What Values of k Give Infinite Solutions in This Homogeneous System?

[Purple94]

Homework Statement

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

Homework Equations

2x – ky + z = 0

-x + y – 3kz = 0

kx – 2y + 2z = 0

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]

 

[Mark44]

Homework Statement

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

Homework Equations

2x – ky + z = 0

-x + y – 3kz = 0

kx – 2y + 2z = 0

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]

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.

 

[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?

 

[Mark44]

But, according to my attempt to this solution, i get this equation is has no solution.

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?

How can to prove this equation is infinetely solution?

 

[RUber]

If anyone 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.

 

[Purple94]

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

 

[Purple94]

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

 

[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

 

[RUber]

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

 

[Mark44]

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

Okay

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

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

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

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.

 

[Purple94]

i ask my lecturer and she said this question is wrong

 

[HallsofIvy]

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

 

[ehild]

Homework Statement

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

Homework Equations

2x – ky + z = 0

-x + y – 3kz = 0

kx – 2y + 2z = 0

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]

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