Problem 15 - Differential Equations and Linear Algebra

Question

Use Gauss-Jordan elimination to determine the solution set to the given system. $$\begin{aligned} x_{1} &-2 x_{3}=-3 \\ 3 x_{1}-2 x_{2}-4 x_{3} &=-9 \\ x_{1}-4 x_{2}+2 x_{3} &=-3 \end{aligned}$$

Step-by-Step Solution

Answer

The general solution of the given system using Gauss-Jordan elimination is: \(\begin{aligned} x_{1} &= -3 + 2t \\ x_{2} &= t \\ x_{3} &= t \end{aligned}\) where \(t\) is any real number.

Step 1: Write the augmented matrix

Write down the augmented matrix for the given system of linear equations: \(A = \left[\begin{array}{ccc|c} 1 & 0 & -2 & -3 \\ 3 & -2 & -4 & -9 \\ 1 & -4 & 2 & -3 \end{array}\right]\)

Step 2: Perform row operations to convert the matrix into RREF

Perform the following row operations on the augmented matrix A: 1. Subtract 3 times Row 1 from Row 2, and name the new matrix B: \(B = \left[\begin{array}{ccc|c} 1 & 0 & -2 & -3 \\ 0 & -2 & 2 & 0 \\ 1 & -4 & 2 & -3 \end{array}\right]\) 2. Subtract Row 1 from Row 3 and name the new matrix C: \(C = \left[\begin{array}{ccc|c} 1 & 0 & -2 & -3 \\ 0 & -2 & 2 & 0 \\ 0 & -4 & 4 & 0 \end{array}\right]\) 3. Add 2 times Row 2 to Row 3 and name the new matrix D: \(D = \left[\begin{array}{ccc|c} 1 & 0 & -2 & -3 \\ 0 & -2 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\) 4. Divide Row 2 by -2 and name the new matrix E: \(E = \left[\begin{array}{ccc|c} 1 & 0 & -2 & -3 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\) Now, we have the RREF of the augmented matrix A.

Step 3: Interpret the RREF and write the general solution

The RREF of the augmented matrix is: \(\left[\begin{array}{ccc|c} 1 & 0 & -2 & -3 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\) It corresponds to the following system of linear equations: \(\begin{aligned} x_{1}-2 x_{3} &=-3\\ x_{2}-x_{3} &=0 \end{aligned}\) Notice that \(x_3\) is a free variable. Let \(x_3 = t\), where \(t\) is any real number. Express \(x_1\) and \(x_2\) in terms of \(t\): \(x_1 = -3 + 2t\) \(x_2 = t\) The general solution of the given system is: \(\begin{aligned} x_{1} &= -3 + 2t \\ x_{2} &= t \\ x_{3} &= t \end{aligned}\) where \(t\) is any real number.

Key Concepts

Linear Equations

Linear equations are mathematical statements that illustrate a relationship where two expressions are set equal, involving variables raised only to the first power. In the given exercise, we have the following system of linear equations:

• \(x_{1} - 2x_{3} = -3\)
• \(3x_{1} - 2x_{2} - 4x_{3} = -9\)
• \(x_{1} - 4x_{2} + 2x_{3} = -3\)

Each of these equations contains variables \(x_1\), \(x_2\), and \(x_3\), and a constant. Our goal is to find values of these variables that satisfy all equations simultaneously. They are called linear because they produce a straight line when graphed on a coordinate plane. Understanding linear equations is crucial as they form the foundation for solving systems of equations using various methods, like Gauss-Jordan elimination.

Augmented Matrix

An augmented matrix is derived from a system of linear equations and is an essential tool in linear algebra. It combines the coefficients of the variables and constant terms in a single matrix format. The augmented matrix allows us to apply matrix operations directly to solve the system. For our system of equations, the augmented matrix becomes:\(A = \left[\begin{array}{ccc|c}1 & 0 & -2 & -3 \3 & -2 & -4 & -9 \1 & -4 & 2 & -3\end{array}\right]\)The vertical bar separates the coefficients of the variables (to the left) from the constants (to the right). Working with an augmented matrix streamlines computations, making methods like Gauss-Jordan elimination more efficient and organized, particularly with more complex systems.

Reduced Row Echelon Form (RREF)

Reduced row echelon form (RREF) of a matrix is a form that is particularly useful because it simplifies identifying the solutions of a linear system. A matrix is in RREF when it meets the following criteria:

• Each leading entry (pivot) is 1.
• Each leading 1 is the only nonzero entry in its column.
• The leading 1 of a lower row is to the right of the leading 1s of any higher rows.
• Rows containing only zeros, if any, are at the bottom.

Applying row operations to our initial augmented matrix, we achieve its RREF:\[\left[\begin{array}{ccc|c}1 & 0 & -2 & -3 \0 & 1 & -1 & 0 \0 & 0 & 0 & 0\end{array}\right]\]This transformation simplifies the system, allowing us to easily read off the solution in terms of any free variables.

Free Variable

The concept of a free variable arises when a row in the RREF contains only zeros or when a variable does not have a leading entry (pivot) in every row. In our system, this occurs with the variable \(x_3\), which means it can take any real value. Free variables offer infinite possibilities for solutions in a system and reveal degrees of freedom. In our example, because \(x_3\) is free, we express the other variables \(x_1\) and \(x_2\) in terms of this confounding variable, making the system dependent on it. Recognizing free variables is crucial because they directly determine the structure of the general solution and indicate systems of linear equations that may have infinitely many solutions.

General Solution

The general solution to a system of equations like ours encompasses all possible solutions, usually expressed in terms of free variables. After converting the system to RREF, we can write down equations that link the variables:

• \(x_{1} = -3 + 2t\)
• \(x_{2} = t\)
• \(x_{3} = t\)

Here, \(t\) represents any real number, highlighting that \(x_3\) is a free variable. This indicates an infinite number of solutions where \(x_1\), \(x_2\), and \(x_3\) vary with changes in \(t\). Understanding the general solution not only helps in grasping how solutions are related within the system but also in appreciating the flexibility conferred by free variables in multivariable systems.