Why, while checking consistency in $3\times3$ matrix with unknowns, I check only last row? I would like to know whether my thinking is right. So, having 3 linear equations,
$$
\begin{align}
x_1 + x_2 + 2x_3 & = b_1 \\
x_1 + x_3 & = b_2 \\
2x_1 + x_2 + 3x_3 & = b_3
\end{align}
$$
I build $3\times 3$ matrix
\begin{bmatrix}1&1&2&b_1\\1&0&1&b_{2}\\2&1&3&b_3\end{bmatrix}
That reduces to
\begin{bmatrix}1&1&2&b_1\\0&1&1&b_1-b_2\\0&0&0&b_3-b_2-b_1 \end{bmatrix}
Now the thing is that if I prove that $b_3 = b_2 + b_1$ (which is obvious) then the system is consistent.
My doubt was why I don't need to prove the same about second row? Is it because there is no way that system of two equations (first and second row) that has contains unknowns ($x_1$ and $x_2$) be inconsistent?

The second row corresponds to the equation $1x_2+1x_3=b_1-b_2$.  This can be solved for any values of $b_1, b_2$.  For example, we can take $x_2=b_1-b_2$, and $x_3=0$.
The third row corresponds to the equation $0x_1+0x_2+0x_3=b_3-b_2-b_1$.  This cannot be solved  if $b_3-b_2-b_1\neq 0$.  No matter what $x_1,x_2,x_3$ are, you can't make 0 equal to 1.
However if $b_3-b_2-b_1=0$ then any values of $x_1,x_2,x_3$ will work.
