Problem 45 - 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. $$\begin{aligned} -x+y &=-1 \\ y-z &=6 \\ x+z &=-1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 3 \), \( y = 2 \), \( z = -4 \).

Step 1: Write the augmented matrix

The given system of equations can be written in matrix form as an augmented matrix. Write each equation as a row in the matrix: \[\begin{bmatrix} -1 & 1 & 0 & | & -1 \ 0 & 1 & -1 & | & 6 \ 1 & 0 & 1 & | & -1 \end{bmatrix} \]

Step 2: Apply row operations to simplify

Start by making the first element of the first column a leading 1. Swap Row 1 and Row 3:\[\begin{bmatrix} 1 & 0 & 1 & | & -1 \ 0 & 1 & -1 & | & 6 \ -1 & 1 & 0 & | & -1 \\end{bmatrix} \]

Step 3: Eliminate -1 in Row 3, Column 1

Add Row 1 to Row 3 to eliminate the -1 in the first column of Row 3:\[\begin{bmatrix} 1 & 0 & 1 & | & -1 \ 0 & 1 & -1 & | & 6 \ 0 & 1 & 1 & | & -2 \\end{bmatrix}\]

Step 4: Subtract Row 2 from Row 3

Subtract Row 2 from Row 3 to eliminate the second column's 1 in Row 3:\[\begin{bmatrix} 1 & 0 & 1 & | & -1 \ 0 & 1 & -1 & | & 6 \ 0 & 0 & 2 & | & -8 \\end{bmatrix}\]

Step 5: Solve for z

In the third row, divide by 2 to solve for z:\[\begin{bmatrix} 1 & 0 & 1 & | & -1 \ 0 & 1 & -1 & | & 6 \ 0 & 0 & 1 & | & -4 \\end{bmatrix}\]So, \( z = -4 \).

Step 6: Substitute z into Row 2 to solve for y

Substitute \( z = -4 \) into Row 2: \[ 0y + 1(-4) = 6 \]Thus, \( y - (-4) = 6 \rightarrow y = 2 \).

Step 7: Substitute z into Row 1 to solve for x

Substitute \( z = -4 \) into Row 1:\[ x + (-4) = -1 \]Thus, \( x - 4 = -1 \rightarrow x = 3 \).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. In our exercise, we are dealing with three equations involving three variables: \(x\), \(y\), and \(z\). The primary goal is to find the values of these variables that satisfy all the equations simultaneously.
Understanding and solving a system of equations can help you model and solve real-world problems. They are foundational in algebra and essential in fields like engineering and physics.
In our given system:

• Equation 1: \(-x + y = -1\)
• Equation 2: \(y - z = 6\)
• Equation 3: \(x + z = -1\)

We are tasked with finding values for \(x\), \(y\), and \(z\) that make each equation true.

Row Operations

Row operations are the key to simplifying an augmented matrix and solving systems of equations. With row operations, you can add, subtract, multiply, or swap entire rows in a matrix to transform it into a more manageable form.
These operations are fundamental as they help in reaching the row echelon form or even reduced row echelon form, making the solution straightforward.
In the exercise you have:

• Swapping rows to position a 1 as the leading entry: Swap Row 1 with Row 3.
• Adding or subtracting rows to eliminate variables: Add Row 1 to Row 3, then subtract Row 2 from Row 3.
• Multiplying a row by a scalar to solve for variables: Divide Row 3 by 2 to find \(z\).

These steps systematically help in simplifying the matrix to uncover solutions to our variables.

Matrix Form

The conversion of a system of equations into a matrix is known as matrix form representation. An augmented matrix includes both the coefficients of the variables and the constants from the equations.
This structured format sets the stage for applying row operations.
For our exercise, the system of equations translates to the following augmented matrix:

\[\begin{bmatrix} -1 & 1 & 0 & | & -1 \ 0 & 1 & -1 & | & 6 \ 1 & 0 & 1 & | & -1 \end{bmatrix}\]
The matrix form allows for more efficient handling and computation, especially with larger systems, using Gaussian elimination or computer algorithms.

Gaussian Elimination

Gaussian elimination is a methodical procedure used to solve systems of equations. The goal is to transform the augmented matrix into an upper triangular form or row echelon form using row operations.
The process includes three main steps:

• Make all entries below the leading coefficient of each row zero.
• Adjust equations (rows) until you can start back-substituting to solve variables.
• Optionally, further simplify to reduced row echelon form for direct reading of variable solutions.

In our exercise, Gaussian elimination helped through:

• Positioning the 1s in diagonal fashion where possible (leading ones).
• Zeroing out elements below these leading 1s through suitable row operations.

This systematic technique ensures you can find accurate solutions for the variables \(x\), \(y\), and \(z\).