91Ó°ÊÓ

A system of equations is a set of two or more equations with multiple variables. In the context of this problem, we have two equations:

• \(7x + 5y = 9\)
• \(-2x + 3y = -7\)

These equations have two variables, \(x\) and \(y\), and we want to find their values.
To achieve this, we can use various methods, such as substitution, elimination, or matrix inversion.
Using matrices introduces a powerful tool for solving larger systems efficiently.
Such systems often appear in scientific computations, economic models, and engineering problems where simultaneous conditions need to be satisfied.
The goal is to find the specific values for the variables that make both equations true at the same time.

A 2x2 matrix is a fundamental concept in linear algebra consisting of two rows and two columns.
In this exercise, we represent the coefficients of the linear equations in a 2x2 matrix:

• Matrix \(A = \begin{bmatrix} 7 & 5 \ -2 & 3 \end{bmatrix}\) holds the coefficients of \(x\) and \(y\).

Matrices allow us to express complex equations in a compact form.
Each number in the matrix corresponds to a specific element in the equations.
The arrangement of these numbers significantly impacts the solutions that can be derived.
For a 2x2 matrix, finding the inverse is straightforward with the formula:\[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]. This inverse, when it exists, opens a path to solve the matrix equation representing a system of equations.
Matrix operations, such as inversion, are shorthand methods that can simplify complex multi-equation systems.

A matrix equation represents a system of linear equations in matrix form: \(A * X = B\).

• Here, \(A\) is the matrix of coefficients, \(X\) is the matrix of variables, and \(B\) is the matrix representing the constant terms.

For this exercise, \(X = \begin{bmatrix} x \ y \end{bmatrix}\) , and \(B = \begin{bmatrix} 9 \ -7 \end{bmatrix}\).To solve for \(X\), we utilize the inverse of matrix \(A\): Multiply both sides by the inverse to isolate \(X\):

• \(A^{-1} * A * X = A^{-1} * B\)
• Simplifying gives \(X = A^{-1} * B\)

Inserting the inversed matrix and the constant matrix gives us the solution for \(x\) and \(y\).
The calculated inverse and multiplication with \(B\) give: \\(X = \begin{bmatrix} 1 \ -1 \end{bmatrix}\), meaning \(x = 1\) and \(y = -1\).
Matrix equations provide clarity, enhancing not just computation, but our understanding of relationships between variables.