Problem 29 - Precalculus: Mathematics for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x-2 y+z &=1 \\ y+2 z &=5 \\ x+y+3 z &=8 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The unique solution is \(x = 1\), \(y = 1\), \(z = 2\).

Step 1: Write the System in Augmented Matrix Form

Convert the system of equations into an augmented matrix.\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \1 & 1 & 3 & | & 8 \\end{bmatrix}\]

Step 2: Perform Row Operations

Subtract the first row from the third row to create a zero in the first column of the third row.\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \0 & 3 & 2 & | & 7 \\end{bmatrix}\]

Step 3: Simplify the Third Row

Divide the third row by 3 to simplify.\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \0 & 1 & \frac{2}{3} & | & \frac{7}{3} \\end{bmatrix}\]

Step 4: Eliminate Second Row First Column Entries

Subtract the third row from the second row to eliminate the entry in the second column of the second row.\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 0 & \frac{4}{3} & | & \frac{8}{3} \0 & 1 & \frac{2}{3} & | & \frac{7}{3} \\end{bmatrix}\]

Step 5: Simplify the Second Row

Multiply the second row by \(\frac{3}{4}\) to simplify.\[\begin{bmatrix}1 & -2 & 1 & | & 1 \0 & 0 & 1 & | & 2 \0 & 1 & \frac{2}{3} & | & \frac{7}{3} \\end{bmatrix}\]

Step 6: Back Substitution to Find Solutions

Use back substitution to find the values of the variables. Substituting from the bottom row upwards:- From Row 2: \(z = 2\)- Substituting \(z = 2\) in Row 3: \( y + \frac{2}{3} \times 2 = \frac{7}{3} \Rightarrow y = 1 \)- Substituting \(y = 1\) and \(z = 2\) in Row 1: \( x - 2 \times 1 + 2 = 1 \Rightarrow x = 1 \)

Step 7: Write the Solution

The solution to the system of equations is:\[ x = 1, \; y = 1, \; z = 2 \]

Key Concepts

Understanding Gauss-Jordan Elimination

Gauss-Jordan elimination is a method used to solve systems of linear equations. It's a variation of Gaussian elimination and simplifies the process by reducing an augmented matrix to reduced row-echelon form (RREF). The goal is to transform the matrix so that every leading coefficient is 1 and ensure that all other elements in its column are 0. This way, the corresponding system of equations will have a direct solution, visible from the matrix.

Gauss-Jordan elimination is particularly useful as it directly provides the solution without needing additional steps. In many cases, it streamlines the elimination of variables by automating the systematic process of row operations, making it easier to solve and understand complex systems of linear equations.

Exploring the Augmented Matrix

An augmented matrix is a powerful tool used in linear algebra to simplify the representation of a system of linear equations. It combines the coefficients of the variables and the constants from the equations into a single matrix, making it easier to manipulate the system.

Consider the system of equations given in the exercise:

• Equation 1: \(x - 2y + z = 1\)
• Equation 2: \(y + 2z = 5\)
• Equation 3: \(x + y + 3z = 8\)

Writing this as an augmented matrix looks like this: \[\begin{bmatrix} 1 & -2 & 1 & | & 1 \0 & 1 & 2 & | & 5 \1 & 1 & 3 & | & 8 \\end{bmatrix}\]
This matrix captures all the crucial information from the system and can be manipulated using row operations without returning to the original equations.

Back Substitution Made Simple

Back substitution is a method used towards the end of solving a system of linear equations, specifically after progressing through elimination techniques such as Gaussian elimination. This method involves solving the equations from the last one to the first, hence the term "back" substitution.

Starting with the system in triangular form (achieved after simplifying all rows), you solve for one variable at a time. For instance, if the reduced matrix yields equations in this form:

• \( z = 2 \)
• \( y + \frac{2}{3}z = \frac{7}{3} \)
• \( x - 2y + z = 1 \)

You would first solve for the simplest variable, \( z \), and use its known value in the equation for \( y \), and subsequently the values for \( z \) and \( y \) in the equation for \( x \).

This reverse, systematic approach allows for constant back-reference and guarantee that each variable is correctly determined.

Basics of a System of Linear Equations

A system of linear equations consists of two or more equations with the same set of variables. The objective is to find the values of these variables that satisfy all equations simultaneously.

In the example from the exercise, we have a system of three linear equations involving three variables \(x\), \(y\), and \(z\):

• \( x - 2y + z = 1 \)
• \( y + 2z = 5 \)
• \( x + y + 3z = 8 \)

Solving this system means finding values for \(x\), \(y\), and \(z\) such that these values satisfy all the given equations. This can be done using various methods, such as substitution, elimination, or matrix operations like Gaussian or Gauss-Jordan elimination. Each method has its advantages, often depending on the number of equations and variables involved.

Systems of linear equations are foundational concepts in algebra, greatly applied in engineering, physics, economics, and various scientific disciplines.