solving a basis with two free variables

Question

I've been trying to solve a basis for this question but nothing seems to be working. I've written the linear system in a matrix which consists of the two equations, put it into reduced row echelon form, found that $z$ and $w$ are free variables. what's next?

$w=-x-y$ then $x-z-2x-2y=0$ so $z=-x-2y$. Now what can you say for the vectors $(1, 0, -1, -1), (0, 1, -2, -1)$ ? — dmtri, Sep 01 '18 at 05:22
they form the basis? how did you arrive at $1,0,-1,-1 and 0,1,-2,-2$ though? — lohboys, Sep 01 '18 at 05:47
You can solve the system of your two equations and you get a solution with 2 free variables. Yes they are independent. — dmtri, Sep 01 '18 at 06:46

score 2 · Answer 1 · answered Sep 01 '18 at 06:16

$x+y+w=0$

$x-z+2w=0$

The first equation leads to $x = -y-w$ and the second leads to $x = z-2w$. Then we have $-y-w = z-2w$, so $y = -z+w$

Then $x = z-2w$ and $y = -z+w$. We have two variables $z,w$ that are free, as expected.

So you have a vector $(x,y,z,w)$ and want to write it as a combination of basis vectors. As the equations showed: $(x,y,z,w) = (z-2w,-z+w,z,w) = (1,-1,1,0)z + (-2,1,0,1)w$, hence a basis would be $\{(1,-1,1,0),(-2,1,0,1)\}$

solving a basis with two free variables

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