Problem 49 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the system of linear equations using the Gauss-Jordan elimination method. \(\begin{array}{rr}x_{1}-x_{2}+3 x_{3}= & 14 \\ x_{1}+x_{2}+x_{3}= & 6 \\ -2 x_{1}-x_{2}+x_{3}= & -4\end{array}\)

Step-by-Step Solution

Answer

The solution to the given system of linear equations using the Gauss-Jordan elimination method is \(x_1 = 4\), \(x_2 = -1\), and \(x_3 = 3\).

Step 1: Set up the augmented matrix from the system of equations

To begin, we rewrite the given system of linear equations as an augmented matrix. Write the coefficients of the variables as the entries in the matrix and use a vertical line to separate the coefficients from the constants: \[ \left[\begin{array}{rrr|r} 1 & -1 & 3 & 14 \\ 1 & 1 & 1 & 6 \\ -2 & -1 & 1 & -4 \end{array}\right] \]

Step 2: Use row operations to transform the matrix into reduced row-echelon form

Perform the following row operations to transform the matrix into its reduced row-echelon form: 1. Add 2 times the first row to the third row to eliminate the first element in the third row: \[ \left[\begin{array}{rrr|r} 1 & -1 & 3 & 14 \\ 1 & 1 & 1 & 6 \\ 0 & -3 & 7 & 24 \end{array}\right] \] 2. Add the first row to the second row to eliminate the first element in the second row: \[ \left[\begin{array}{rrr|r} 1 & -1 & 3 & 14 \\ 0 & 2 & -2 & -8 \\ 0 & -3 & 7 & 24 \end{array}\right] \] 3. Multiply the second row by 1/2 to have a leading 1 in the second row: \[ \left[\begin{array}{rrr|r} 1 & -1 & 3 & 14 \\ 0 & 1 & -1 & -4 \\ 0 & -3 & 7 & 24 \end{array}\right] \] 4. Add the second row to the first row and add 3 times the second row to the third row to eliminate the second element in the first and third rows: \[ \left[\begin{array}{rrr|r} 1 & 0 & 2 & 10 \\ 0 & 1 & -1 & -4 \\ 0 & 0 & 4 & 12 \end{array}\right] \] 5. Divide the third row by 4 to have a leading 1 in the third row: \[ \left[\begin{array}{rrr|r} 1 & 0 & 2 & 10 \\ 0 & 1 & -1 & -4 \\ 0 & 0 & 1 & 3 \end{array}\right] \] 6. Subtract 2 times the third row from the first row and add the third row to the second row to eliminate the third element in the first and second rows: \[ \left[\begin{array}{rrr|r} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \]

Step 3: Extract the solutions from the reduced row-echelon form

The matrix is now in reduced row-echelon form, and we can read the solutions directly from the matrix: \[ x_1 = 4, \quad x_2 = -1, \quad x_3 = 3 \] So the solution to the given system of linear equations is \(x_1 = 4\), \(x_2 = -1\), and \(x_3 = 3\).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. In the context of the problem, we have a system with three variables: \(x_1\), \(x_2\), and \(x_3\). Each equation describes a plane in three-dimensional space, and the solution to the system is the point where these planes intersect.

Let's break down the problem:

• First equation: \(x_1 - x_2 + 3x_3 = 14\)
• Second equation: \(x_1 + x_2 + x_3 = 6\)
• Third equation: \(-2x_1 - x_2 + x_3 = -4\)

Solving this system means finding the values of \(x_1\), \(x_2\), and \(x_3\) that satisfy all three equations simultaneously. Techniques such as substitution and elimination are common, but in this exercise, we use the Gauss-Jordan elimination method.

This systematic approach allows us to solve for each variable by reducing the equation system to a simpler one, eventually isolating each variable.

Augmented Matrix

An augmented matrix is a compact way to handle systems of linear equations. It includes the coefficients of the variables and the constants from the equations written in matrix form. This allows us to apply row operations easily, which are a series of arithmetic operations used to simplify the equations.

For the equations given:
The augmented matrix looks like this: \[\left[\begin{array}{rrr|r}1 & -1 & 3 & 14 \1 & 1 & 1 & 6 \-2 & -1 & 1 & -4\end{array}\right]\]

The columns before the vertical line represent the coefficients of \(x_1\), \(x_2\), and \(x_3\), while the column after the line represents the constants. This matrix form is crucial for implementing row operations and transforming the system into its reduced form.

Reduced Row-Echelon Form

The reduced row-echelon form (RREF) is a final, simplified version of a matrix that makes it straightforward to extract the solutions of a system of linear equations. In RREF, each leading entry in a row is 1, and it is the only non-zero entry in its column.

For the provided system, we performed a series of row operations to achieve:\[\left[\begin{array}{rrr|r}1 & 0 & 0 & 4 \0 & 1 & 0 & -1 \0 & 0 & 1 & 3\end{array}\right]\]

This matrix form has been transformed so that:

• Each leading coefficient (first non-zero number from the left in a non-zero row) is clear and isolated.
• Every leading 1 is to the right of any leading 1s in the rows above it, ensuring a clear cascade downwards.
• All other values in the leading 1's columns are zeros.

From this form, the solutions \(x_1 = 4\), \(x_2 = -1\), and \(x_3 = 3\) can be directly read from the matrix, revealing the intersection point of the planes described by the original system of equations.