91Ó°ÊÓ

A system of linear equations consists of two or more equations involving the same set of variables. For example, in the given problem, we have:

• \(10x - 6y = 2\)
• \(-5x + 8y = -1\)

These equations are linear because each term is either a constant or a product of a constant and a single variable. Linear equations are used to determine variable values that satisfy all equations simultaneously.
To solve a system of linear equations, we often aim to find the values of \(x\) and \(y\) that satisfy both equations at once. This can be represented graphically as the point where the lines intersect each other.
When using algebraic methods like Cramer's Rule, systems are expressed in terms of matrices to utilize matrix arithmetic for solutions.

Matrix determinants are scalars that provide important information about a matrix. They are central when solving systems of linear equations using methods such as Cramer's Rule. For a 2x2 matrix, the determinant is calculated as follows for a matrix \(A\):

• Suppose \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\).
• The determinant, expressed as \(\text{det}(A)\), is calculated by \(ad - bc\).

In the provided exercise, the coefficient matrix is:\[\begin{bmatrix} 10 & -6 \ -5 & 8 \end{bmatrix}\]Plugging into the determinant formula, we find \( \text{det}(A) = 10 \times 8 - (-6 \times -5) = 80 - 30 = 50 \).
The determinant indicates whether the system of equations has a unique solution, no solution, or infinitely many solutions. If the determinant is non-zero, as in our case, a unique solution exists.

Solving algebraic equations involves finding the values that satisfy the equation. With Cramer's Rule, this involves a systematic approach using determinants. Let’s break down how this rule applies:

• First, calculate the determinant of the coefficient matrix, \(\text{det}(A)\), to ensure a unique solution.
• Create matrices \(A_x\) and \(A_y\) by replacing their respective columns with the constant terms from the equations.
• Calculate the determinants \(\text{det}(A_x)\) and \(\text{det}(A_y)\).
• Apply Cramer's Rule: \( x = \frac{\text{det}(A_x)}{\text{det}(A)} \) and \( y = \frac{\text{det}(A_y)}{\text{det}(A)} \).

Using the problem's data, \( x \) was computed as \( \frac{22}{50} = 0.44 \) and \( y \) as \( \frac{0}{50} = 0 \).
This shows how linear algebra techniques can solve equations efficiently, giving precise solutions where other methods might be cumbersome.