I had to prove the following theorem:
Suppose that $A\mathbf x=\mathbf b$ is consistent for some given $\mathbf b$, and let $\mathbf p$ be a solution. Then the solution set of $A\mathbf x=\mathbf b$ is the set of all vectors of the form $\mathbf s=\mathbf p +\mathbf w_h$ where $\mathbf w_h$, is any solution of the homogeneous equation $A\mathbf x=\mathbf 0$.
where the boldface characters are vectors.
Can you please let me know if my proof is correct and if you're upto it is there a more shorter way. Thanks in advance.
There is a shorter way.
Let $s$ a solution of $Ax=b$. We have $As=b$ and $Ap=b$. So $A(s-p)=0$, which implies that $s-p$ is equal to some $w_h$, solution of $Ax=0$. Hence $s=p+w_h$.
On the other hand, if $w_h$ is a solution of $Ax=0$, then $s=p+w_h$ is a solution of $Ax=b$ because $As=A(p+w_h)=Ap+Aw_h=b+0=b$.
This shows that the set of solutions is exactly $\{p+w_h, w_h\text{ solution of } Ax=0\}.$