Set of linear equation question

transgalactic
Messages
1,386
Reaction score
0
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

??
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
i wrote a matrix
and inputed alpha=1 and beta=2
and this is my answer
is that ok?
 
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.
 
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??
 
transgalactic said:
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

??
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:
how to interpret this thing :

http://img244.imageshack.us/img244/5226/26420883un6.gif

into equation?
 
Last edited by a moderator:
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??
 
transgalactic said:
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??
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.
 
Back
Top