Homogeneous Equation (easier question)

[n0_3sc]

Can someone explain what the homogeneous equation is :redface: and how do you find the 'null vectors' and hence the general solution.
Eg.

AX =
[6]
[8]
[4]

A =
[1 2 4]
[3 1 2]
[0 2 4]

X =
[2]
[0]
[1]

Find the null vectors of A and general solution.

[matt grime]

Something is homogeneous if, in this case, it is equal to zero. I already explained that you're just doing subsitution to find general solutions, possibly having to parametrize them to describe the line or plane of the solution. Seeing as you won't be satisfied until you get a (partially) worked example:

Let A be as above, set Aw=0 and let w=(x,y,z)

then 2y+4z=0, so y=-2z
then 3x+y+2z=0=3x, thus x=0

So a necessary condition for w to be a null vector is w is a multiple of (0,-2,1)
using the first equation we see that it is sufficient let w=(0,2,-1) all null vectors are of the form tw for some t in R and every such is a null vector

Now suppose that Av=X is a particular solution (one that we find by inspection) and so is Av', then A(v-v')=0 so v'=v+tw for some t.

Hence a general solution is a particular solution plus some combination of the null vector(s).

So find one particular solution.

[n0_3sc]

Oh...Right so that makes sense, I think thats what my 1/2 page of notes from my lecturer was "trying" to say. Thanks :biggrin: