Problem 79 - Precalculus

Question

Solve the system of linear equations using Gauss-Jordan elimination. $$\begin{array}{r} x+2 y+z=3 \\ 2 x-y+3 z=7 \\ 3 x+y+4 z=5 \end{array}$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{17}{5} - \frac{4}{5}z \), \( y = -\frac{1}{5} + \frac{1}{5}z \), \( z = z \), parameterized by \( z \).

Step 1: Write the Augmented Matrix

The first step in solving the system of equations using Gauss-Jordan elimination is to write the augmented matrix. For the given system \( \begin{array}{r} x+2 y+z=3 \ 2 x-y+3 z=7 \ 3 x+y+4 z=5 \end{array} \), the augmented matrix is: \[\begin{bmatrix} 1 & 2 & 1 & | & 3 \2 & -1 & 3 & | & 7 \3 & 1 & 4 & | & 5 \\end{bmatrix}\]

Step 2: Create a Leading 1 in Row 1, Column 1

The leading entry in Row 1, Column 1 is already 1, so no changes are needed for this step. If it weren't 1, you would need to divide the entire first row by that value.

Step 3: Eliminate Column 1 Below the Leading 1

To zero out the elements below the leading 1 in the first column, perform the following row operations:1. \( R_2 = R_2 - 2R_1 \) 2. \( R_3 = R_3 - 3R_1 \)This updates the matrix to:\[\begin{bmatrix} 1 & 2 & 1 & | & 3 \0 & -5 & 1 & | & 1 \0 & -5 & 1 & | & -4 \\end{bmatrix}\]

Step 4: Create a Leading 1 in Row 2, Column 2

The leading value in Row 2, Column 2 is -5, so divide Row 2 by -5: \( R_2 = -\frac{1}{5}R_2 \)This updates to:\[\begin{bmatrix} 1 & 2 & 1 & | & 3 \0 & 1 & -\frac{1}{5} & | & -\frac{1}{5} \0 & -5 & 1 & | & -4 \\end{bmatrix}\]

Step 5: Eliminate Column 2 Above and Below the Leading 1

To zero out the elements above and below the leading 1 in column 2, perform the following operations:1. \( R_1 = R_1 - 2R_2 \)2. \( R_3 = R_3 + 5R_2 \)This updates the matrix to:\[\begin{bmatrix} 1 & 0 & \frac{4}{5} & | & \frac{17}{5} \0 & 1 & -\frac{1}{5} & | & -\frac{1}{5} \0 & 0 & 0 & | & 0 \\end{bmatrix}\]

Step 6: Create a Leading 1 in Row 3, Column 3

Since Row 3 has become all zeros, there is no further row reduction needed. The solution to the system is in reduced row-echelon form, which represents consistent equations.

Step 7: Interpret the Results

The matrix now represents the equations:- \( x + \frac{4}{5}z = \frac{17}{5} \)- \( y - \frac{1}{5}z = -\frac{1}{5} \)- The third equation is redundant, as it simplifies to \( 0 = 0 \).From these, express \( x \) and \( y \) in terms of \( z \).

Step 8: Write the Solution

Expressing the variables in terms of the free variable \( z \), we have:\[x = \frac{17}{5} - \frac{4}{5}z \y = -\frac{1}{5} + \frac{1}{5}z \z = z \]The solution is a parametric representation depending on the free variable \( z \).

Key Concepts

Augmented Matrix

To solve a system of linear equations using the Gauss-Jordan elimination method, the first step is to convert the system into something called an "augmented matrix." An augmented matrix combines the coefficients of the variables as well as the constants from each equation into a compact, matrix format.

For the given system of equations:

• \( x + 2y + z = 3 \)
• \( 2x - y + 3z = 7 \)
• \( 3x + y + 4z = 5 \)

This corresponds to an augmented matrix:\[\begin{bmatrix}1 & 2 & 1 & | & 3 \2 & -1 & 3 & | & 7 \3 & 1 & 4 & | & 5 \end{bmatrix}\] Here, the columns before the vertical line represent the coefficients of each variable \( x, y, \) and \( z \), while the column after the line represents the constants from each equation.

Converting the system into an augmented matrix is useful because it streamlines the process of elimination and helps in finding solutions systematically, leading us towards something called the reduced row-echelon form.

Reduced Row-Echelon Form

During Gauss-Jordan elimination, the goal is to manipulate the augmented matrix until it's in something called the "reduced row-echelon form" (RREF). This means performing a series of row operations to get a matrix where:

• Leading coefficients are 1 (sometimes called pivot elements).
• Each leading 1 appears to the right of the leading 1 in the row above it.
• The leading 1 is the only non-zero entry in its column.

To achieve this, we go through several steps, such as swapping rows, multiplying a row by a non-zero scalar, or adding/subtracting a multiple of one row from another.

In our exercise, the last matrix in reduced row-echelon form looks like this:\[\begin{bmatrix}1 & 0 & \frac{4}{5} & | & \frac{17}{5} \0 & 1 & -\frac{1}{5} & | & -\frac{1}{5} \0 & 0 & 0 & | & 0 \end{bmatrix}\] This form allows us to directly get the solutions of the system as it presents the variables in a clear hierarchy, with respect to a free variable (in this case, \( z \)). RREF simplifies interpreting the solutions to the system of equations.

System of Linear Equations

A system of linear equations consists of multiple linear equations that share several variables. Solving these systems involves finding the value of each variable that satisfies all equations simultaneously.

In our problem, we have a system with three equations and three variables (\( x, y, \) and \( z \)). There are different methods to solve such systems, like substitution, elimination, or using matrices. Using the matrix method via Gauss-Jordan elimination is efficient for larger systems and provides insights into the nature of the solutions.

• **Unique solution**: If the system has exactly one set of variable values fulfilling all equations.
• **No solution**: If no possible set of values satisfies all equations (indicating the lines or planes don't converge).
• **Infinitely many solutions**: Occurs if the equations are dependent, meaning they represent the same plane or line.

In the provided example, by reaching the reduced row-echelon form, we can clearly see the system yields infinitely many solutions, expressed in terms of a free variable, \( z \). This format is particularly useful for understanding the relationship between the variables and how changes in one affect the others.