Problem 14 - Contemporary Precalculus

Question

In Exercises \(13-16,\) use Gaussian elimination to solve the system. $$\begin{aligned} 2 x-3 y+2 z &=8 \\ 3 x+2 y+z &=3 \\ x+2 y+3 z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

Answer: The values of x, y, and z are \(x = \frac{369}{34}, y = \frac{17}{26}, z = -\frac{9}{17}\).

Step 1: Set up the augmented matrix

Write the given system of linear equations as an augmented matrix. An augmented matrix combines the coefficients and constants from the linear system of equations. $$\left[\begin{array}{ccc|c} 2 & -3 & 2 & 8 \\ 3 & 2 & 1 & 3 \\ 1 & 2 & 3 & 1 \end{array}\right]$$

Step 2: Row reduction

Perform row operations to get a Row Echelon Form (REF) following Gaussian elimination. First, use Row 1 to eliminate the x-terms in Rows 2 and 3. Add \(-\frac{3}{2}\) times Row 1 to Row 2: $$\left[\begin{array}{ccc|c} 2 & -3 & 2 & 8 \\ 0 & 6.5 & -2 & -9 \\ 1 & 2 & 3 & 1 \end{array}\right]$$ Subtract \(\frac{1}{2}\) times Row 1 from Row 3: $$\left[\begin{array}{ccc|c} 2 & -3 & 2 & 8 \\ 0 & 6.5 & -2 & -9 \\ 0 & 3.5 & 4 & -3 \end{array}\right]$$

Step 3: Row reduction (continued)

Next, use Row 2 to eliminate the y-terms in Rows 3. Add \(\frac{-3.5}{6.5}\) times Row 2 to Row 3: $$\left[\begin{array}{ccc|c} 2 & -3 & 2 & 8 \\ 0 & 6.5 & -2 & -9 \\ 0 & 0 & \frac{34}{13} & -\frac{18}{13} \end{array}\right]$$

Step 4: Solve for the variables

Now that the matrix is in Row Echelon Form (REF), we can easily solve for the variables using back-substitution. From the third row, we have \(\frac{34}{13}z = -\frac{18}{13}\). Dividing by \(\frac{34}{13}\), we get $$z = -\frac{18}{34} = -\frac{9}{17}.$$ From the second row, we have \(6.5y - 2(-\frac{9}{17}) = -9\). Substituting \(z\) value and solving for \(y\), we get: $$y = \frac{17}{26}$$ From the first row, we have \(2x - 3(\frac{17}{26}) + 2(-\frac{9}{17}) = 8\).Substituting \(y\) and \(z\) values and solving for \(x\), we get: $$x = \frac{369}{34}$$ Thus, the solution to the system is \(x = \frac{369}{34}, y = \frac{17}{26}, z = -\frac{9}{17}\).

Key Concepts

Augmented Matrix

An augmented matrix is an essential tool when solving systems of linear equations using methods like Gaussian elimination. This matrix is a simple way to organize the coefficients of the variables and the constants from each equation in the system.
Imagine you have a system of linear equations like the one in the exercise. You write down the coefficients of each variable in a row, and at the end, you include the constants. This forms a rectangular array of numbers called an augmented matrix.
In the exercise, the system of equations is represented by:

• First row: Coefficients are from the first equation.
• Second row: Coefficients come from the second.
• Third row: Coefficients are taken from the third line.

An augmented matrix helps visually organize your work and sets the stage for the steps of Gaussian elimination.

Row Operations

Row operations are the backbone of Gaussian elimination. They help transform the augmented matrix into a format that's easier to solve. There are three main types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero constant.
• Adding or subtracting a multiple of one row to another row.

These operations aim to simplify the system by creating zeros below the pivots (leading coefficients), forcing the matrix into a Row Echelon Form (REF).
In the exercise, row operations were used to eliminate the x- and y-terms, like modifying Row 2 and Row 3 to create zeros beneath the pivot in Row 1. It helps in simplifying further calculations.

Back-Substitution

Once you have transformed your augmented matrix into a Row Echelon Form, it's time for back-substitution. This process helps to find the values of the variables step by step.
Back-substitution generally starts from the last row, where typically one variable is isolated (like z in the example). You then sequentially substitute this value back into preceding rows to solve for other variables.
In the provided solution, starting from the last row, we solve for z first. Then, moving upwards, substituting the known value of z into the second row solves for y. Lastly, substituting both y and z into the first row allows us to find x. Back-substitution is the final step to retrieving the specific solutions from the simplified matrix.

Systems of Linear Equations

Systems of linear equations involve multiple equations that share a set of variables. These systems can have unique solutions, no solutions, or infinitely many solutions.
Each equation represents a hyperplane in n-dimensions, and the solution represents the intersection point of all these hyperplanes.
Gaussian elimination is one effective method to find these intersection points. By manipulating the augmented matrix of the system into an easier form, we can navigate the complexities of multiple equations more seamlessly.
In the exercise, the system involved three equations and three unknowns. Successfully applying Gaussian elimination gave the exact meeting point of these equations as the values of x, y, and z, solving the entire system. Understanding systems of linear equations can unlock numerous mathematical and real-world applications.