Problem 37 - College Algebra

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l}{0.1 x-0.2 y+0.3 z=2} \\ {0.5 x-0.1 y+0.4 z=8} \\ {0.7 x-0.2 y+0.3 z=8}\end{array} $$

Step-by-Step Solution

Answer

The solution is \(x=10\), \(y=10\), and \(z=10\).

Step 1: Write the Augmented Matrix

First, write the given system of linear equations as an augmented matrix: \[\begin{bmatrix}0.1 & -0.2 & 0.3 & | & 2 \0.5 & -0.1 & 0.4 & | & 8 \0.7 & -0.2 & 0.3 & | & 8\end{bmatrix}\]

Step 2: Scale the First Row

To simplify the arithmetic, let's eliminate fractions and decimals by multiplying the first row by 10: \[\begin{bmatrix}1 & -2 & 3 & | & 20 \0.5 & -0.1 & 0.4 & | & 8 \0.7 & -0.2 & 0.3 & | & 8\end{bmatrix}\]

Step 3: Eliminate x from Second Row

To eliminate the variable \(x\) from the second row, subtract \(0.5\) times the first row from the second row:- New Second Row = (Second Row) - \(0.5\times\) (First Row)\[\begin{bmatrix}1 & -2 & 3 & | & 20 \0 & 0.9 & -1.1 & | & -2 \0.7 & -0.2 & 0.3 & | & 8\end{bmatrix}\]

Step 4: Eliminate x from Third Row

To eliminate the variable \(x\) from the third row, subtract \(0.7\) times the first row from the third row:- New Third Row = (Third Row) - \(0.7\times\) (First Row)\[\begin{bmatrix}1 & -2 & 3 & | & 20 \0 & 0.9 & -1.1 & | & -2 \0 & 1.2 & -1.8 & | & -6\end{bmatrix}\]

Step 5: Simplify Third Row

Divide the entire third row by 1.2 to simplify it:- New Third Row = \(\frac{1}{1.2}\times\) (Third Row)\[\begin{bmatrix}1 & -2 & 3 & | & 20 \0 & 0.9 & -1.1 & | & -2 \0 & 1 & -1.5 & | & -5\end{bmatrix}\]

Step 6: Eliminate y from Second Row

To simplify the system further, eliminate \(y\) from the second row by subtracting \(0.9\) times the third row from the second row:- New Second Row = (Second Row) - \(0.9\times\) (Third Row)\[\begin{bmatrix}1 & -2 & 3 & | & 20 \0 & 0 & 0.25 & | & 2.5 \0 & 1 & -1.5 & | & -5\end{bmatrix}\]

Step 7: Solve for z

From the second row, solve for \(z\):- \(0.25z = 2.5\)- \(z = \frac{2.5}{0.25} = 10\)

Step 8: Back Substitute to Find y

Use \(z=10\) in the third row equation to solve for \(y\):- \(y - 1.5(10) = -5\)- \(y - 15 = -5\)- \(y = 10\)

Step 9: Back Substitute to Find x

Use \(y=10\) and \(z=10\) in the first row equation to solve for \(x\):- \(x - 2(10) + 3(10) = 20\)- \(x - 20 + 30 = 20\)- \(x + 10 = 20\)- \(x = 10\)

Step 10: Verify the Solution

Verify the solution by substituting \(x=10\), \(y=10\), and \(z=10\) back into the original equations to ensure consistency.

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool to solve systems of linear equations. It is a compact way of representing the coefficients and constant terms of a system of equations. By organizing these elements into rows and columns, it simplifies the process of identifying and applying operations needed to find the solutions.

The augmented matrix is created by writing down the coefficients of each variable in the equations, adding a vertical line, and then placing the constants on the right-hand side. Each row corresponds to an individual equation, while each column corresponds to a variable, except for the last column which contains the constants. For example, in the exercise, the augmented matrix looks like this:

• The coefficients of the variables:
  • First row: 0.1, -0.2, 0.3
  • Second row: 0.5, -0.1, 0.4
  • Third row: 0.7, -0.2, 0.3
• The constants: 2, 8, and 8, forming a separate column

Understanding how to form an augmented matrix is crucial as it sets the foundation for applying Gaussian elimination to solve the system effectively.

System of Linear Equations

A system of linear equations consists of multiple linear equations that are solved together to find a common solution. The primary aim is to determine the values of variables that satisfy all the given equations simultaneously.

These systems can be expressed in matrix form, like the augmented matrix, to simplify manipulation. Linear systems can have:

• A single solution, where the lines or planes represented by the equations intersect at one point
• No solution, when the lines or planes are parallel and never intersect
• Infinite solutions, when the equations correspond to the same line or plane

In the exercise provided, we dealt with three equations with three variables, forming a classic representation for a 3-dimensional linear system. By using Gaussian elimination, we efficiently streamline the process of rearranging the system into a simpler form that displays the solutions directly through back substitution.

Back Substitution

Back substitution is the final step in solving systems of equations using Gaussian elimination. Once the system is transformed into an upper triangular or row-echelon form through elimination steps, back substitution allows us to find the solutions for each variable.

This process involves starting from the last row of the matrix, where usually one variable is already isolated. By substituting the value of this variable into the preceding rows, another variable gets isolated. This process continues upward row by row, until all variables are solved.

In our exercise:

• The second to last row helps in finding the value of the variable \(z\).
• The calculated \(z\) is used in the previous row to solve for \(y\).
• Finally, both \(y\) and \(z\) are used in the top-most row to solve for \(x\).

Back substitution is straightforward as it iteratively builds upon previously found solutions, providing a complete solution set efficiently for the given system of equations.