Finding the straight line from an equation system. I have this problem:
the matrix below is an equation system:
$$
        \begin{matrix}
        1 & 1 & -1 & 19 \\
        5 & 4 & -6 & 43 \\
        7 & -1 & 7 & 80 \\
        \end{matrix}
$$
It simplifies to 
$$
        \begin{matrix}
        1 & 0 & 0 & 0 \\
        0 & 1 & 0 & 35.5 \\
        0 & 0 & 1 & 16.5 \\
        \end{matrix}
$$
My book claims that the solution set of this problem is a straight line that goes through the point $(11, -3, 0)$ and has the direction vector $(-2, 7, 3)$
How can you get this point and vector from the matrix above?
 A: The book is wrong, and as one of the commenters said. The solution set is the single point $\begin{bmatrix} 0 \\ 35.5 \\ 16.5 \end{bmatrix}$.
The general rule when row reducing $[A |\mathbf{b}]\rightarrow [RREF(A)|\mathbf{P}]$ where $\mathbf{P}\in\mathbb{R}^n$.


*

*$0$ free variables implies the solution set is only the point $\mathbf{P}$. (Like your example)

*$1$ free variable implies the solution set is a line passing through the coordinates of $\mathbf{P}$ corresponding to fixed variables

*$2$ free varaibles implies the solution set is a 2-dimensional plane passing through coordinates of $\mathbf{P}$ corresponding to fixed variables

*etc. 


(It is kind of a weird explanation, because the solution set doesn't necessarily pass through $\mathbf{P}$ but the point constructed from each fixed coordinate(s) of $\mathbf{P}$ assigned to its corresponding variables. I will show by example to make it more clear.)
Example: Let $A\mathbf{x}=\mathbf{b}=\begin{bmatrix} -3 && 0 && -1 \\ 2 && 0 && 2 \\ 2 && 0 && 0\end{bmatrix}\mathbf{x}=\begin{bmatrix} 1 \\ -6 \\ 2 \end{bmatrix}$
and the row equivalent version $RREF(A)\mathbf{x}=\mathbf{P}=\begin{bmatrix} 1 && 0 && 0 \\ 0 && 0 && 1 \\ 0 && 0 && 0\end{bmatrix}\mathbf{x}=\begin{bmatrix} 1 \\ -4 \\ 0 \end{bmatrix}$
The general solution is $\begin{bmatrix} 1 \\ t \\ -4 \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \\ -4 \end{bmatrix}+\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}t$   ($t\in\mathbb{R}$)
which is clearly a line spanned by $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ that passes through $\begin{bmatrix} 1 \\ 0 \\ -4 \end{bmatrix}$    (at $t=0$)
Also, $\begin{bmatrix} 1 \\ 0 \\ -4 \end{bmatrix}$ is a particular solution to  $A\mathbf{x}=\mathbf{b}$ while $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}t$ is the general solution to $A\mathbf{x}=\mathbf{0}$.
Example: Let $A\mathbf{x}=\mathbf{b}=\begin{bmatrix} 0 && 0 && 1 \\ 0 && 0 && -2 \\ 0 && 0 && 6\end{bmatrix}\mathbf{x}=\begin{bmatrix} 3 \\ -6 \\ 18 \end{bmatrix}$
and the row equivalent version $RREF(A)\mathbf{x}=\mathbf{P}=\begin{bmatrix} 0 && 0 && 1 \\ 0 && 0 && 0 \\ 0 && 0 && 0\end{bmatrix}\mathbf{x}=\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}$
The general solution is $\begin{bmatrix} s \\ t \\ 3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}+\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}s+\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}t$    ($s,t\in\mathbb{R}$)
which is clearly the plane spanned by $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ which passes through $\begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}$ (at $s,t=0$)
Again, $\begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}$ is a particular solution to  $A\mathbf{x}=\mathbf{b}$ while $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}s+\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}t$ is the general solution to $A\mathbf{x}=\mathbf{0}$.
