 down vote accepted

I have taken the liberty of editing your post, writing the system of linear equations explicitly. Please check to see whether I did it right, according to your intentions.

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.

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Thanks very much for your time, you did well rewriting by the system. Can you know by just having the vectorial equation of the line if the points (2,1,0) and (0,1,2) belong to that line? With the points (2,1,0) and (0,1,2) I got this vector equation : (2,1,0)+t(-2,0,0), if I have it done it right , do I have to compare it somehow with (2,1,0)+t(-1,0,1) to know if those points belong to the first equation? – Maximilian1988 Feb 13 '13 at 12:33
@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
oh ok! I thought that given two points, in this case, lets call A(2,1,0) and B(0,2,1) you made the equation of the line by seeting and origin point, let's say A and then B-A to get the director vector, so (x,y,z)=(x1,y1,z1)+ t(x2-x1, y2-y1, z2-z1) So A(x1,y1,z1) and B (x1, y2,z2) but it seems that you did it the other way, when it comes to stablish the vector director – Maximilian1988 Feb 14 '13 at 1:48

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