Problem 10 - Differential Equations and Linear Algebra

Question

Use Gaussian elimination to determine the solution set to the given system. $$\begin{aligned} x_{1}+2 x_{2}-x_{3}+x_{4} &=1 \\ 2 x_{1}+4 x_{2}-2 x_{3}+2 x_{4} &=2 \\ 5 x_{1}+10 x_{2}-5 x_{3}+5 x_{4} &=5 \end{aligned}$$

Step-by-Step Solution

Answer

The general solution of the given system can be written as: \[\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 1 - r + s - 2t \\ t \\ s \\ r \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + t \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + s \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + r \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\] The solution set can have infinitely many solutions for \(x_1, x_2, x_3\), and \(x_4\) depending on the values of t, s, and r.

Step 1: Write the augmented matrix

To begin the Gaussian elimination, we will first write the given system as an augmented matrix. The augmented matrix for the given system will be: \[\left[\begin{array}{cccc|c} 1 & 2 & -1 & 1 & 1 \\ 2 & 4 & -2 & 2 & 2 \\ 5 & 10 & -5 & 5 & 5 \end{array}\right]\]

Step 2: Start with the first pivot

Apply the row operations to make the elements below the diagonal equal to zero. Here, we will use the element in the first row and the first column as the pivot. R2 = R2 - 2 * R1: \[\left[\begin{array}{cccc|c} 1 & 2 & -1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 5 & 10 & -5 & 5 & 5 \end{array}\right]\] R3 = R3 - 5 * R1: \[\left[\begin{array}{cccc|c} 1 & 2 & -1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

Step 3: Analyze the resulting matrix

After performing the Gaussian elimination, we are left with this augmented matrix: \[\left[\begin{array}{cccc|c} 1 & 2 & -1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\] From the above matrix, we can generate the linear equation as: \[x_1 + 2x_2 - x_3 + x_4 = 1\] Since we have only one equation and three variables (x_2, x_3, x_4), there can be infinitely many solutions. Therefore, we will express the variables x_2, x_3, and x_4 in terms of a parameter.

Step 4: Express free variables in terms of a parameter

Let's express the free variables (x_2, x_3, and x_4) in terms of a parameter t: \[x_2 = t\] Now, express x_3 and x_4 in terms of another parameter, say s: \[x_3 = s\] From the first equation, we have: \[x_1 = 1 - x_4 + x_3 - 2x_2\] Substitute the values of x_2, x_3, and x_4 in terms of t and s: \[x_1 = 1 - x_4 + x_3 - 2t\] \[x_4 = x_3 - 1 + 2t\] Since x_4 is also a free variable, let's express it in terms of another parameter, say r: \[x_4 = r\]

Step 5: Find the general solution

Now we have the expressions for all four variables in terms of the parameters t, s, and r: \[x_1 = 1 - r + s - 2t\] \[x_2 = t\] \[x_3 = s\] \[x_4 = r\] The general solution of the given system can be written as: \[\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 1 - r + s - 2t \\ t \\ s \\ r \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + t \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + s \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + r \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\] The solution set can have infinitely many solutions for x_1, x_2, x_3, and x_4 depending on the values of t, s, and r.

Key Concepts

System of Linear Equations

Understanding a system of linear equations is crucial for tackling many problems in algebra and applies widely in fields such as economics, physics, and engineering. A system of linear equations consists of two or more equations involving the same set of variables. For example, the equations provided illustrate a system with four variables, \( x_1, x_2, x_3, x_4 \).

When solving such a system, the goal is to find the values of the variables that satisfy all the equations simultaneously. In some cases, the system may have a single unique solution, but it can also have infinitely many solutions or no solution at all. The provided exercise is an example where, after applying Gaussian elimination, we end up realizing that the system does not have a unique solution but rather infinitely many.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations by including the coefficients of the variables along with the constants from the right-hand side of the equations within a single matrix. The vertical bar divides the coefficients from the constants, which helps to maintain clarity when performing row operations.

For the given problem, the augmented matrix dramatically simplifies the visualization of the system and provides a pathway to manipulation using row operations. It also helps to quickly identify the number of equations and the number of unknowns in the system, making it easier to determine the potential number of solutions.

Row Operations

Row operations are transformations that can be applied to an augmented matrix without changing the solutions to the system it represents. These operations include swapping rows, multiplying a row by a nonzero scalar, and adding multiples of one row to another row.

In the Gaussian elimination process, row operations are used to obtain zero coefficients below (and sometimes above) the leading coefficients (pivots) in each column, aiming to reach a form where the solution to the system is evident. In the example provided, row operations revealed that two equations in the system were actually multiples of the first, reducing the system to a single essential equation.

Free Variables

Free variables refer to variables in a system of linear equations that can take on any value; they are not leading variables (those with a leading coefficient in any row when the augmented matrix is in row-echelon form). In our exercise, after using Gaussian elimination, the variables \(x_2, x_3, x_4\) emerge as free variables.

This means that they are not bound to specific values and can be set to arbitrary parameters, which is essential for expressing the general solution of the system. By assigning parameters to free variables, we demonstrate the infinite nature of the solution set.

General Solution

The general solution of a system of linear equations is an expression that captures all possible solutions. It incorporates the free variables, represented by parameters, to describe an infinite set of solutions. When a system has more free variables than equations, as in our example, there is an infinite number of ways to satisfy the given equations.

In the final step of the provided exercise, the general solution was expressed as a vector equation that linearly combines the constant solution vector with the vectors that represent free variables. Each parameter corresponds to one degree of freedom in the solution space, allowing the representation of an infinite set of solutions. This is an elegant way to comprehend and communicate the multiplicity of solutions in such a system.