Problem 44 - College Algebra

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{l} \frac{1}{4} x-\frac{2}{3} z=-\frac{1}{2} \\ \frac{1}{5} x+\frac{1}{3} y=\frac{4}{7} \\ \frac{1}{5} y-\frac{1}{3} z=\frac{2}{9} \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = 2\), \(y = \frac{5}{14}\), \(z = \frac{3}{2}\).

Step 1: Convert system to augmented matrix

First, represent the given system of equations as an augmented matrix. The coefficients matrix and the constants matrix are represented together.\[\begin{bmatrix}\frac{1}{4} & 0 & -\frac{2}{3} & | & -\frac{1}{2} \\frac{1}{5} & \frac{1}{3} & 0 & | & \frac{4}{7} \0 & \frac{1}{5} & -\frac{1}{3} & | & \frac{2}{9}\end{bmatrix}\]

Step 2: Clear first column entries below pivot

Use row operations to make the elements below the first pivot (in the first column) equal to zero. The first pivot is \(\frac{1}{4}\).Multiply row 1 by \(5\) and subtract from row 2 to eliminate \(x\) from the second equation:\[\begin{bmatrix}\frac{1}{4} & 0 & -\frac{2}{3} & | & -\frac{1}{2} \0 & \frac{1}{3} & \frac{2}{3} & | & \frac{25}{14} \0 & \frac{1}{5} & -\frac{1}{3} & | & \frac{2}{9}\end{bmatrix}\]

Step 3: Clear second column entries below pivot

Next, eliminate the element below the second pivot (in the second column). The second pivot is \(\frac{1}{3}\) from row 2.Multiply row 2 by \(\frac{5}{3}\) and subtract row 3 from it to eliminate \(y\) from row 3:\[\begin{bmatrix}\frac{1}{4} & 0 & -\frac{2}{3} & | & -\frac{1}{2} \0 & \frac{1}{3} & \frac{2}{3} & | & \frac{25}{14} \0 & 0 & \frac{1}{3} & | & \frac{1}{2}\end{bmatrix}\]

Step 4: Solve for variables starting from the last row

Begin by solving the third equation for \(z\):\[\frac{1}{3}z = \frac{1}{2}\]So, \(z = \frac{3}{2}\).Next, substitute \(z\) into the second equation:\[\frac{1}{3}y + \frac{2}{3}\frac{3}{2} = \frac{25}{14}\]\(\Rightarrow y = \frac{5}{14}\).Finally, substitute \(z\) into the first equation and solve for \(x\):\[\frac{1}{4}x - \frac{2}{3}\frac{3}{2} = -\frac{1}{2}\]\(\Rightarrow x = 2\).

Step 5: Conclude with solution

Having solved the system, conclude that the solution is:\(x = 2\), \(y = \frac{5}{14}\), \(z = \frac{3}{2}\).

Key Concepts

Augmented Matrix

An augmented matrix is a crucial layout used in linear algebra to help solve systems of equations. It simplifies the solution process by combining both the coefficients of the variables and the constants from a system of linear equations into one comprehensive matrix. This format is a compact and efficient way to organize the system.
- Each row of the augmented matrix represents one of the equations. - The columns before the vertical line represent the coefficients of the variables. - The column after the vertical line consists of the constants from the right-hand sides of the equations.
For example, with the system in the exercise, we convert it into an augmented matrix as follows:\[\begin{bmatrix}\frac{1}{4} & 0 & -\frac{2}{3} & | & -\frac{1}{2} \\frac{1}{5} & \frac{1}{3} & 0 & | & \frac{4}{7} \0 & \frac{1}{5} & -\frac{1}{3} & | & \frac{2}{9}\end{bmatrix}\]This matrix helps us to apply row operations systematically to solve the system.

Pivot Element

The pivot element is a foundational concept in Gaussian elimination, used to streamline the process of solving systems of equations by transforming the matrix into a row-echelon form. Essentially, the pivot is the leading non-zero entry in each row of a matrix after performing row operations.
- It guides the elimination of other entries below it in the same column, turning them into zeros.- Choosing the right pivot is crucial as it ensures stability and simplifies computations.
In our example, the pivot for the first step is \(\frac{1}{4}\), which is the first non-zero element in the first row. By choosing the pivot carefully and applying appropriate row operations, we can clear the column below it, simplifying the system progressively towards a solution.

Row Operations

Row operations are the manipulative tools used to alter the rows of a matrix without changing the solution set of the corresponding system of equations. In Gaussian elimination, these operations help in transforming the matrix into a form that is easier to solve. There are three primary row operations: - **Swap two rows:** Sometimes needed to bring a desired pivot into position. - **Multiply a row by a non-zero scalar:** Useful for adjusting a pivot or scaling equations. - **Add or subtract a multiple of one row to another:** This is the most common operation for eliminating entries below pivot elements. In the exercise, these operations are essential. For instance: - The second row was manipulated by multiplying a row by 5 and subtracting from another to clear an element. - Similarly, another step involved a mix of multiplying and subtracting to create zeros below pivot points. These operations efficiently bring the augmented matrix into row-echelon form, from which the solution can be easily read.