Set of linear equations with 3 unknown with Delta = DeltaX = DeltaY = 0, Prove DeltaZ = 0 For a system of linear equations on x, y and z, is it possible that $\Delta z \neq 0$, when $\Delta = \Delta x = \Delta y = 0 $ ? (where $\Delta$ is the coefficient determinant and $\Delta x, \Delta y, \Delta z$ are numerator determinants (in crammer's rule) of x, y and z correspondingly)
Is there a theorem/proof for the same, or at least an example?
 A: Yes, it is possible (kind of). Take for example
$$\matrix{
x&&&=1\\
x&&&=0\\
&y&&=0
}$$
Where is $z$ in this example? All coefficients of $z$ happen to be zero. Indeed, this is the only case in which this is possible, as I will show in a minute, but even though it feels like cheating you will easily check that here $\Delta=\Delta_x=\Delta_y=0$ and $\Delta_z\ne0$.
Now to see that this can only happen if all coefficients of $z$ are zero, I ask you to recognise the symmetry of the situation: You are really just looking at four column vectors and are asking under which circumstances it can happen that of the four possible matrices obtained by taking three of the columns, exactly one has non-zero determinant. (Note that changing the order of the columns of a matrix can only change the sign of its determinant, it can therefore not change whether it is zero.) We can therefore swap the right hand side and the $z$-column. The question then becomes: Under what circumstances can it happen that $\Delta\ne0$ but $\Delta_x=\Delta_y=\Delta_z=0$? You already know the answer to this one by Cramer's rule: This means that the system has a unique solution, and that solution is zero. But then the right hand side must have been zero! This solves our little problem.
