Problem 14 - Differential Equations and Linear Algebra

Question

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

Step-by-Step Solution

Answer

The solution set to the given system using Gauss-Jordan elimination is \(\{(-1, 3, 2)\}\), with \(x_{1}=-1, x_{2}=3, x_{3}=2\).

Step 1: Write down the augmented matrix

First, let's write down the augmented matrix for the given system of equations: \[ \begin{bmatrix} 3 & 1 & 5 & \vline & 2 \\ 1 & 1 & -1 & \vline & 1 \\ 2 & 1 & 2 & \vline & 3 \end{bmatrix} \]

Step 2: Apply row operations to get rid of elements under the pivot

We will start with the first pivot, which is the \(3\) in the top left corner of the matrix. We will apply row operations to get rid of all elements in the first column below the pivot: R2 = R2 - (1/3)R1 and R3 = R3 - (2/3)R1: \[ \begin{bmatrix} 3 & 1 & 5 & \vline & 2 \\ 0 & 2/3 & -8/3 & \vline & 1/3 \\ 0 & 1/3 & -1/3 & \vline & 5/3 \end{bmatrix} \]

Step 3: Apply row operations to get zeros below the next pivot

Now, we will focus on the second pivot, which is the \(2/3\) in the second row. We want to get a zero in the second column below the pivot: R3 = R3 - (1/2)R2: \[ \begin{bmatrix} 3 & 1 & 5 & \vline & 2 \\ 0 & 2/3 & -8/3 & \vline & 1/3 \\ 0 & 0 & 1 & \vline & 2 \end{bmatrix} \]

Step 4: Normalize the pivot rows

Now, we will normalize the pivot rows (or make the pivot coefficients equal to 1) by dividing each row by its respective pivot coefficient: R1 = R1/3, R2 = (3/2)R2, and R3 is already normalized: \[ \begin{bmatrix} 1 & 1/3 & 5/3 & \vline & 2/3 \\ 0 & 1 & -4 & \vline & 1/2 \\ 0 & 0 & 1 & \vline & 2 \end{bmatrix} \]

Step 5: Apply row operations to obtain the reduced row echelon form of the matrix

Finally, we will apply row operations to obtain the reduced row echelon form of the matrix: R1 = R1 - (5/3)R3 and R2 = R2 + 4R3: \[ \begin{bmatrix} 1 & 1/3 & 0 & \vline & -2 \\ 0 & 1 & 0 & \vline & 3 \\ 0 & 0 & 1 & \vline & 2 \end{bmatrix} \] And R1 = R1 - (1/3)R2: \[ \begin{bmatrix} 1 & 0 & 0 & \vline & -1 \\ 0 & 1 & 0 & \vline & 3 \\ 0 & 0 & 1 & \vline & 2 \end{bmatrix} \]

Step 6: Write the solution set

Now, the matrix is in reduced row echelon form, and we can read off the solution set: \(x_{1}=-1, x_{2}=3, x_{3}=2\) So, the solution set is \(\{(-1, 3, 2)\}\).

Key Concepts

Augmented Matrix

The concept of an augmented matrix is foundational in understanding how to work with systems of linear equations using the Gauss-Jordan elimination method. An augmented matrix is a compact and efficient way to represent a system of equations. It combines the coefficients of the variables with the constants from the equations.
Consider the system of equations provided:

• 3x₁ + x₂ + 5x₃ = 2
• x₁ + x₂ - x₃ = 1
• 2x₁ + x₂ + 2x₃ = 3

These can be represented by the following augmented matrix:\[\begin{bmatrix}3 & 1 & 5 & \vline & 2 \1 & 1 & -1 & \vline & 1 \2 & 1 & 2 & \vline & 3\end{bmatrix}\]
In this matrix, each row corresponds to one equation, while each column before the vertical line corresponds to the coefficients of the variables. The column after the vertical line contains the constants from the right-hand side of each equation. Augmented matrices are particularly useful for applying systematic row operations, which are the key steps in transforming the matrix to solve the system of equations.

Row Operations

Row operations are the backbone of Gauss-Jordan elimination, a process used to simplify matrices to solve systems of linear equations. There are three types of row operations that can be applied:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row to another row

By strategically applying these operations, we can manipulate the matrix to reach reduced row echelon form (RREF). Let's see how row operations function by looking at our example, starting from the first matrix provided:\[\begin{bmatrix}3 & 1 & 5 & \vline & 2 \1 & 1 & -1 & \vline & 1 \2 & 1 & 2 & \vline & 3\end{bmatrix}\]
The goal is to create zeros below each pivot (leading 1) while maintaining equivalent equations. The operations we apply consecutively aim to clear non-zero elements under the pivots in each column. For example, by using row subtraction or addition, one can create zeros below the main diagonal to simplify the matrix further.

Reduced Row Echelon Form

Reduced row echelon form (RREF) is a specific form of a matrix from which the solution of a linear system can be easily determined. A matrix is in RREF if:

• Each leading entry in a row is 1
• Leading 1's are the only non-zero entries in their column
• The leading 1 of any row is to the right of the leading 1 of the row above it
• Any rows containing only zeros are at the bottom of the matrix

Achieving this form allows for direct reading off of solutions from the matrix. For example, for the problem at hand, after a series of row operations, we reduced the matrix to:\[\begin{bmatrix}1 & 0 & 0 & \vline & -1 \0 & 1 & 0 & \vline & 3 \0 & 0 & 1 & \vline & 2\end{bmatrix}\]
This matrix is in RREF, where each pivot is a 1, and all elements above and below each pivot are zeros. This greatly simplifies the task of determining individual solutions associated with each variable.

Solution Set of Linear Equations

The ultimate goal of Gauss-Jordan elimination is to find the solution set of the linear equations, which contains all possible solutions of the system. Once a matrix is in reduced row echelon form, it’s straightforward to extract the solutions.
In the example, the RREF matrix:\[\begin{bmatrix}1 & 0 & 0 & \vline & -1 \0 & 1 & 0 & \vline & 3 \0 & 0 & 1 & \vline & 2\end{bmatrix}\]
corresponds to the equations:

• x₁ = -1
• x₂ = 3
• x₃ = 2

Thus, the solution set is expressed as a set of coordinates: \(\{(-1, 3, 2)\}\).
This set contains just one point, which indicates that the system has a unique solution. Gauss-Jordan elimination simplifies determining such solutions by transforming the matrix into a form where the answer is directly visible, eliminating guesswork and ensuring accuracy.