Question

For the following system.
kx + y + z = 1
x + ky + z = 1
x+ y + kz = 1
Determine for what values of k the system has:
a) No solutions
b) One solution
c) A lot of solutions

Answer 1

Sorry i translated the problem from the spanish. I hope its clear

Answer 2

Determinants!

Answer 3

El sistema tendrá solamente una solución cuando las tres líneas se intersecten en un solo punto. El sistema no tendrá solución si las líneas son paralelas. Y el sistema tendrá infinitas soluciones si las líneas son lo mismo. :3

Answer 4

I wanted and idead like that one! thanks across

Answer 5

By the way, your spanish is great!

Answer 6

The system will have only one solution when the three lines intersect at a single point. The system does not have a solution if the lines are parallel. And the system will have infinitely many solutions if the lines are the same.

Answer 7

Oh - ha, I realized it was asking for *what* values of k. ignore my previous answer.

Answer 8

thanks Moneybird

Answer 9

yes ktlown

Answer 10

I was going to say if k=1 then the system has an infinite number of solutions since all 3 equations are the same

Answer 11

I think for any other value of k there's 1 solution.

Answer 12

That's pretty clever ktklown

Answer 13

I got the determinant of the system
k^3-3k+2;

Answer 14

k^3-3k+2=0

Answer 15

so k = -2 and k = 1

Answer 16

For those values of k there are multiple solutions. For every other value of k, the determinant is non-zero and therefore invertible and therefore the system has a unique solution.

Answer 17

Eso habría escrito en español, pero no tengo el vocabulario de matemáticas para hacerlo

Answer 18

haha oh crap, you're right, my cowboy solution led you astray. k=-2 means no solutions.

Answer 19

Mmm so there is no way to get an incompatible system?

Answer 20

with this system, no.

Answer 21

James, chat, NOW.

Answer 22

uh oh james is in trouble

Answer 23

doesn't k=0 have no solutions?

Answer 24

I thougt k = 1, k = -2 meant multiple solutions

Answer 25

Ack, death from a thousand typos. k=1 means an infinite number of solutions

Answer 26

k=-2 means no solutions. at least according to mathematica.

Answer 27

but, what is the reason?

Answer 28

yes, sorry. k = 1, multiple solutions; you can see that by inspection as every equation is the same.

Answer 29

k= - 2, no solutions. You can see that by inspection because you can add up all three equations and have 0 on the left-hand side but 3 on the other: 0 = 3.

Answer 30

james: that was my original answer (for that reason), but since I wasn't using determinants I failed to find that other interesting value of k, -2

Answer 31

So once i get the values of k such that the determinant is zero, how can i conclude that there is no solutions, without substitution

Answer 32

with the value k = -2

Answer 33

for the get = 0 case, you need to look at the system and see which case it is

Answer 34

ok, Thank U James!

Answer 35

It's interesting what's going on here. Think about the matrix of coefficients, A, as a linear map R^3 -> R^3

In the case det(A) is not zero, that means the map is an isomorphism because it's invertible and has rank 3. So the 'target' column vector which is the inhomogeneous values of the equations--in this case (1 1 1)^t--is in the range and therefore there is one unique value in the domain which gives it.

Answer 36

In the case det(A) = 0, that means the null space of A is non trivial and therefore if the target vector is in the range of A, then any one of those solutions plus an arbitrary member of the null space is also a solution; hence an infinite number of solutions. This is what happens here with k = 1.

But it could also be that the target vector is not in the range of A; as happens here with
k = -2. In that case there's no solution.

Hence when get of the coefficient matrix is zero, you need to look at the system further to figure out which case you're in. There's no simple test I can think of right now that tells you which it is.

Answer 37

What are the requisites to have an isomorphism?

Answer 38

Oh, he leaved

Answer 39

For a square matrix, or for a linear map between two vector spaces of the same (finite) dimension; it is sufficient that det of the matrix is not zero. That implies the null space is trivial and that the dimension of the range is equal to the dimension of the domain, which will necessarily imply the range is the entire target vector space.

Answer 40

¿Dónde estás? Espero que no estás en España dado la hora

Answer 41

across, ya sé.

Answer 42

Ahh, thank you!

Answer 43

Yo estoy en México

Answer 44

Bien. ¿En qué parte?

Answer 45

En la Ciudad de México

Answer 46

Me gusta DF

Answer 47

Bueno. Nos hablamos. Me tengo que ir.

Answer 48

Thanks!

Answer 49

de nada