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Show that the solution vectors of a consistent nonhomogeneous system of m linear equations in n unknowns do not form a subspace of $\mathbb{R}^n$.

I'm not really sure how to go about this problem. I know that I'm suppose to check if the vectors are with the subspace by addition and scalar multiplication but I'm not really sure how to set it up.

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There is one (and only one) vector that is in every subspace of $\mathbb{R}^n$. Is that vector in the solution set?

The statement is also true if the system is inconsistent, because the empty set is not a subspace, for the same reason.

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If we have a consistent nonhomogeneous system $Ax = y$ with a solutions $x_0$ and $x_1$, then $A(x_0 - x_1) = Ax_0 - Ax_1 = y - y = 0$. But if we assume that the solutions to the nonhomogeneous system form a subspace, then $x_0 - x_1$ must also be a solution to the nonhomogeneous system, thus $0 = A(x_0 - x_1) = y$. This is clearly false, so the solutions to the nonhomogeneous system do not form a subspace.

Or, even easier, $0$ must be an element of any subspace (since $x - x = 0$) but $A0 = 0 \ne y$.

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