Can Non-Homogeneous Linear Systems Have Multiple Solutions?

[strugglinginmat]

(123)x=a
(456)y=b
(789)z=c
For which values (a,b,c) does the above equation have a solution where a,b,c,x,y,z belong to R?
Initially I found the determinant of the above matrix and it was O. From this I know that there will be nontrivial solution solutions for (a,b,c) = (0,0,0).
But I am confused as to the possibility of having any other solutions for nonhomogenous system. I know that it will not have a uniques solution. Can anybody tell me if there is going to be any other solutions for values of (a,b,c) other than the homogenous one.

 

[fopc]

Any multiple of (1,-2,1) is sent to (0,0,0) by your matrix.
In other words, <{(1,-2,1)}> represents its one-dimensional nullspace.

The rhs has to lie in the two-dimensional columnspace for there to be any solutions.
The constraint on a,b,c for existence of solutions is a-2b+c=0.

For any rhs that satisfies the constraint, there exits a one-dimensional linear manifold that
constitutes the solution set.

Simple example:

Let (a,b,c) = (-1,-1,-1). Then <{(1,-2,1)}> + (1,-1,0) is the solution set.

 

[tacman]

Yes, there can be other solutions for values of (a,b,c) other than the homogeneous one. This is because the system of equations is not fully determined and can have an infinite number of solutions. For example, if we let x=1, y=2, and z=3, then the equations become 123=a, 912=b, and 2367=c. These values of (a,b,c) satisfy the equations and are not the homogeneous solution of (0,0,0). In general, as long as the determinant of the matrix is not equal to 0, there will be other solutions for (a,b,c) in addition to the homogeneous solution. However, it is important to note that the solutions may not be unique and there may be an infinite number of solutions.