Problem 40 - Advanced Engineering Mathematics

Question

Use a CAS to solve the given system. \(x_{1}+2 x_{2}-2 x_{3}=0\) \(2 x_{1}-2 x_{2}+x_{3}=0\) \(3 x_{1}-6 x_{2}+4 x_{3}=0\) \(4 x_{1}+14 x_{2}-13 x_{3}=0\)

Step-by-Step Solution

Answer

The solution is parameterized: \(x_1 = \frac{2}{3}t\), \(x_2 = \frac{5}{6}t\), \(x_3 = t\) for \(t \in \mathbb{R}\).

Step 1: Set up matrix representation

Represent the system of equations as an augmented matrix. For this system:\[\begin{bmatrix}1 & 2 & -2 & | & 0 \2 & -2 & 1 & | & 0 \3 & -6 & 4 & | & 0 \4 & 14 & -13 & | & 0 \\end{bmatrix}\] This matrix captures the coefficients of each variable and the constants on the right-hand side of the equations.

Step 2: Use Gaussian Elimination

Apply row operations to transform the matrix into row-echelon form. The goal is to have leading ones and zeros below them:1. Swap Row 1 and Row 2 to bring a larger pivot.2. Zero elements below the first pivot (Row 1).3. Perform further row operations to get leading zeros below each pivot. The process should simplify down to a clear form.The row operations will yield:\[\begin{bmatrix}1 & 2 & -2 & | & 0 \0 & -6 & 5 & | & 0 \0 & 0 & 0 & | & 0 \0 & 18 & -17 & | & 0 \\end{bmatrix}\]

Step 3: Interpret the simplified system

From the row-echelon form, translate back to equations:1. The second row becomes \(0x_1 - 6x_2 + 5x_3 = 0\).2. The fourth row shows dependency, allowing for a free variable approach if needed.Since the third row (\(0 = 0\)) is dependent, this indicates that there will be free variables in the solution set.

Step 4: Solve for variables

Choose a free variable (e.g., let \(x_3 = t\)). Solve the non-zero rows for the other variables in terms of \(t\):1. Solve the second row for \(x_2\) in terms of \(t\):\(-6x_2 + 5t = 0 \implies x_2 = \frac{5}{6}t\).2. Substitute into the first row to find \(x_1\):\(x_1 + 2\left(\frac{5}{6}t\right) - 2t = 0\ \implies x_1 = \frac{2}{3}t\).Thus, the solution is \(x_1 = \frac{2}{3}t\), \(x_2 = \frac{5}{6}t\), \(x_3 = t\).

Step 5: Express the solution set

The solution can be written as a parameterized vector:\[\begin{bmatrix}x_1 \x_2 \x_3 \\end{bmatrix} = t \begin{bmatrix}\frac{2}{3} \\frac{5}{6} \1 \\end{bmatrix}, \text{ where } t \in \mathbb{R}\] This describes an infinite set of solutions along a line in 3D space with \(t\) as the parameter.

Key Concepts

Understanding Linear Algebra

Linear algebra deals with the study of vectors, vector spaces (also known as linear spaces), linear mappings, and systems of linear equations. In the context of this exercise, we're solving a system of linear equations. These equations are algebraic expressions that involve unknown variables and constants. Linear algebra offers various methods and tools, like Gaussian elimination, to find solutions for these systems. This branch of mathematics is not only essential in theoretical contexts but also has significant practical applications in fields like computer science, physics, and engineering. Learning linear algebra helps you understand concepts such as vector addition, scalar multiplication, and matrix operations, all of which are crucial for solving complex problems involving multiple interconnected elements.

What is an Augmented Matrix?

An augmented matrix is a compact and systematic way of writing a system of linear equations. It combines the coefficients of the variables and the constants from the equations into a single matrix. For example, consider the simple system:

• Equation 1: \( x_1 + 2x_2 = 3 \)
• Equation 2: \( 4x_1 - 5x_2 = 6 \)

The augmented matrix would be:\[ \begin{bmatrix} 1 & 2 & | & 3 \ 4 & -5 & | & 6 \end{bmatrix} \]Here, the vertical line separates the coefficients of the variables from the constants on the right side of the equations. Using this representation simplifies the application of Gaussian elimination, making it easier to manipulate the system during the solution process.

Free Variables and Their Role

In the context of systems of linear equations, free variables are those variables that are not leading variables in any of the row-reduced equations. They emerge when a system has more variables than independent equations. Free variables suggest that the system has infinitely many solutions. In a matrix reduced to row-echelon or reduced row-echelon form, these free variables can take any value, typically represented by a parameter like \( t \). In the exercise, \( x_3 \) is chosen as the free variable. This freedom allows us to express other variables \( x_1 \) and \( x_2 \) in terms of \( t \), revealing the parameterized form of solutions and indicating that these solutions form a line, plane, or even higher-dimensional space.

The Solution Set Explained

A solution set is the collection of all possible solutions that satisfy a system of equations. For the given system, after Gaussian elimination, the solution set can often be expressed in terms of one or more parameters. In this exercise, the solution set is represented using a parameter \( t \), showing all possible values \( x_1, x_2, \) and \( x_3 \) can take to satisfy all the equations simultaneously. The solution set:\[ \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = t \begin{bmatrix} \frac{2}{3} \ \frac{5}{6} \ 1 \end{bmatrix} \]represents an infinite number of solutions that form a straight line through the origin in the three-dimensional space. This happens because the system is dependent, involving free variables, which leads to parametric solutions rather than a single unique solution. Understanding the structure of the solution set is pivotal in grasping the nature of solutions in linear algebra problems.