I am having some difficulty thinking about a certain concept in my linear notes.
I will post below the theorem that is involved,
Theorem: Let X$_0$ be a particular solution of the system AX = B (that is, assume AX$_0$= B).
(1) If X$_0$ is any solution of AX=0, then X'+X$_0$ is a solution of AX=B.
(2) Every solution of AX = B can be written as X = X' + X$_0$where AX' = 0.
It then gives an example,
Let A= $$\begin{pmatrix}1 & -2 & 2 & -1\\ 2 & -4 & 3 & 1\\ 3 & -6 & 5 & 0 \\ \end{pmatrix} $$
and B= $$\begin{pmatrix}1 \\ 3 \\ 4 \\ \end{pmatrix} $$
" Use Gaussian elimination to express every solution to the system AX=B as the sum of a particular solution and the general solution to the homogenous equation AX=0"
I will skip the steps of reducing but after doing row reductions we come to
$$\begin {pmatrix} 1 & -2 & 0 & 5 & 3 \\ 0 & 0 & 1 & -3 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix} $$
Up to here I am still following along, and now I understand we can write
X= $$\begin {pmatrix} 3 \\ 0 \\ -1 \\ 0 \ \end{pmatrix} $$ + s $$\begin {pmatrix} 2 \\ 1 \\ 0 \\ 0 \ \end{pmatrix} $$ + t $$ \begin {pmatrix} -5 \\ 0 \\ 3 \\ 1 \end{pmatrix} $$
Where s and t are the two parameters, x2=s and x4=t
The part I really dont understand and am looking for help with is what the book states after this. It then says The particular solution X0 is $$\begin {pmatrix} 3 \\ 0 \\ -1 \\ 0 \ \end{pmatrix} $$ and the general solution to the homogenous equation is
s$$\begin {pmatrix} 2 \\ 1 \\ 0 \\ 0 \ \end{pmatrix} $$ + t $$ \begin {pmatrix} -5 \\ 0 \\ 3 \\ 1 \end{pmatrix} $$
My main question is how from this theorem were we able to know that the solution without the parameters was a solution in itself, are the parameters not part of the solution? And especially how did we know that the solutions containing the s and t only and not the constants were solutions to the homogenous equation.
I hope what I am asking makes some sense, thanks all in advance for any help.
First, I think you may have misstated the first part of the theorem slightly. It should be: If $X_{0}$ is a solution to $AX= B$, and $X'$ a solution to $AX = 0$, then $X_{0} + X'$ is also a solution to $AX = B$, since $A(X_{0} + X') = AX_{0} + AX' = B + 0.$
Now we can write the general solution you computed as \begin{equation} X = X_{0} + sX_{1} + tX_{2}, \end{equation} where $X_{0} = \left[ {\begin{array}{c} 3 \\ 0 \\ -1 \\ 0\\ \end{array} } \right]$, $X_{1} = \left[ {\begin{array}{c} 2 \\ 1 \\ 0 \\ 0\\ \end{array} } \right]$, and $X_{2} = \left[ {\begin{array}{c} -5 \\ 0 \\ 3 \\ 1\\ \end{array} } \right]$. This solution satisfies $AX = B$ for all choices of $s$ and $t$. In particular then, if we set $s = t= 0$, then we have $AX_{0} = B$, meaning that $X_{0}$ is a particular solution (note there is not a unique particular solution).
So to answer your question, the parameters are part of the general solution; $X_{0}$ is not the solution without parameters, rather it is the solution at particular values of the parameters ($s= t= 0$).
Next, notice that $A(sX_{1} + tX_{2}) = AX - AX_{0} = 0$. So $sX_{1} + tX_{2}$ is a solution to the homogeneous equation for all values of $s$ and $t$.
I think this example was meant to demonstrate the various aspects of the theorem, but you do not need the theorem to deduce all the above facts.