0

In this article (near the end), they say that we can not apply Cramer's rule to find our unknowns if the determinant is 0.

In this stack, I saw an answer which tries to explain using Cramer's rule but I don't think it's very clear. I am looking for a geometric intuition of how to test the nature of the solution using Cramer's rule.

Clemens Bartholdy
  • 14,222

2 Answers2

1

Imho, most easy consideration of possible cases for system linear equations is trough Kronecker–Capelli theorem (also known by several other names - see link), which holds even when quantity of variables differs from quantity of equations:

system have solution, i.e. is consistent, iif rank of coefficient matrix equal to rank of augmented matrix.

$$\exists x,\ Ax=b \Leftrightarrow \text{rank}[A]=\text{rank}[A|b]$$

So, if we have quadratic case system with $n$ linear equations and $n$ variables, then in case of non zero determinant, when $n$ equals to rank of coefficient matrix i.e rank of coefficient matrix equal to rank of augmented matrix and we have unique solution, otherwise infinitely many solutions. System have no solution when determinant is zero and rank of coefficient matrix differs from rank of augmented matrix.

Geometric interpretation here

zkutch
  • 14,313
0

When the determinant vanishes it means that at least one of the given equations is duplicated. (row or column is/are same). There is linear dependence among the system of equations.

The system so formed is underdetermined, there are no solutions.

A graph for plot of the variables in the plane or 3-space reveals that at least there is one pair of parallel lines.

Jacobian vanishes in case of functional dependence.

Narasimham
  • 42,260
  • I meant something to figure out if system has infinite or no solutions – Clemens Bartholdy Aug 08 '20 at 18:23
  • What about system $x+y=1,\ 2x+2y=2$? it has infinite number of solutions yet determinant vanishes. So your second paragraph isn't quite right. I think OP wants a criterion, which for vanishing determinant, tells whether no or infinitely many solutions., – coffeemath Aug 08 '20 at 19:32
  • They are not linearly independent. The second equation is the same as the first, got by doubling both RHS and LHS. – Narasimham Aug 08 '20 at 19:52
  • Yes, agreed they are not independent. But in your second paragraph you say "no solutions" rather than the correct infinitely many. – coffeemath Aug 08 '20 at 22:19
  • A set of linear equations should have the same number of solutions with non-zero determinant. If there is an infinite set of wrong solutions that also means there is no one right solution. – Narasimham Aug 08 '20 at 23:47