Problem 44 - College Algebra

Question

Solve each system by Gaussian elimination. $$ \begin{array}{l} 0.3 x+0.3 y+0.5 z=0.6 \\ 0.4 x+0.4 y+0.4 z=1.8 \\ 0.4 x+0.2 y+0.1 z=1.6 \end{array} $$

Step-by-Step Solution

Answer

The solution is \( x = -35 \), \( y = 62 \), \( z = -15 \).

Step 1: Write the augmented matrix

First, transform the given system of equations into an augmented matrix. The system is: \[ \begin{align*} 0.3x + 0.3y + 0.5z &= 0.6 \ 0.4x + 0.4y + 0.4z &= 1.8 \ 0.4x + 0.2y + 0.1z &= 1.6 \end{align*} \] The augmented matrix is: \[ \begin{bmatrix} 0.3 & 0.3 & 0.5 & | & 0.6 \ 0.4 & 0.4 & 0.4 & | & 1.8 \ 0.4 & 0.2 & 0.1 & | & 1.6 \end{bmatrix} \]

Step 2: Row Reduction to Row-Echelon Form

Perform row operations to convert the matrix into row-echelon form. Start by using row operations to make the first element of the second row zero by subtracting a suitable multiple of the first row from the second row: Subtract \( \frac{4}{3} R_1 \) from \( R_2 \): \[ R_2 = R_2 - \frac{4}{3} R_1 \]New matrix:\[ \begin{bmatrix} 0.3 & 0.3 & 0.5 & | & 0.6 \ 0 & 0 & -0.0667 & | & 1.0 \ 0.4 & 0.2 & 0.1 & | & 1.6 \end{bmatrix} \] Next, clear the first column of the third row by subtracting \( R_1 \) scaled by \( \frac{4}{3} \). Subtract \( \frac{4}{3} R_1 \) from \( R_3 \): \[ R_3 = R_3 - \frac{4}{3} R_1 \]New matrix:\[ \begin{bmatrix} 0.3 & 0.3 & 0.5 & | & 0.6 \ 0 & 0 & -0.0667 & | & 1.0 \ 0 & -0.1 & -0.4667 & | & 0.8 \end{bmatrix} \]

Step 3: Further Simplification to Upper Triangular Form

Continue with row operations to make the augmented matrix as upper triangular as possible. Make the second element of the third row zero by adding a suitable multiple of the second row to the third row. As second rows has already only 0 in first and second element this step is not needed.

Step 4: Back Substitution

Use back substitution to solve for the variables. Start from the third variable: From the second row, solve for \( z \): \[ -0.0667z = 1.0 \ z = -15 \]With \( z \) known, substitute back into the remaining equations to solve for \( y \) and \( x \). Plugging \( z = -15 \) into the third row equation to find \( y \): \[ -0.1y - 0.4667(-15) = 0.8 \ -0.1y + 7 = 0.8 \ -0.1y = 0.8 - 7 \ y = 62 \]Finally, use the first equation: \[ 0.3x + 0.3(62) + 0.5(-15) = 0.6 \ 0.3x + 18.6 - 7.5 = 0.6 \ 0.3x + 11.1 = 0.6 \ 0.3x = 0.6 - 11.1 \ 0.3x = -10.5 \ x = -35 \]

Step 5: Final Solution: Write the Solution of the System

The values for \( x \), \( y \), and \( z \) were found through the steps of Gaussian elimination and back substitution. Thus, the solution to the system of equations is: \[ x = -35, \; y = 62, \; z = -15 \]

Key Concepts

Augmented Matrix

When solving a system of linear equations using Gaussian Elimination, the first step is to represent the system as an augmented matrix. An augmented matrix is an efficient way to display coefficients and constants of equations. It combines the coefficient matrix and the constants from each equation into a single matrix. For example, if you have the system:

• \(0.3x + 0.3y + 0.5z = 0.6\)
• \(0.4x + 0.4y + 0.4z = 1.8\)
• \(0.4x + 0.2y + 0.1z = 1.6\)

You write the augmented matrix as:\[\begin{bmatrix} 0.3 & 0.3 & 0.5 & | & 0.6 \0.4 & 0.4 & 0.4 & | & 1.8 \0.4 & 0.2 & 0.1 & | & 1.6 \end{bmatrix}\]Each row represents an equation, and each column represents the coefficients of one variable and the constants from the equations. This matrix form is a foundation for performing row operations to simplify the system.

Row Reduction

Row reduction is a process aimed at transforming a matrix into a simpler form for solving a system of equations. The goal is to change the matrix into row-echelon form, where each successive row has more leading zeros. Row operations include:

• Adding or subtracting multiples of rows
• Swapping rows
• Multiplying a row by a non-zero scalar

In the example, row operations are used to clear variables step-by-step, such as subtracting a multiple of one row from another to eliminate coefficients under leading coefficients. This is crucial for creating zeros below the pivot positions in each column. Once the matrix reaches this upper triangular form, you can then solve for variables through back substitution.

Back Substitution

Back substitution is the final step after achieving row-echelon form from row reduction. This method involves solving for the variables starting from the last row and moving upward. Consider a matrix arranged for back substitution:

• \(0.3x + 0.3y + 0.5z = 0.6\)
• \(0 = 0 + 0 - 0.0667z = 1.0\)
• \(0 = 0 - 0.1y - 0.4667z = 0.8\)

Start solving from the bottom row upward. First, find \(z\), then substitute back into previous equations to solve for \(y\), and continue to solve for \(x\). This process systematically identifies the values of each variable, using substitution and arithmetic simplifications.

System of Linear Equations

A system of linear equations consists of multiple linear equations with common variables. The solution to the system is the point(s) where all equations are satisfied simultaneously, which is typically represented in n-dimensional space. The given system involves three equations with three variables \(x, y, z\). The objective is to find a unique combination of these variables that solves all the equations concurrently. Using methods like Gaussian Elimination:

• First, use augmented matrices to set the scene.
• Apply row reduction to simplify the system.
• Finally utilize back substitution to deduce the values of the variables.

Such systems are foundational in various fields like physics, engineering, economics, and more, where they are used to model relationships and constraints in real-world situations.