Thread: Checking a solution to a homogenous system

1. Checking a solution to a homogenous system

I have the following system:

2x - y + 7z = 0
-x + 2y -3z = 0
x + y + 4z = 0

I have determined the solution to be:

x = -y -4t
y = -1/3t
z = t

what I need to know now, is if (0, 7, -1) is a solution to the system?

2. Originally Posted by blackhug
I have the following system:

2x - y + 7z = 0
-x + 2y -3z = 0
x + y + 4z = 0

I have determined the solution to be:

x = -y -4t
y = -1/3t
z = t

what I need to know now, is if (0, 7, -1) is a solution to the system?
x = 0 , y = 7 , and z = -1 makes the first equation of the system false ... so I'd say no.

3. lol yeah, I figured as much. I was just confused about it having infinite solutions.
Thought it might have been a little more complicated than just putting it into the equation.

4. Originally Posted by blackhug
I have the following system:

2x - y + 7z = 0
-x + 2y -3z = 0
x + y + 4z = 0

I have determined the solution to be:

x = -y -4t
y = -1/3t
z = t

what I need to know now, is if (0, 7, -1) is a solution to the system?
Why is the red y part of your solution for x?

5. Originally Posted by mr fantastic
Why is the red y part of your solution for x?
Well because I was told to solve this system by reducing it to echelon form.
So I was left with:

1 1 4
0 1 1/3
0 0 0

If i had my solution for x as $11/3t$ wouldn't that be solving by reduced echelon form?

6. Originally Posted by mr fantastic
Why is the red y part of your solution for x?
Your solution for x, y and z needs to be in terms of the parameter t only.