How do you find conditions on constants a,b,c so that a system is inconsistent?

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So I have the these two equations that make up a system:

$x + ay = 1$

$bx+cy =2 $

And I'm supposed to find the constants a,b, and c so that the system is inconsistent. What criteria am I suppose to look for here?

For exactly one solution, I would try to make it so that a row has an nonzero element that matches up with a real number

For infinitely many solutions, I would try to make it so that there's zero across the row.

For no solutions, I would have a row with all zeros that matched with a real number.

But these aren't what I'm trying to find and I am lost on where to go with this problem.

Now I know that for a system to be inconsistent, rk(A) =/= rk (A|b) would I have to use this fact in some way?

EDIT: Attempt with Gaussian Elimination

$$ \left[ \begin{array}{cc|c} 1&a&1\\ b&c&2\\ \end{array} \right]$$

R2 = (1/b)R2

R2 = R2-1R1

$$ \left[ \begin{array}{cc|c} 1&a&1\\ 0&c/b -a&2/b -1\\ \end{array} \right]$$

At a loss at what to do with 3 variables here.

EDIT: Using Dr. Sonnhard Graubner's hint:

$ x = 1 - ay $

$ y(c-ab) = 2 -b $

=> $y = \frac{2-b}{c-ab}$

=> $x = 1 - a(\frac{2-b}{c-ab})$

then the solution set is $ \left[ \begin{array}{c} x\\ y\\ \end{array} \right]$ = $ \left[ \begin{array}{c} 1 - a(\frac{2-b}{c-ab})\\ \frac{2-b}{c-ab}\\ \end{array} \right]$

So I see that c-ab cannot be equal to zero.

Therefore: If c-ab = 0 then the system is inconsistent.

Is this a correct conclusion?

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Gaussian elimination can be done more simply: \begin{align} \left[\begin{array}{cc|c} 1 & a & 1 \\ b & c & 2 \end{array}\right] &\to \left[\begin{array}{cc|c} 1 & a & 1 \\ 0 & c-ab & 2-b \end{array}\right] && R_2\gets R_2-bR_1 \end{align} Now the necessary and sufficient condition for the system to be inconsistent is that the third column is dominant (a pivot column). This translates into $$ c-ab=0 \qquad 2-b\ne0 $$ This is just the same as saying that $\operatorname{rk}(A)\ne\operatorname{rk}(A|b)$.

Note that just $c-ab=0$ does not guarantee inconsistency, because $b=2$ would give a system with infinitely many solutions.

If you're interested in particular values, take $b=1$, $a=1$ and $c=1$.

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with $$x=1-ay$$ we get in the second equation $$y(c-ab)=2-b$$ can you finish?