anonymous 3 years ago For which scalars a, b, c, d do the solutions to the equation ax+by+cz=d form a subspace of R(3).

1. anonymous

In $$ax + by + cz = d$$, if x and y are any number in R(1), then z is constrained to $z=\frac{d-ax-by}{c}$ The subspace S = { any (x, y, z)} is a plane which does not pass through (0,0,0) and is therefore not a subspace.

2. anonymous

Actually, I failed to address all the trivial cases. If a = b = c = 0, there is no solution. The empty set is not a subspace. If one or two of a,b, or c equals 0, then you can rotate your axes such that the one that is not 0 is the constant for z, in which case the theorem holds.

3. anonymous

so your saying that if atleast a, b, or c is 0 then the others can be non zero numbers and be a subspace of R(3)?

4. anonymous

No, I'm saying that if d = 0 then the solution set (x, y, z) is a subspace of R(3) UNLESS all three are equal to 0.

5. anonymous

ll three a, b, c****

6. anonymous

and that would be the same thing if it was a, b, c to the equation ax+by=c form a subspace of R(2)... then c=0 in this case.

7. anonymous

also, if d=0, then as long as atleast one of a, b, or c is non zero then it IS a subspace right?

8. helder_edwin

actually the set u r given is a subspace if and only if d=0.

9. anonymous

but why? and does it matter what a, b, and c are?

10. helder_edwin

no.

11. helder_edwin

u have $\large D=\{(x,y,z):ax+by+cz=d\}$

12. helder_edwin

if d=0 then u have $\large D=\{(x,y,z):ax+by+cz=0\}.$ So let $$u=(x_0,y_0,z_0)$$, $$v=(x_1,y_1,z_1)$$ both in $$D$$ and $$\alpha\in\mathbb{R}$$ so (i) $$u+v=(x_0+x_1,y_0+y_1,z_0+z_1)$$ so $\large a(x_0+x_1)+b(y_0+y_1)+c(z_0+z_1)=$ $\large =(ax_0+by_0+cz_0)+(ax_1+by_1+cz_1)=0+0=0$ (ii) $$\alpha v=(\alpha x_1,\alpha y_1,\alpha z_1)$$ so $\large a(\alpha x_1)+b(\alpha y_1)+c(\alpha z_1)=\alpha(ax_1+by_1+cz_1= \alpha0=0$ this proves that D is a subspace of R^3.

13. helder_edwin

For the converse. Let's suppose that $\large D=\{(x,y,z):ax+by+cz=d\}$ is a subspace of R^3. then the zero vector MUST be in D so $\large 0=a0+b0+c0=d$

14. anonymous

I do understand that for sure. but how did you know to make d=0, or how do you know it doesn't work for any non zero number?

15. anonymous

ohhh okay.

16. helder_edwin

therefore D is a subspace if and only if d=0

17. anonymous

so that same idea applies if the question is ax+by=c, for scalars a, b, c ... subspace of R(2) right

18. helder_edwin

yes

19. anonymous

It seems like the book is trying to trick me... why give one question stating for ax+by=c, and then the question right after that be ax+by+cz=d, when c has to be 0 for the first and d has to be 0 for the second. Its the same question isnt it, or is there more to it that I am missing.

20. helder_edwin

yes it is. the book is trying to stress that for a subset of a given to be a subspace (i.e., a space on its own) the zero vector (which is unique) must also belong to the subset in question.

21. helder_edwin

that's what u r supposed to get from this two exercises.

22. helder_edwin

*given space*

23. helder_edwin

i hope i was helpful gotta go

24. anonymous

it most definitely was... i hope to see you around again, you have been very helpful. thank you.