Originally Posted by

**Ackbeet** (i) I'm a little confused by the matrix you've exhibited there. What size is the original problem? A nonhomogeneous 4 x 4? Or a homogeneous 4 x 5?

In general, the columns that don't have leading 1's in the rref (the second and fourth, in your case) correspond to the variables that will be the arbitrary parameters of your solution.

[EDIT]: See HallsofIvy's post for a correction here. I would say that the way I do back substitution, my statement stands.

(ii) You need also to show that the set is nonempty. Easiest way is usually to show that the zero vector is in your set (which is necessary, anyway, to have closure under scalar multiplication and vector addition). The zero vector is in the set. Why?

Take scalar multiplication. Start with the vector $\displaystyle \langle x,y,z\rangle$ in your set. That implies $\displaystyle ax+by+cz=0.$ Multiply the vector by an arbitrary scalar $\displaystyle r.$ Is the result still in the set?

Take vector addition. Start with two vectors $\displaystyle \langle x_{1},y_{1},z_{1}\rangle$ and $\displaystyle \langle x_{2},y_{2},z_{2}\rangle$ that are both in the set. That implies two conditions. Add the two vectors together and check if the new vector still satisfies the necessary condition.

Using the plane condition probably amounts to recognizing that the equation for the set is also a dot product:

$\displaystyle \mathbf{n}\cdot\mathbf{x}=0,$ where

$\displaystyle \mathbf{n}=\langle a,b,c\rangle$ and

$\displaystyle \mathbf{x}=\langle x,y,z\rangle.$

This is also the equation for a plane.

You can use the properties of the dot product to prove the necessary result. Do you see how?