91Ó°ÊÓ

A system of linear equations consists of two or more equations with multiple variables. The goal is to find the values of these variables that satisfy all the equations simultaneously. In the given exercise, the system consists of two equations with two variables, \(x\) and \(y\). Here's what this looks like:

• Equation 1: \(2x - 4y = c\)
• Equation 2: \(x + 3y = d\)

These equations are interdependent, meaning the solution to this system is the set of \(x\) and \(y\) values that make both equations true at the same time. There are several methods for solving such systems, including substitution, elimination, and, as we explore here, the inverse matrix method.

Matrix algebra is a powerful mathematical tool used to solve systems of equations, among other applications. In matrix algebra, numbers, variables, or equations are often arranged in a rectangular array known as a matrix. In this context of solving linear systems, we represent the coefficients of the variables in a square matrix termed as matrix \(A\).

• For our system: \(A = \begin{bmatrix} 2 & -4 \ 1 & 3 \end{bmatrix}\), where each number corresponds to the coefficients of \(x\) and \(y\) in the system.
• The variables \(x\) and \(y\) are also represented in a matrix form as a column vector \(\mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}\).

Finally, the constants from the right side of the equations are organized in another column matrix \(\mathbf{b} = \begin{bmatrix} c \ d \end{bmatrix}\). Thus, our system turns into the concise matrix equation \(A\mathbf{x} = \mathbf{b}\). The goal is to solve for the matrix \(\mathbf{x}\).

Determinants are special numbers calculated from square matrices and play a key role in various matrix operations, including finding the inverse of a matrix. The determinant of a 2x2 matrix, \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), is given by the formula \(ad - bc\). This scalar value provides key insights, for instance:

• If the determinant of a matrix is zero, the matrix is singular, meaning it cannot be inverted.
• A non-zero determinant implies that the matrix can be inverted, which is crucial for solving systems using matrix inversion.

In our exercise, the determinant of matrix \(A\) is calculated as \(2 \times 3 - (-4) \times 1 = 10\). Since it is not zero, matrix \(A\) has an inverse, crucial for the inverse matrix method used here.

Matrix inversion is a process where you find a new matrix, called the inverse, that "undoes" the effect of the original matrix. For a 2x2 matrix \(A\), the inverse \(A^{-1}\) is calculated using the determinant. The formula is:\[A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]In our problem, the matrix \(A = \begin{bmatrix} 2 & -4 \ 1 & 3 \end{bmatrix}\) has an inverse:\[A^{-1} = \frac{1}{10}\begin{bmatrix} 3 & 4 \ -1 & 2 \end{bmatrix}\]This means by multiplying the inverse of matrix \(A\) by matrix \(\mathbf{b}\), we can solve for \(\mathbf{x}\). The solutions in both parts of the exercise were found using this method. By applying the inverse, we transformed the complex task of solving simultaneous equations into simpler matrix multiplications, demonstrating the power and efficiency of matrix inversion in linear algebra.