# 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?

• Those are not 3x3 matrices – GTX OC Nov 26 '13 at 4:57
• Yeah, really? How does it refer to the real problem here. – Mike Nov 26 '13 at 4:59

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.