# Can Cramer's Rule really distinguish between infinite no. of solutions and no solution?

This is a question which was asked in a high-school exam held in India(JEE ADVANCED). Going by Cramer's rule, for infinite solutions, I should get $$D=D_1=D_2=D_3=0$$ (where $$D$$ is the original determinant and $$D_1, D_2, D_3$$ are the respective coefficient determinants). Using these, I arrive at $$\alpha^2$$=1, so that $$\alpha$$=1,-1. But, $$\alpha=1$$ yields no solution here (if I write down the system of equations using $$\alpha=1$$). Why does it happen so? Is this a rare failure of Cramer's rule? How should we explain this unexpected result?

• It is not really a failure of Cramer's rule because this usually assumes $D\ne 0$ giving a unique solution. Commented Jun 19, 2020 at 18:26
• The determinant is $(\alpha^2-1)^2$ so you must have $\alpha = \pm 1$. If you choose $\alpha = 1$ then the right hand side would have to have all the same values. So just by elimination we have $\alpha = -1$ and so $1+ \alpha+\alpha^2 = 1$. Commented Jun 19, 2020 at 18:29
• Cramer’s rule can’t distinguish between incompatible and underdetermined systems of more than two equations; see here. Commented Jun 19, 2020 at 18:29
• A valid way to distinguish the three cases $1)$ no solution $2)$ a unique solution $3)$ infinite many solutions is to determine the rank of $A$ and the rank of the matrix if we concatenate $A$ with $b$ (if $Ax=b$ is given). Then, we have case $1)$ if the ranks do not coincide , case $2)$ if the ranks coincide and are equal to the number of columns and $3)$ if the ranks coincide and are smaller than the number of columns. Commented Jun 19, 2020 at 18:35

For Cramer's rule for 3 equations, if $$D\ne 0$$, then you have a single solution. IF $$D=0$$ and any one of $$D_1, D_2, D_3$$ is non-zero, then the system is inconsistent. If all of those are zero, you still have two possibilities, one where the system is indeterminate and one where the system is inconsistent.

Cramer's rule applies to the case where the coefficient determinant is nonzero. In the 2×2 case, if the coefficient determinant is zero, then the system is inconsistent (no solutions) if the numerator determinants are nonzero. And, indeterminate (infinitely many solutions) if the numerator determinants are zero.

For 3×3 or higher systems, the only thing one can say when the coefficient determinant equals zero is that if any of the numerator determinants are nonzero, then the system must be inconsistent (no solution). However, having all determinants zero does not imply that the system is indeterminate (infinitely many solutions).

An example of this would be your own case:

x + y + z = 1,

x + y + z = -1,

x + y + z = 1

This has the coefficient determinant as zero, as well as the numerator determinants are zero, but this is inconsistent.