Problem 37 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the system of linear equations using the Gauss-Jordan elimination method. \(\begin{aligned} 2 x-3 y &=-8 \\ 4 x+y &=-2 \end{aligned}\)

Step-by-Step Solution

Answer

The solution to the system of linear equations using the Gauss-Jordan elimination method is \(x = -2.8\) and \(y = 1.2\).

Step 1: 1. Write the augmented matrix of the system.

: The augmented matrix is obtained by writing the coefficients of the variables and the constants in the system of equations. The augmented matrix A is written as: A = \(\begin{bmatrix} 2 & -3 & |-8 \\\\ 4 & 1 & |-2 \end{bmatrix}\)

Step 2: 2. Perform the necessary row operations to reduce it to the reduced row echelon form (RREF).

: First, we need to make the element in the first row and first column equal to 1. We will divide the entire first row by 2: \(\frac{1}{2}R_1 \rightarrow R_1\) A = \(\begin{bmatrix} 1 & -\frac{3}{2} & |-4 \\\\ 4 & 1 & |-2 \end{bmatrix}\) Now, we will make the element in the second row and first column equal to 0. To do this, we will subtract 4 times the first row from the second row: \(R_2 - 4R_1 \rightarrow R_2\) A = \(\begin{bmatrix} 1 & -\frac{3}{2} & |-4 \\\\ 0 & 5 & |6 \end{bmatrix}\) Next, we need to make the element in the second row and second column equal to 1. We will divide the entire second row by 5: \(\frac{1}{5}R_2 \rightarrow R_2\) A = \(\begin{bmatrix} 1 & -\frac{3}{2} & |-4 \\\\ 0 & 1 & |1.2 \end{bmatrix}\) We will now make the element in the first row and second column equal to 0. We will add \(\frac{3}{2}\) times the second row to the first row: \(R_1 + \frac{3}{2}R_2 \rightarrow R_1\) A = \(\begin{bmatrix} 1 & 0 & |-2.8 \\\\ 0 & 1 & |1.2 \end{bmatrix}\)

Step 3: 3. Read the solutions from the RREF of the augmented matrix.

: We have now obtained the RREF of the augmented matrix, and we can read the solutions directly from it: \(x = -2.8\) \(y = 1.2\) The solution to the system of linear equations is x = -2.8 and y = 1.2.

Key Concepts

Linear Equations

The system of linear equations involves determining values for variables that make each equation in the system true. In this exercise, we're given:\[\begin{aligned} 2x - 3y &= -8 \ 4x + y &= -2 \end{aligned}\]A linear equation is one where each term is either a constant or the product of a constant and a single variable. This property allows these equations to be graphed as straight lines, with the solutions being the intersection point(s) of these lines.
Understanding linear equations is foundational because they form the basis of algebra and are used in various applications such as economics, engineering, and the physical sciences. Solving these systems efficiently is possible through multiple methods, including substitution, elimination, or matrix techniques like Gauss-Jordan elimination.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It condenses all the coefficients and constants from the equations into a single matrix. For our given problem, the augmented matrix is constructed by aligning the coefficients of variables and appending the constants:\[\begin{bmatrix} 2 & -3 & |-8 \ 4 & 1 & |-2 \end{bmatrix}\]Here, each row corresponds to an equation, and the vertical bar separates the coefficients of the variables from the constant terms. This is especially useful in matrix operations, as it allows for direct application of mathematical operations to simplify the system of equations.
Utilizing an augmented matrix in solving systems of equations is beneficial because it provides a visual and manipulative tool, making complex systems easier to handle, especially for multiple equations and variables.

Row Operations

Row operations are maneuvers applied to the rows of a matrix to simplify and eventually solve a system of equations. For Gauss-Jordan elimination, three types of elementary row operations are used:

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting a multiple of one row from another row

In the given solution, these operations help us transform the matrix into a reduced form: 1. We start by making the leading coefficient of the first row 1 by dividing it. 2. Next, we eliminate the numbers below and above this leading 1. 3. This is repeated for other rows until the matrix reaches the reduced row echelon form.
The beauty of these operations lies in their ability to retain the integrity of the original system while simplifying it to an easily interpretable form. This is crucial for solving efficiently without altering the system's solutions.

Reduced Row Echelon Form

Reduced row echelon form (RREF) is a simplified form of a matrix achieved through row operations. In RREF, the matrix has distinct characteristics:

• Each leading entry of a row is 1, and is the only non-zero entry in its column.
• Leading entries in consecutive rows appear to the right of those in the previous row.
• Any row containing only zeros appears at the bottom of the matrix.

In this exercise, the matrix finally becomes:\[\begin{bmatrix} 1 & 0 & |-2.8 \ 0 & 1 & |1.2 \end{bmatrix}\]From this form, we directly read the variables' values: \(x = -2.8\) and \(y = 1.2\).
The RREF is pivotal in linear algebra as it allows us to effortlessly extract the solution of a system, even for large and complex matrices, making it a universal standard for matrix equations.