How does the rightmost column vector tell which row combinations have been taken? [Strang P193 3.6.19(b)] Following the method of Problem 18, reduce A to echelon form and look at zero rows.
The $\mathbf{b}$ column/resource vector tells which combinations you have taken of the rows.
From the $\mathbf{b}$ column after elimination, read off $m - r$ basis vectors in the left nullspace. Those $y$s are combinations of rows that give zero rows.
$\begin{bmatrix}
    1 & 2 & R1 \\
    2 & 3 & R2 \\
    2 & 4 & R3 \\
    2 & 5 & R4 \\
    \end{bmatrix} \rightarrow REF \rightarrow  \begin{bmatrix}
    1 & 2 & R1 \\
    \color{#009900}0 & \color{#009900}0 & \color{#009900}{-4} + \color{#009900}1R2 + \color{#009900}{0}R3 + \color{#009900}1R4 \\
    \color{#FF4F00}0 & \color{#FF4F00}0 & \color{#FF4F00}{-2}R1 + \color{#FF4F00}{0}R2 + \color{#FF4F00}{1}R3 + \color{#FF4F00}{0}R4\\
    0 & 0 & -2R1 + R4 \\
    \end{bmatrix}$
$ \implies \color{#009900}{\begin{bmatrix} 4 & 1 & 0 & 1 \\
    \end{bmatrix}}\begin{bmatrix}
    R1 & R2 & R3 & R4 \\
    \end{bmatrix}^T = \color{#009900}{\begin{bmatrix}
    0 & 0 \\
    \end{bmatrix}} $ and $  \color{#FF4F00}{\begin{bmatrix} -2 & 0 & 1 & 0 \\
    \end{bmatrix}}
\begin{bmatrix}
    R1 & R2 & R3 & R4 \\
    \end{bmatrix}^T = \color{#FF4F00}{\begin{bmatrix}
    0 & 0 \\
    \end{bmatrix}} $.
I perceive the above and thus see : $\color{#009900}{\begin{bmatrix} 4 & 1 & 0 & 1 \\
    \end{bmatrix}},\color{#FF4F00}{\begin{bmatrix} -2 & 0 & 1 & 0 \\
    \end{bmatrix}}$ are in $A$'s left nullspace.
I recollect that during REF, row operations are performed which means taking linear combinations of the rows. Nonetheless, I still don't apprehend how and why the resource vector divulges which row combinations you've taken? Would someone please explain? 
Please omit the following which succeed this question: Orthogonality, Determinants, eigenvalues and eigenvectors, and linear maps.
 A: Consider the (4x2) matrix, 
$A = \begin{bmatrix}
   1 & 2 \\
   2 & 3 \\
   2 & 4 \\
   2 & 5 \\
   \end{bmatrix}$
Augment $A$ with the "resource" vector, $b = \begin{pmatrix} R1, R2, R3, R4 \end{pmatrix}^T$ and then do elementary row operations on the augmented matrix to compute the row echelon form. I'll show the individual steps...so that it is clear that the last column contains a handy short hand notation of the row operations performed. 
$\begin{bmatrix}
    1 & 2 & R1 \\
    2 & 3 & R2 \\
    2 & 4 & R3 \\
    2 & 5 & R4 \\
    \end{bmatrix} \rightarrow \begin{bmatrix}
    1 & 2 & R1 \\
    0 & -1 & R2 -2R1 \\
    0 & 0 & R3 -2R1 \\
    0 & 1 & R4 -2R1 \\
    \end{bmatrix} \rightarrow \begin{bmatrix}
    1 & 2 & R1 \\
    0 & -1 & R2 -2R1 \\
    0 & 0 & R3 -2R1 \\
    0 & 0 & R4 -4R1 + R2 \\
    \end{bmatrix}$
Now, the left null space of A are those vectors y such that 
\begin{equation}y^TA = 0\end{equation}
So, the left null space is the space spanned by the vectors 
$y_1 = \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0\end{bmatrix}$ , 
$y_2 = \begin{bmatrix} -4 \\ 1 \\ 0\\ 1\end{bmatrix}$
The components of these vectors are easily determined by inspection of the zero rows of the row echelon form above.
Why is it that the row operations that produce zero rows in the row echelon matrix reveal exactly the vectors in the left null space?
Recognize that elementary row operations are simply linear combinations of the rows and recall that vector-matrix multiplication can be viewed as specifying a linear combination of the rows. 
Let's form a matrix $E$ whose rows represent the elementary row operations used to find the REF of $A$. Having already found the REF of a conveniently  augmented $A$, we can form $E$ by inspection. 
$$E = \begin{bmatrix}
   1 & 0 & 0 & 0 \\
  -2 & 1 & 0 & 0 \\
  -2 & 0 & 1 & 0 \\
  -4 & 1 & 0 & 1 \\
\end{bmatrix}$$ 
I claim that $EA = REF(A)$. Consider the last row of $E$ and the last row of the resultant, $EA$. 
$$\begin{bmatrix}-4 & 1 & 0 & 1\end{bmatrix} \begin{bmatrix}
   1 & 2 \\
   2 & 3 \\
   2 & 4 \\
   2 & 5 \\
   \end{bmatrix} = \begin{bmatrix}0 & 0\end{bmatrix}
$$
This is precisely $y_2^TA = 0$ and so, we see that which was to be shown. The vector representation of the elementary row operation which results in aa zero row is $\in$ the left null space of $A$. 
