Problem 50 - A Graphical Approach to Precalculus with Limits

Question

Use row operations on an augmented matrix to solve each system of equations. Round to nearest thousandth when appropriate. $$\begin{array}{l} 3 x-2 y+z=2 \\ x+2 y-2 z=1 \\ x-4 y+2 z=-1 \end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{2}\), \(y = -\frac{1}{4}\), \(z = 0\).

Step 1: Write the Augmented Matrix

First, write the system of equations as an augmented matrix. The matrix will be:\[\begin{bmatrix}3 & -2 & 1 & | & 2 \1 & 2 & -2 & | & 1 \1 & -4 & 2 & | & -1 \end{bmatrix}\]

Step 2: Row Operations to Simplify

Perform row operations to simplify. Start by subtracting Row 2 from Row 1 to eliminate the first element of the second row:\[R_2 = R_2 - R_1 \]Now, apply the operation:\[\begin{bmatrix}3 & -2 & 1 & | & 2 \0 & 4 & -3 & | & -1 \1 & -4 & 2 & | & -1 \end{bmatrix}\]

Step 3: Continue Reducing Rows

Next, eliminate the first element of the third row by subtracting the third row from the first row:\[R_3 = R_3 - \frac{1}{3}R_1 \]This results in:\[\begin{bmatrix}3 & -2 & 1 & | & 2 \0 & 4 & -3 & | & -1 \0 & -\frac{10}{3} & \frac{5}{3} & | & -\frac{5}{3} \end{bmatrix}\]

Step 4: Simplify Third Row Further

Simplify the third row by attempting to clear out additional elements. Add \(\frac{5}{4}\) times Row 2 to Row 3 to ideally start the back-substitution phase:\[ R_3 = R_3 + \frac{5}{4}R_2\]Resulting:\[\begin{bmatrix}3 & -2 & 1 & | & 2 \0 & 4 & -3 & | & -1 \0 & 0 & \frac{1}{4} & | & 0 \end{bmatrix}\]

Step 5: Solve for Z

Row 3 allows us to easily solve for \(z\):\[\frac{1}{4}z = 0 \z = 0\]

Step 6: Substitute Z and Solve for Y

Substitute \(z = 0\) in Row 2 to solve for \(y\):\[4y - 3(0) = -1 \y = -\frac{1}{4}\]

Step 7: Substitute Y, Z and Solve for X

Substitute \(y = -\frac{1}{4}\) and \(z = 0\) in Row 1 to solve for \(x\):\[3x - 2\left(-\frac{1}{4}\right) + 0 = 2 \3x + \frac{1}{2} = 2 \3x = 2 - \frac{1}{2} \3x = \frac{3}{2} \x = \frac{1}{2}\]

Step 8: Solution

The solution to the system of equations is \(x = \frac{1}{2}\), \(y = -\frac{1}{4}\), and \(z = 0\).

Key Concepts

Row Operations

Row operations are mathematical procedures used to simplify matrices, especially augmented matrices which help solve systems of equations. When we write a system of equations as an augmented matrix, row operations become tools to transform it into a simpler form.
- **Types of Row Operations**: There are three fundamental operations you can perform on the rows of a matrix:

• Swapping two rows.
• Multiplying a row by a nonzero scalar.
• Adding or subtracting the multiples of one row to another.

These operations help us achieve an easier-to-handle form called row-echelon form or even further to the reduced row-echelon form, bringing us closer to solving for variables like "x," "y," and "z." In this exercise, they are used to eliminate variables systematically from the equations.

Row operations are key to simplifying the process, ensuring each step moves towards isolating the variables and finding the solution.

System of Equations

A system of equations consists of multiple equations that use the same set of variables. The goal is to find values for these variables that satisfy all equations in the system simultaneously.

In a system like
\(\begin{align*}3x - 2y + z &= 2 \x + 2y - 2z &= 1 \x - 4y + 2z &= -1 \end{align*}\),
each equation restricts the potential values of "x," "y," and "z". The solution to the system must be a single set of values that satisfies all these constraints.

Using Augmented Matrices

An augmented matrix is a compact form of writing down systems of equations, separating the coefficients of variables from the constants. Through this method:

• Variables are aligned on one side.
• Constants are aligned on the right, separated by a vertical bar.

This setup allows for a systematic approach using row operations to navigate through the system to find the solution, which in our example ended up being \(x = \frac{1}{2}, y = -\frac{1}{4}, z = 0\).

Back-Substitution

Back-substitution is a process used particularly with row-echelon form, where you solve for the variables starting from the bottom row of an upper-triangular matrix.
In the given system, once we simplified via row operations to: \[ \begin{bmatrix} 3 & -2 & 1 & | & 2 \ 0 & 4 & -3 & | & -1 \ 0 & 0 & \frac{1}{4} & | & 0 \end{bmatrix}\], the state of the matrix allowed for back-substitution:

• From the last row, \(\frac{1}{4}z = 0\) gives \(z = 0\).
• Plug \(z = 0\) into the second row, solving \(4y = -1\), yields \(y = -\frac{1}{4}\).
• Substitute \(y\) and \(z\) into the first row, and solve for \(x\): \(3x + \frac{1}{2} = 2\), leading to \(x = \frac{1}{2}\).

This process is methodical and essential for transforming a simplified matrix back into its variable solutions.

Linear Algebra

Linear algebra is the branch of mathematics that deals with vector spaces, matrices, and solving systems of linear equations. Central to linear algebra is the study of lines, planes, and subspaces but it also extends into transformations and solving more complex real-world problems.

This subject provides the tools to:

• Achieve systematic solutions to linear equations through matrices.
• Understand complex systems by converting them into manageable forms.
• Analyze relationships and dependencies between equations and their respective variables.

The problem we're examining showcases all these elements effectively, as using the augmented matrix effectively applies linear algebra concepts. Working through this process of row operations and back-substitution not only provides the solution to the given system but also deepens one's understanding of manipulating systems within the linear algebra framework.