Question

From the 8th lecture.
The solution of Ax = b is
x = x particular + x nullspace
When solving for x particular, is it always necessary to set the free variables to 0? or can we use some other random numbers?
If yes, wouldn't the solutions be different depending on what you set the free variables to?
If no, why not?

Answer 1

I had the same question.
My answer is NO.

I don't have general proof, but an example was enough to convince me.
(I wonder how one can prove a statement like this)

What I did was the following:
A={ (x1,x2,x3,x4) | x1+x2+2x3+3x4 = 0 and 4x3+4x4=0 x1,x2,x3,x4 in R}
B={ t(-1,1,0,0) + s(1,0,-1,1) | t,s in R}

They are the null space for a specific matrix(in the text book's example)
I can prove A = B.

Similarly I think of A', B' when b != 0.
And I did the same thing to show A'=B'
(You set B'={Xparticular + t(-1,1,0,0) +s(1,0,-1,1) | t,s in R})

You will see that you can set Xparticular to anything Xparticular + any of Xnull and
A'=B' still checks out.

Answer 2

I meant NO the solution sets don't differ no matter what you choose what you choose for Xparticular.
But yes, you can set random number to get Xparticular provided it produces b.

Answer 3

You can also think of this way.
You have one Xparticular such that AXparticular = b
Now, we assume there's another Xparticular2.

AX1=b Ax2=b
=> A(X1-X2)=0

X1-X2 is in null space.
X1-X2 = Xnul
X2 = Xnull + X1
In other words, any solution to Ax=b will be of the form Xnull + X1.

Hence whatever random value you put to get your Xparticular, it will be Xnull + Xparticular(which you get when you set free variable to zeroes).

Xparticular + Xnull is a solution set to Ax = B
and there's no solution which is not covered by Xparticular+Xnull.