The solution set of Homogeneous System is the null space.
Non-Homogeneous System solution set is not a vector space, there for if $x_1,x_2$ are solutions: $x_1+x_2$, $\alpha x$, $\alpha_1x_1+\alpha_2x_2$ will not be solutions.
It can be proven be examples on a simple linear equation like $x+2y=6$.
Is there a better way?
[Math] Solution Set Of Non-Homogeneous System
linear algebra
Best Answer
If you are asking for a good way to prove that the solution set isn't a vector space, I think the best way is to notice that it doesn't contain the zero vector.
EDIT: On re-reading the question, I think you are saying that you know the solution set isn't a vector space, you want to prove that if $x_1$ and $x_2$ are solutions then $x_1+x_2$, $ax_1$, and $a_1x_1+a_2x_2$ aren't solutions. Perhaps the easiest way to handle this is to look at the non-homogeneous system as a matrix equation $Ax=b$ with $b$ not the zero vector. Then we are assuming $$Ax_1=b,\qquad Ax_2=b$$ Thus, we have $$A(x_1+x_2)=Ax_1+Ax_2=b+b\ne b$$ since we are assuming $b\ne0$, so this proves $x_1+x_2$ is not a solution. We have $$A(ax_1)=aAx_1=ab\ne b$$ again since $b\ne0$ (of course, we have to assume here $a\ne1$). Finally, $$A(a_1x_1+a_2x_2)=a_1Ax_1+a_2Ax_2=a_1b+a_2b=(a_1+a_2)b$$ So $a_1x_1+a_2x_2$ can be a solution; indeed, it will be a solution, if and only if $a_1+a_2=1$. There's a name for this: if $a_1+a_2=1$, then $a_1x_1+a_2x_2$ is called a convex linear combination of $x_1$ and $x_2$, and the solution set of a non-homogeneous system is closed under taking convex linear combinations.