Problem 45 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{aligned} &0.8 x+0.8 y+0.8 z=2.4\\\ &0.3 x-0.5 y+0.2 z=0\\\ &0.1 x+0.2 y+0.3 z=0.6 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = 1\), \(y = 1\), and \(z = 1\).

Step 1: Write the System as an Augmented Matrix

Represent the given system of equations as an augmented matrix:\[\begin{bmatrix}0.8 & 0.8 & 0.8 & | & 2.4 \0.3 & -0.5 & 0.2 & | & 0 \0.1 & 0.2 & 0.3 & | & 0.6 \end{bmatrix}\]

Step 2: Normalize the First Row

Normalize the first row by dividing all entries of the first row by 0.8 to make the leading coefficient 1:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0.3 & -0.5 & 0.2 & | & 0 \0.1 & 0.2 & 0.3 & | & 0.6 \end{bmatrix}\]

Step 3: Eliminate x From the Second Row

Subtract 0.3 times the first row from the second row to eliminate the variable x from the second equation:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & -0.8 & -0.1 & | & -0.9 \0.1 & 0.2 & 0.3 & | & 0.6 \end{bmatrix}\]

Step 4: Eliminate x From the Third Row

Subtract 0.1 times the first row from the third row to eliminate the variable x from the third equation:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & -0.8 & -0.1 & | & -0.9 \0 & 0.1 & 0.2 & | & 0.3 \end{bmatrix}\]

Step 5: Simplify the Second Row

Divide the entire second row by -0.8 to make the leading coefficient 1:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & 1 & 0.125 & | & 1.125 \0 & 0.1 & 0.2 & | & 0.3 \end{bmatrix}\]

Step 6: Eliminate y From the Third Row

Subtract 0.1 times the second row from the third row to eliminate the variable y from the third equation:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & 1 & 0.125 & | & 1.125 \0 & 0 & 0.1875 & | & 0.1875 \end{bmatrix}\]

Step 7: Solve for z

From the third row, solve for z by dividing by 0.1875:\[z = \frac{0.1875}{0.1875} = 1\]

Step 8: Back Substitute to Find y

Back substitute z = 1 into the second row:\[1 + 0.125z = 1.125 \y + 0.125(1) = 1.125 \y = 1.125 - 0.125 = 1\]

Step 9: Back Substitute to Find x

Back substitute z = 1 and y = 1 into the first row:\[x + y + z = 3 \x + 1 + 1 = 3 \x = 3 - 2 = 1\]

Step 10: Solution of the System

The solution to the system of equations is:\[x = 1, \quad y = 1, \quad z = 1\]

Key Concepts

System of Equations

A system of equations consists of multiple equations that involve the same set of variables. The main objective is to find values for these variables that satisfy all the equations simultaneously. In the given exercise, we have three equations:

• \( 0.8x + 0.8y + 0.8z = 2.4 \)
• \( 0.3x - 0.5y + 0.2z = 0 \)
• \( 0.1x + 0.2y + 0.3z = 0.6 \)

In this set, the variables \( x, y, \text{ and } z \) are interrelated through the coefficients given in the equations. Solving the system means finding the specific values for \( x, y, \text{ and } z \) that make all these equations true simultaneously. By representing it in a matrix form and using Gaussian Elimination, which involves systematic row operations, we can simplify and solve this system effectively.

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 equations' right-hand side into a matrix format. For the given system, our augmented matrix looks like this:\[ \begin{bmatrix} 0.8 & 0.8 & 0.8 & | & 2.4 \ 0.3 & -0.5 & 0.2 & | & 0 \ 0.1 & 0.2 & 0.3 & | & 0.6 \end{bmatrix} \]Here, each row corresponds to an equation, and each column corresponds to the coefficients of \( x, y, \text{ and } z \). The vertical bar separates the coefficients from the constants, providing a clear visualization of the system.
This format is essential for applying Gaussian elimination, as it allows us to perform row operations directly on the matrix to systematically simplify and solve the system.

Back Substitution

Back substitution is a method used after transforming a matrix into an upper triangular form using Gaussian elimination. In this form, all elements below the main diagonal are zero. The process starts at the last equation and uses known values to find solutions for the remaining variables.
In our example:

• The last equation reduces to \( z = 1 \).
• Then in the second-to-last equation, by substituting \( z = 1 \), we find \( y = 1 \).
• Finally, substituting \( y = 1 \) and \( z = 1 \) into the first equation gives \( x = 1 \).

By starting from the simplest equation, we use known values to solve backwards through the system, ensuring a step-by-step resolution to all variables. This method confirms the solution’s correctness by using previously calculated values.

Leading Coefficient

The leading coefficient is the first non-zero number from the left in a row of a matrix. In the context of Gaussian elimination, ensuring the leading coefficient is 1 simplifies further calculations and is crucial for systematic row reductions.
In our example, the process involves adjusting each leading coefficient to 1:

• Normalizing the first row: Dividing all elements by 0.8 made the leading coefficient 1 in the first row.
• Similarly, the second row was adjusted by dividing it by its initial leading coefficient of -0.8.

Achieving a leading 1 helps create zeros beneath this number, systematically turning the matrix into an upper triangular form. This simplifies back substitution and ultimately leads to finding the solution for each variable. Recognizing and adjusting the leading coefficient ensures clarity and precision during matrix operations.