In this publication, we will consider the formulation and formula of the Cramer method, as well as analyze an example of a practical problem to consolidate the theoretical material.
Kramer’s theorem
can be solved in several ways, and one of them is Kramer’s method (or theorem), which is so named after the Swiss mathematician Gabriel Cramer.
Statement of the theorem:
If the matrix corresponding to the square SLAE is non-zero, then the system is compatible and has one solution, which can be found as follows:
- Δ is the system matrix determinant;
- Δi is the determinant in which the column position i the column of the right parts is located.
Note: if the determinant of the matrix corresponding to the system is equal to zero, then it can be both consistent and inconsistent.
Example of a problem
Let’s use Cramer’s method to solve the system of linear equations below:
Solution
1. To begin with, we represent the given SLAE in the form of an expanded matrix A.
2. The determinant of the matrix (without taking into account the column of free members) is not equal to zero, which means that the system has a solution, and the only one.
3. We consider the determinant, replacing the first column with the third (i.e. for the root x).
4. Now we calculate the determinant in a similar way, substituting the third column instead of the second one (for y).
5. It remains only to use the formula above to find x и y.
Answer: x = 2, y = 5.