# Homogeneous system of linear equation:

1. Aug 30, 2005

### TonyC

I am having trouble finding the solution to the homogeneous system of linear equations:
2x-2y+z=0
-2x+y+z=0

2. Aug 30, 2005

### TonyC

I guess I should have also put:

How can I break this down?

3. Aug 30, 2005

### FredGarvin

Perhaps I have been out of the rigid math stuff for a while, but how are you going to solve a system with three unknowns and only two equations?

4. Aug 30, 2005

### TD

Choose one variable (e.g. z) as "t" and solve the system for x and y in function of t.
You'll get an infinite number of solutions, for every t (so z), you have a couple (x,y).

5. Aug 30, 2005

### TonyC

Hence my problem..
The answers to choose from are:
x=3/4t, y=-t,z=1/2t
x=-3/4t,y=-t,z=1/2t
z=-3/4t,y=t,z=1/2t
z=3/4t,y=t,z=1/2t

This is why I am stumped.

6. Aug 30, 2005

### TD

In your case, y was substituted for t. Then solve it as if t was a parameter for x and z.

7. Aug 30, 2005

### TonyC

? I am still confused.

8. Aug 30, 2005

### TD

$$\left\{ \begin{gathered} 2x - 2y + z = 0 \hfill \\ - 2x + y + z = 0 \hfill \\ \end{gathered} \right$$

Substitute y = t, t is now a parameter, and solve the following (2x2)-system for x and z

$$\left\{ \begin{gathered} 2x + z = 2t \hfill \\ - 2x + z = - t \hfill \\ \end{gathered} \right$$

9. Aug 30, 2005

### HallsofIvy

Staff Emeritus
You might also want to review your list of possible answers. "x" got changed to "z" in some of them!

(Am I the only person who hates multiple choice questions in mathematics?)

10. Aug 30, 2005

(No :yuck:)