Problem 45 - College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Write the system of equations as an augmented matrix

First, express the system of equations as an augmented matrix. The given system is:\[\begin{array}{l}{0.8x + 0.8y + 0.8z = 2.4} \{0.3x - 0.5y + 0.2z = 0} \{0.1x + 0.2y + 0.3z = 0.6}\end{array}\]The augmented matrix representation is:\[\begin{bmatrix}0.8 & 0.8 & 0.8 & | & 2.4 \0.3 & -0.5 & 0.2 & | & 0 \0.1 & 0.2 & 0.3 & | & 0.6\end{bmatrix}\]

Step 2: Use row operations to get a leading 1 in the first row, first column

Divide the entire first row by 0.8 to simplify the matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0.3 & -0.5 & 0.2 & | & 0 \0.1 & 0.2 & 0.3 & | & 0.6\end{bmatrix}\]

Step 3: Eliminate the x-coefficient from the second and third rows

Subtract 0.3 times the first row from the second row and 0.1 times the first row from the third row:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & -0.8 & -0.1 & | & -0.9 \0 & 0.1 & 0.2 & | & 0.3\end{bmatrix}\]

Step 4: Simplify the second row for further elimination

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

Step 5: Eliminate the y-coefficient from the third row

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

Step 6: Solve for z, then back substitute to find y and x

From the third row, solve for z: \[0.1875z = 0.1875 \rightarrow z = 1\]Substitute \(z = 1\) into the second row: \[y + 0.125(1) = 1.125 \rightarrow y = 1\]Substitute \(y = 1\) and \(z = 1\) into the first row:\[x + 1 + 1 = 3 \rightarrow x = 1\]The solution is \(x = 1\), \(y = 1\), \(z = 1\).

Step 7: Conclusion

The solution to the system of equations is \(x = 1\), \(y = 1\), and \(z = 1\).

Key Concepts

Augmented Matrix

When solving a system of linear equations using Gaussian elimination, an important first step is to write the system as an augmented matrix. An augmented matrix is a compact way to represent a system of equations by displaying only the coefficients and constant terms. Instead of rewriting all the variables repeatedly, the augmented matrix puts the coefficients into a rectangular grid-like format, which is easier to manipulate during further calculations.

To transform a system of equations into an augmented matrix, align each equation in the same way and extract the coefficients into a matrix. Below each coefficient goes into its column corresponding to its variable, and the last column shows the constants. For example, for the given system:

• The coefficients of the equations are written in rows.
• Each variable's coefficients make up a column.
• Constants from the right side of the equations are put in the final column of the matrix, separated by a vertical line as a visual aid (though not mathematically necessary).

This compact representation allows for easy application of various matrix operations to solve the equations.

Row Operations

Row operations are crucial in simplifying augmented matrices to solve systems of equations. They help us manipulate the matrix into a more easily solvable form called row-echelon form. There are three types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting multiples of one row from another row to create zeros below pivot positions (leading coefficients).

By following a sequence of row operations, you gradually develop leading 1s down the diagonal of the matrix and zeros below these numbers. This systematic approach simplifies your task of solving the system of equations. In our solution scenario, row operations efficiently created leading coefficients and sequentially eliminated variables below these positions.

The result is an easier, more consistent path towards obtaining the solution with fewer operations. When done correctly, row operations don't alter the solution set but simplify finding it.

Back Substitution

Once the matrix reaches row-echelon form, back substitution becomes the tool to find the solutions for the variables. Row-echelon form brings the matrix into a shape where you can easily pinpoint specific values for each variable, starting from the bottommost row.

In principle:

• First, solve the bottom row for its corresponding variable.
• Use this solution to substitute into the row above to solve for the next variable.
• Continue this pattern upwards until all variables have been resolved.

For example, if the last row of your simplified matrix is in the form \[ az = c \] then solve for \( z \). Once \( z \) is found, substitute this value into the equation from the next upper row to find \( y \). Repeat this step for the top row to determine \( x \).

Back substitution allows a step-by-step unraveling of each variable's value by leveraging the simplified relationships between them, ensuring clarity and accuracy.

System of Equations

A system of equations is simply a collection of two or more equations that are solved together. Each equation represents a condition or relationship between several variables. Solving these systems means finding values for the variables that simultaneously satisfy all the equations.

• Linear systems have only straight-line equations, and they are frequently represented in matrix form to facilitate easy manipulation.
• The equations usually involve finding an intersection point where all the equations hold true together.

Gaussian elimination is a widely-used method for solving systems of equations because it provides a systematic approach by simplifying the system into row-echelon form before utilizing back substitution.

In essence, converting the system into an augmented matrix, applying row operations to simplify it, and using back substitution to find the solutions, represents a robust strategy to tackle these systems. Efficiently solving systems of equations plays prominent roles across many fields, including physics, engineering, computer science, and economics, where they model real-world scenarios for better decision-making.