# Homogenous linear system

Tags:
1. Oct 28, 2014

### Purple94

1. The problem statement, all variables and given/known data
1. Determine the values of k such that the following homogeneous linear system has infinitely number of solutions.
My problem is: I cannot to find the value of k

2. Relevant equations
2xky + z = 0

-x + y – 3kz = 0

kx – 2y + 2z = 0

3. The attempt at a solution
After I did a reduced form, I get this :
[1 -1 3k]
[0 -k+2 1-6k]
[0 0 2-3k^2]

2. Oct 28, 2014

### Staff: Mentor

The first thing you need to do is understand how this matrix could represent a linear system that had an infinite number of solutions. Once you understand that, it's fairly easy to answer the other question, which is finding the values of k for this to happen.

3. Oct 28, 2014

### Purple94

But, according to my attempt to this solution, i get this equation is has no solution. How can to prove this equation is infinetely solution?

4. Oct 28, 2014

### Staff: Mentor

Why do you think this?
It might be helpful to write the system of equations that your reduced matrix represents. For example, the first equation would be x - y + 3kz = 0. What are the other two equations?

5. Oct 28, 2014

### RUber

If any one of your variables is not explicitely defined, then the system may have infinitely many solutions.
Similarly, if your matrix is not full rank, i.e. a 3x3 matrix representing 3 linearly independent vectors, you will have an underconstrained system with infinitely many solutions.

6. Oct 28, 2014

### Purple94

the other 2 equation is (-k+2)y + (1-6k)z=0, (2-3k^2)z=0

7. Oct 28, 2014

### Purple94

I try, i get the value of k is not equal to zero

8. Oct 28, 2014

### Purple94

I get this:
x-y+3kz=0
(-k+2)y +(1-6k)z=0
(2-3k^2)z=0

let z=u,
(2-3k^2)u=0
2u-3k^2=0
-3k^2u=-2u
k^2u=2/3u
k^2=2/3
k=2/3,-2/3

9. Oct 28, 2014

### RUber

You forgot the square roots on the values for k.
Do you see why there would be infinitely many solutions to the system?

10. Oct 28, 2014

### Staff: Mentor

Okay

That's not the point. How would you find z?

This is roughly the right idea, but some of your work isn't necessary and some is not even correct. For example, why do you replace z with u?

Also, if (2 - 3k2)z = 0, you can say right away that either z = 0 or 2 - 3k2 = 0.

In one of your steps, you "cancel" u, which is a very bad idea.

11. Oct 29, 2014

### Purple94

i ask my lecturer and she said this question is wrong

12. Oct 29, 2014

### HallsofIvy

What does that mean? A "question" is not "right" or wrong.

13. Nov 1, 2014

### ehild

The last line is wrong. It should be 0 k-2 2-3k^2