1. ## Homogeneous Linear System

Hello, I believe I have most of a homogeneous linear system solved. The coefficient matrix is:

$\left(\begin{array}{cccc}1&0&0&0\\0&1&1&0\end{arra y}\right)$

That amounts the the following system:
1x +0y + 0w + 0z = 0
0z + 1y + 1w + 0z = 0

Which is:
x = 0
y + w = 0

I let the parameter t = w and got:

x = 0
y = -t
w = t
z = ?

I think z equals another parameter, perhaps s, but I am not certain.

2. Correct. w and z would be free variables in this case. So you can define it as anything you want.

You can also think of the matrix like this (where the last two variables, in this case w and z, are free variables):

$\left(\begin{array}{cccc}1&0&0&0\\0&1&1&0\\0&0&0&0 \\0&0&0&0\end{array}\right)$

3. Ok, that's what I figured, I just wanted to be sure. Is y not a free variable because it depends on w? Which in turn is free and can depend on some parameter t which can be any real number?

4. In this case yes, y is not a free variable.