91Ó°ÊÓ

A system of equations involves two or more equations with the same set of variables. Each equation represents a relationship between these variables. In this case, we are working with a system of linear equations:

$$2x + 5y = 7$$
$$3x - 2y = 1.$$

Our goal is to find the values of the variables, x and y, that satisfy both equations simultaneously. Systems of equations can be solved using various methods such as substitution, elimination, or matrix methods like Cramer's rule. Each method follows logical steps to find the common solution.

A matrix determinant is a special number calculated from a square matrix. The determinant helps in solving systems of linear equations and understanding the matrix properties. For a 2x2 matrix: $$A = \begin{bmatrix} a & b \ c & d \end{bmatrix},$$
its determinant (denoted as det(A) or |A|) is computed as:

$$|A| = ad - bc.$$

In the context of Cramer's rule, we use the determinant of the coefficient matrix to find the solutions for the variables. For the given system: $$\begin{bmatrix} 2 & 5 \ 3 & -2 \end{bmatrix},$$
the determinant is calculated by: $$|A| = (2)(-2) - (5)(3) = -4 - 15 = -19.$$

This determinant value is crucial for applying Cramer's rule.

The coefficient matrix is a matrix that consists solely of the coefficients of the variables in a system of linear equations. For the system:

$$2x + 5y = 7$$
$$3x - 2y = 1,$$

the coefficient matrix A is: $$A = \begin{bmatrix} 2 & 5 \ 3 & -2 \end{bmatrix}.$$

Replacing columns in the coefficient matrix with the constants from the equations helps us find solutions for x and y respectively. We first replace the first column with the constant terms and calculate the new determinant (denoted as $$D_x$$):
$$A_x = \begin{bmatrix} 7 & 5 \ 1 & -2 \end{bmatrix},$$

then we get: $$|A_x| = (7)(-2) - (5)(1) = -14 - 5 = -19.$$

For y, we replace the second column: $$A_y = \begin{bmatrix} 2 & 7 \ 3 & 1 \end{bmatrix},$$
then we get: $$|A_y| = (2)(1) - (7)(3) = 2 - 21 = -19.$$

Finally, applying Cramer's rule:
$$x = \frac{|A_x|}{|A|} = \frac{-19}{-19} = 1$$
$$y = \frac{|A_y|}{|A|} = \frac{-19}{-19} = 1.$$

Hence, the solutions are x = 1 and y = 1.