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Why is $Ax=B$ only solvable if and only if the last $m-r$ equations reduces to $ 0 = 0$? Why can't each row be independent of each other and still be solvable? r = rank, m = rows. Also how does this relate to $B$ being in $C(A)$. Why is B only in the column space if one row is a combination of the others? Isn't that what the initial solvability condition is stating? Picture added of where confusion is arising. How does does the last paragraph conditions relate? Why must there be a row that is a combination of the other rows for there to be a solution?

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    Keep in mind that even for singular matrices the system is still solvable if $b\in\mathcal{R}(A)$. – 2017-01-17
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    Changed what I'm trying to ask. – 2017-01-17
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    What are $m$ and $r$? – 2017-01-17
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    Is the matrix supposed to be square? If so, then then it would be vice versa: The system is only solvable for all $B$ if $m=r$. Note that $C(A)$ is the space spanned by the columns of $A$, i.e. the set $\{Ax:x\in\mathbb{R}^n\}$, so the system is clearly not solvable if $b\ne C(A)$. – 2017-01-17
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    Updated Again... – 2017-01-17

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Say $A\in\mathbb{R}^{m\times n}$. Then $\mathbb{R}^m=\mathcal{N}(A^T)\oplus\mathcal{C}(A)$. Thus, $Ax=b$ is solvable iff $b\in\mathcal{C}(A)$, i.e., iff $b\perp\mathcal{N}(A^T)$. Since the third row of $A$ is the sum of the first two, it is clear that $(1,1,-1)^T\in\mathcal{N}(A^T)$, and since $A$ has rank two, we have in fact that $\mathcal{N}(A^T)=\text{span}\{(1,1,-1)^T\}$. Hence $b\perp\mathcal{N}(A^T)$ iff $(1,1,-1)^T\cdot b=0$, i.e., iff $b_1+b_2-b_3=0$ as stated in the text you quoted.

EDIT: Note that the entries of $b$ might have to fulfill a different relation if the matrix $A$ had different entries.

EDIT 2: $\mathcal{N}(A^T)$ denotes the null space of $A^T$, and the fundamental theorem of linear algebra says that $\mathbb{R}^m=\mathcal{N}(A^T)\oplus\mathcal{C}(A)$, i.e., that every $x\in\mathbb{R}^m$ can uniquely be decomposed into $x=x_C+x_N$, where $x_C\in\mathcal{C}(A)$, $x_N\in\mathcal{N}(A^T)$, and $x_C\perp x_N$.