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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?

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  • $\begingroup$ Those are not 3x3 matrices $\endgroup$
    – GTX OC
    Commented Nov 26, 2013 at 4:57
  • $\begingroup$ Yeah, really? How does it refer to the real problem here. $\endgroup$
    – Mike
    Commented Nov 26, 2013 at 4:59

1 Answer 1

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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.

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  • $\begingroup$ Oh. Please correct me if I'm still not following. Second equation can be solved for any values since we operate on unknowns on both sides which basically gives us all the possibilities whereas in last equation we have fixed value (0) and therefore certain conditions apply. $\endgroup$
    – Mike
    Commented Nov 26, 2013 at 5:03

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