Problem 41 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the system of linear equations using the Gauss-Jordan elimination method. \(\begin{aligned} 2 x+2 y+z=& 9 \\ x+z=& 4 \\ 4 y-3 z=& 17 \end{aligned}\)

Step-by-Step Solution

Answer

The solution to the given system of linear equations using the Gauss-Jordan elimination method is \(x = -11\), \(y = 8\), and \(z = 15\).

Step 1: Convert the system of linear equations into an augmented matrix

Write the given system of linear equations as an augmented matrix: \( \left[ \begin{array}{ccc|c} 2 & 2 & 1 & 9 \\ 1 & 0 & 1 & 4 \\ 0 & 4 & -3 & 17 \end{array} \right] \)

Step 2: Perform elementary row operations to reach reduced row-echelon form (RREF)

Using the Gauss-Jordan elimination method, perform elementary row operations as follows: - Multiply row 1 by 1/2 to have a leading 1: \( \left[ \begin{array}{ccc|c} 1 & 1 & 1/2 & 9/2 \\ 1 & 0 & 1 & 4 \\ 0 & 4 & -3 & 17 \end{array} \right] \) - Subtract row 1 from row 2 and replace row 2 with the result: \( \left[ \begin{array}{ccc|c} 1 & 1 & 1/2 & 9/2 \\ 0 & -1 & 1/2 & -1/2 \\ 0 & 4 & -3 & 17 \end{array} \right] \) - Multiply row 2 by -1 to have a leading 1: \( \left[ \begin{array}{ccc|c} 1 & 1 & 1/2 & 9/2 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 4 & -3 & 17 \end{array} \right] \) - Subtract row 2 from row 1 and replace row 1 with the result: \( \left[ \begin{array}{ccc|c} 1 & 0 & 1 & 4 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 4 & -3 & 17 \end{array} \right] \) - Subtract 4 times row 2 from row 3 and replace row 3 with the result: \( \left[ \begin{array}{ccc|c} 1 & 0 & 1 & 4 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 0 & 1 & 15 \end{array} \right] \)

Step 3: Convert the RREF matrix back into a system of linear equations

The reduced row-echelon form of the augmented matrix is: \( \left[ \begin{array}{ccc|c} 1 & 0 & 1 & 4 \\ 0 & 1 & -1/2 & 1/2 \\ 0 & 0 & 1 & 15 \end{array} \right] \) Now, convert it back into a system of linear equations: \(\begin{aligned} x+z=& 4 \\\ y-\frac{1}{2}z=& \frac{1}{2} \\\ z=& 15 \end{aligned}\)

Step 4: Solve for x, y, and z.

Using the values obtained in step 3, we can easily find the values of x, y, and z: - From the third equation, we get \(z = 15\) - Substitute the value of z into the second equation to find y: \(y-\frac{1}{2}(15) = \frac{1}{2}\) \(\Rightarrow y = \frac{1}{2} + \frac{1}{2}(15)\) \(\Rightarrow y = 8\) - Substitute the value of z into the first equation to find x: \(x + 15 = 4\) \(\Rightarrow x = 4 - 15\) \(\Rightarrow x = -11\) The solution to the system of linear equations is \(x = -11\), \(y = 8\), and \(z = 15\).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations involving the same set of variables. The goal is to find the values of the variables that satisfy all the equations simultaneously. For example, the system from our textbook exercise is composed of three equations with variables x, y, and z. The solution to a system like this can result in a single set of values for the variables (a unique solution), many solutions (infinitely many solutions), or no solution at all (inconsistent).

The Gauss-Jordan elimination method is a systematic process that can be used to solve such a system. By converting the equations into an augmented matrix and then applying a series of elementary row operations, we can find the solution in a clear and concise manner, transforming the matrix into what is known as reduced row-echelon form.

Row-echelon Form

Row-echelon form is a specific arrangement of an augmented matrix which aids in solving systems of linear equations. A matrix is in row-echelon form if it satisfies the following conditions:

• All nonzero rows are above any rows of all zeros.
• Each leading entry of a row is in a column to the right of the leading entry of the row above it.
• All entries in a column below a leading entry are zeros.

A reduced row-echelon form (RREF) goes a step further by ensuring that the leading entry in each non-zero row is 1, and that it is the only non-zero entry in its column. This form is unique for a given matrix and directly leads to the solution of the system when it's consistent. In the solution provided, the objective was to reach the RREF to find the values for x, y, and z.

Elementary Row Operations

Elementary row operations are tools used to manipulate an augmented matrix to obtain row-echelon form or reduced row-echelon form. They are critical in the process of solving a system of equations using the Gauss-Jordan elimination method. There are three types of operations:

• Type I: Swapping two rows.
• Type II: Multiplying all entries in a row by a nonzero constant.
• Type III: Adding or subtracting the multiples of one row to another row.

These operations are used to simplify the augmented matrix systematically without changing the solutions of the original system. The provided solution applies these operations step by step - for example, multiplying a row by a scalar to obtain a leading one, and adding subtracting rows from each other to obtain zeros in specific positions.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficients of the variables and the constants from the right-hand side of the equations into a single matrix. The left section of this matrix corresponds to the coefficients of the variables in the system, and the rightmost column (after the vertical bar) contains the constants from each equation.

For instance, the exercise provided started by translating the system of linear equations into an augmented matrix format before applying Gauss-Jordan elimination. This matrix is then manipulated, maintaining equivalency with the original system, eventually revealing the solution through the transformations. The technique of converting a verbal or symbolic system into an augmented matrix form is foundational in linear algebra and greatly simplifies the process of solving the equations.