# Set of linear equation question

1. Dec 9, 2008

### transgalactic

give an example for a set of linear equation (from field R) that has the following
general solution:

http://img244.imageshack.us/img244/5226/26420883un6.gif [Broken]

??

Last edited by a moderator: May 3, 2017
2. Dec 9, 2008

### Defennder

You need to write out equations for x,y,z such that alpha, beta cancels out. Eg. Suppose x = a + 1 and y = a, where a is the free variable, then a possible eg of an equation would be x - y = 1.

3. Dec 9, 2008

### transgalactic

i wrote a matrix
and inputed alpha=1 and beta=2
is that ok?

4. Dec 9, 2008

### Defennder

What matrix did you write? And I don't think you can arbitrarily set values of alpha, beta. Remember that these are free variables, so if you set some particular values for them you are in effect saying they are not parameters which may be varied.

5. Dec 9, 2008

### transgalactic

i need to give an example so the set that i was given
would be its general solution(i guess i need to present one case)
i have constructed this general matrix:

1 a b
2 a b
3 a 0

and inputed some a=1 b=2

but i am not sure if this the correct way of solving it??

6. Dec 9, 2008

### Staff: Mentor

transgalactic,
I find that it is frequently difficult for me to understand what you are trying to do, which makes it hard to help you.

From what you posted on imageshack, it appears to me that you are supposed to find a 3 x 3 matrix A so that <what you posted> is the general solution of Ax = b.

The vectors with alpha and beta multipliers, (1, 1, 1)^T and (1, 1, 0)^T appear to me to be all of the solutions of Ax = 0. That is, these vectors form a basis for the nullspace of A.

The other vector, (1, 2, 3)^T seems to be the particular solution of Ax = b, for some unknown vector b.

Possibly there is enough information here so that you can find the coefficients of matrix A.

Last edited by a moderator: May 3, 2017
7. Dec 10, 2008

### transgalactic

how to interpret this thing :

http://img244.imageshack.us/img244/5226/26420883un6.gif [Broken]

into equation?

Last edited by a moderator: May 3, 2017
8. Dec 10, 2008

### transgalactic

i got a way of solving this question:
(x,y,z)=(1+a+b) + (2+a+b) +(3+a)

i say that z=0 because (3+a) is not dependent on b.
i say that a+b=t

x=1+t y=2+t
x-y=-1
y-x=1

is that a prove??

9. Dec 10, 2008

### Defennder

No, that doesn't satisfy the required general solution. z is not zero here, since in the general solution z=3+a where a is arbitrary.