Problem 42 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l} 0.5 x-0.5 y-0.3 z=0.13 \\ 0.4 x-0.1 y-0.3 z=0.11 \\ 0.2 x-0.8 y-0.9 z=-0.32 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x \approx 0.251\), \(y \approx 0.183\), \(z \approx 0.336\).

Step 1: Write the Augmented Matrix

Start by writing the given system of equations as an augmented matrix. The system is: \(\begin{aligned} 0.5x - 0.5y - 0.3z &= 0.13 \ 0.4x - 0.1y - 0.3z &= 0.11 \ 0.2x - 0.8y - 0.9z &= -0.32 \end{aligned}\). The augmented matrix is: \[\begin{bmatrix} 0.5 & -0.5 & -0.3 & | & 0.13 \ 0.4 & -0.1 & -0.3 & | & 0.11 \ 0.2 & -0.8 & -0.9 & | & -0.32 \end{bmatrix}\].

Step 2: Eliminate First Column Lower Entries

Eliminate the entries below the first pivot (0.5) in the first column using row operations. Replace the second row with \((R_2 - \frac{0.4}{0.5}R_1)\) and the third row with \((R_3 - \frac{0.2}{0.5}R_1)\). After calculations, the matrix becomes:\[\begin{bmatrix} 0.5 & -0.5 & -0.3 & | & 0.13 \ 0 & 0.3 & 0 & | & 0.055 \ 0 & -0.6 & -0.78 & | & -0.372 \end{bmatrix}\].

Step 3: Eliminate Second Column Lower Entry

Now, eliminate the non-zero entry in the second column below the second pivot (0.3). Replace the third row with \((R_3 + 2R_2)\) to eliminate the -0.6. The matrix becomes:\[\begin{bmatrix} 0.5 & -0.5 & -0.3 & | & 0.13 \ 0 & 0.3 & 0 & | & 0.055 \ 0 & 0 & -0.78 & | & -0.262 \end{bmatrix}\].

Step 4: Solve for Variables

Start with the last row to solve for variable \(z\). Since \(-0.78z = -0.262\), solve to get \(z = \frac{-0.262}{-0.78} \approx 0.336\). Use this value to substitute back into the second row to solve for \(y\): \(0.3y = 0.055\), giving \(y = \frac{0.055}{0.3} \approx 0.183\). Finally, substitute \(y\) and \(z\) into the first row to solve for \(x\): \(0.5x - 0.5(0.183) - 0.3(0.336) = 0.13\), resulting in \(x \approx 0.251\).

Step 5: Verify Solution

Substitute \(x = 0.251\), \(y = 0.183\), and \(z = 0.336\) back into the original equations to verify correctness. All three equations should hold true, confirming the solution is valid.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a shared set of variables. The goal is to find values for these variables that satisfy all the equations in the system. Systems of equations can arise in various contexts, such as solving real-world problems like determining costs or predicting trends.
In this exercise, the system consists of three equations involving three variables: \(x\), \(y\), and \(z\). These equations are:

• \(0.5x - 0.5y - 0.3z = 0.13\)
• \(0.4x - 0.1y - 0.3z = 0.11\)
• \(0.2x - 0.8y - 0.9z = -0.32\)

The solution involves finding the specific values for \(x\), \(y\), and \(z\) that make all three equations true at the same time.

Augmented Matrix

An augmented matrix is a powerful tool for solving systems of equations, especially when using methods like Gaussian elimination. It combines the coefficients of the variables and the constants from each equation into a single matrix, making calculations more straightforward.
To create the augmented matrix for a given system:

• List the coefficients of each variable in the equations.
• Include the constants from the right side of the equations.

For the given system, the augmented matrix is: \[\begin{bmatrix} 0.5 & -0.5 & -0.3 & | & 0.13 \0.4 & -0.1 & -0.3 & | & 0.11 \0.2 & -0.8 & -0.9 & | & -0.32 \end{bmatrix}\]This matrix is a compact representation of the system and becomes the foundation for applying row operations to solve for the variables.

Row Operations

Row operations are essential when working with augmented matrices to solve systems of equations. They manipulate the rows of the matrix to simplify the system, leading us towards a solution. There are three types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting rows from each other.

In Gaussian elimination, the goal is often to create a matrix in row-echelon form, where:

• The first non-zero number from the left (called the "pivot") in each row is 1.
• The pivots must move downwards to the right as you move down the matrix.
• Zeros are beneath each pivot.

For our system, row operations are used to eliminate the entries below the pivots, transforming the matrix step by step to eventually solve for the variables.

Solving for Variables

After applying row operations and transforming the augmented matrix, the next step is solving for the variables. This stage involves using the simplified matrix, typically in row-echelon form, to find the precise values of the variables involved.
Starting with the last row of the matrix, these values are calculated using back substitution:

• Start solving for \(z\) using the last row, which often has only one variable left after elimination.
• Once \(z\) is found, substitute its value into the previous rows to find \(y\).
• Finally, substitute the known values of \(y\) and \(z\) into the first row to solve for \(x\).

For this exercise, this methodically leads us to the solutions: \(x \approx 0.251\), \(y \approx 0.183\), and \(z \approx 0.336\). It's crucial to verify these values by substituting them back into the original system to ensure they satisfy all the equations.