ACOBI ITERATION How do Jacobi iterations work in linear algebra?

[EvLer]

I am a bit confused with this problem:

Given AX = B, B != 0; X and Y satisfy the system. Find constants a and b such that aX + bY also satisfy the system.

The hint was: does (1/3)X+(2/3)Y work? Which would mean (1/3 + 2/3)(X + Y), which means X + Y. So, then I have X+Y as a potential solution, which is a solution? So, then does it mean that if a system has solutions, sum of those solutions is still a valid solution, but is any linear combination a solution? Then it would make a,b either such that a+b=1 or a,b any number.
Could someone explain, please?

Thank you very much.

 

[Nylex]

The hint was: does (1/3)X+(2/3)Y work? Which would mean (1/3 + 2/3)(X + Y), which means X + Y.

No, you mean (1/3)(X + 2Y), (1/3 + 2/3)(X + Y) isn't the same as X/3 + 2Y/3.

 

[diegojco]

Is there some extradata in the problem, like a relation between X and Y. Because if you don't have one you can prove that these constants exist

 

[EvLer]

No, you mean (1/3)(X + 2Y), (1/3 + 2/3)(X + Y) isn't the same as X/3 + 2Y/3.

Ooops, sorry you are right...basic algebra...
I am still not clear if a system has several solutions, when combination of those solutions is a valid solution...if ever.

No, there is no extra info.
But maybe if one of (a,b) is zero then it's possible?

 

[diegojco]

If X is solution and Y is solution, then X=cY where c is a constant. then

c*AY=B
AY=B

or: ((c+1)/2)*AY=B

By linearity:

A(X/2)+A(Y/2)=B

then a=1/2 b=1/2

 

[EvLer]

If X is solution and Y is solution, then X=cY where c is a constant.

That's always true? Is solution set a vector space?

or: ((c+1)/2)*AY=B

Where does that follow from?

Thanks.

 

[diegojco]

this condition is true if the X and the Y are lineary dependent, then there is unique solution, the X or the Y. and that's the problem. for that system, exist unique solution or infinite.

The expression comes from summing the eqs:

cAY+AY=2B
((c+1)/2)*AY=B

 

[diegojco]

note the expression comes from summing the eqs i initially post. The eqs. i post latter is the reductions

 

[Justin Lazear]

Z = aX + bY

AZ = B
A(aX + bY) = B
AaX + AbY = B
a(Ax) + b(AY) = B
aB + bB = B
a + b = 1

--J