A linear system of equations $A\mathbf{x} = \mathbf{b}$ has two different solutions $\mathbf{x} = \mathbf{u}$ and $\mathbf{x} = \mathbf{v}$

Question

A linear system of equations $A\mathbf{x} = \mathbf{b}$ has two different solutions $\mathbf{x} = \mathbf{u}$ and $\mathbf{x} = \mathbf{v}$.
1. Show that $\frac{1}{3}\mathbf{u}+\frac{2}{3}\mathbf{v}$ is also a linear solution of the linear system.
So we do $\frac{1}{3}\mathbf{u}+\frac{2}{3}\mathbf{v} = \frac{1}{3}\mathbf{x}+\frac{2}{3}\mathbf{x} = \mathbf{x}$. What next? Is it $A(\mathbf{x}-\mathbf{x}) = \mathbf{0}$? Thanks!

Answer 1

No. We do$$A\left(\frac13\mathbf u+\frac23\mathbf v\right)=\frac13(A\mathbf u)+\frac23(A\mathbf v)=\frac13\mathbf b+\frac23\mathbf b=\mathbf b.$$