Can Non-Homogeneous Linear Systems Have Multiple Solutions?

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The discussion centers on the conditions under which the non-homogeneous linear system represented by the equations (123)x=a, (456)y=b, and (789)z=c has solutions. It is established that the determinant of the coefficient matrix is zero, indicating that the system does not have a unique solution. The key finding is that for the system to have solutions, the right-hand side values (a, b, c) must satisfy the constraint a - 2b + c = 0, allowing for a one-dimensional linear manifold of solutions. An example is provided where (a, b, c) = (-1, -1, -1) yields a specific solution set.

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strugglinginmat
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(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.
 
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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.
 
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