In a system of equations AX = B. by Cramer's rule if det(A)=0, then the system will have no solution but on the other hand, in non-homogeneous system of equations suppose the rank of square matrix nxn is (n-1) and rank of the augmented matrix[A|B] is also having the same rank but then we say that system is consistent, having rank < no. of variables implies infinitely many solutions.

how is this correct? I mean if the rank of A is n-1 < n and rank(A) = rank(A|B), then it also implies that determinant of square matrix nxn is zero but then we say it is consistent.

pls correct me if my theory is wrong too.


Cramer rule doesn't says that when determinant is zero, it has no solution. When the determinant is zero, the system can either have no solution (inconsistent) or infinitely many solutions (consistent).

Cramer rule gives us a way to compute the solution when the solution is unique.

  • $\begingroup$ thanks , doing a question got mistaken somewhere $\endgroup$ – ashwani yadav Aug 10 '18 at 5:19

For the system of two equations:

$ax+by=c$ and $Ax+By=C$

compute $D=\begin{vmatrix}a&b\\A&B\end{vmatrix}$, $D_x=\begin{vmatrix}c&b\\C&B\end{vmatrix}$ and $D_y=\begin{vmatrix}a&c\\A&C\end{vmatrix}$. Now you have two cases:

  1. If $D\ne 0$ then the unique solution is $x=\frac{D_x}{D}$ and $y=\frac{D_y}{D}$.
  2. If $D=0$ then Cramer's rule fails to give the solution (if it exists).
  • $\begingroup$ Since when could you divide matrices? $\endgroup$ – Chase Ryan Taylor Aug 10 '18 at 6:12
  • $\begingroup$ Oh sorry! I used wrong brackets for determinats. $\endgroup$ – Mathlover Aug 10 '18 at 6:13
  • $\begingroup$ $\ddot\smile{}{}{}{}{}{}{}{}{}{}$ $\endgroup$ – Chase Ryan Taylor Aug 10 '18 at 6:14
  • $\begingroup$ Edited it @Chase Ryan Taylor. Thanks a lot! $\endgroup$ – Mathlover Aug 10 '18 at 6:15

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.