Problem 51 - Differential Equations and Linear Algebra

Question

Determine the solution set to the system \(A \mathbf{x}=0\) for the given matrix \(A\). $$A=\left[\begin{array}{rrrr} 1 & 1 & 1 & -1 \\ -1 & 0 & -1 & 2 \\ 1 & 3 & 2 & 2 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution set for the given system is: \[ \mathrm{Solution Set} = \left\{ \begin{pmatrix} -2x_4-x_3 \\ x_4+x_3 \\ -x_4+1 \end{pmatrix} \, | \, x_4 \in \mathbb{R} \right\} \]

Step 1: Write down the augmented matrix

First, write down the augmented matrix for the given system. Since \(A\mathbf{x}=0\), we can omit writing the augmented column of zeros. \[ \left[\begin{array}{rrrr} 1 & 1 & 1 & -1 \\ -1 & 0 & -1 & 2 \\ 1 & 3 & 2 & 2 \end{array}\right] \]

Step 2: Perform row-reduction (Gaussian elimination) until reaching row-echelon form

Perform Gaussian elimination to reach the row-echelon form. We start by adding the first row to the second row to eliminate the -1 in the position (2,1): \[ \left[\begin{array}{rrrr} 1 & 1 & 1 & -1 \\ 0 & 1 & 0 & 1 \\ 1 & 3 & 2 & 2 \end{array}\right] \] Now, eliminate the 1 in position (3,1) by subtracting the first row from the third row: \[ \left[\begin{array}{rrrr} 1 & 1 & 1 & -1 \\ 0 & 1 & 0 & 1 \\ 0 & 2 & 1 & 3 \end{array}\right] \] Finally, eliminate the 2 in position (3,2) by subtracting two times the second row from the third row: \[ \left[\begin{array}{rrrr} 1 & 1 & 1 & -1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right] \] Now we have the row-echelon form of the augmented matrix.

Step 3: Write down the corresponding linear system for the row-echelon form

Now, write down the corresponding linear system for the row-echelon form of the augmented matrix: \[ \begin{cases} x_1 + x_2 + x_3 = -x_4 \\ x_2 = x_4 + x_3 \\ x_3 = -x_4 + 1 \end{cases} \]

Step 4: Express the basic variables in terms of the free variables

From the linear system above, we can express the basic variables, \(x_1\), \(x_2\), and \(x_3\) in terms of the free variable \(x_4\): \[ x_1 = -x_4 - x_2 = -x_4 - (x_4+x_3) \] \[ x_2 = x_4 + x_3 \] \[ x_3 = -x_4 + 1 \]

Step 5: Write the solution set as the set of all possible solutions

Now we can write the solution set as the set of all possible solutions, using the expressions found in Step 4: \[ \mathrm{Solution Set} = \left\{ \begin{pmatrix} -x_4-(x_4+x_3) \\ x_4+x_3 \\ -x_4+1 \end{pmatrix} = \begin{pmatrix} -2x_4-x_3 \\ x_4+x_3 \\ -x_4+1 \end{pmatrix} \, | \, x_4 \in \mathbb{R} \right\} \] The solution set represents the set of all solutions to the system \(A\mathbf{x}=0\).

Key Concepts

Row-echelon form

Gaussian Elimination is a method used to solve systems of linear equations. One of its key goals is to transform a given matrix into a special form called row-echelon form (REF). This form is crucial because it simplifies solving the equations.
To reach the row-echelon form:

• The leading entry in each nonzero row (after the first) is 1, typically called a "pivot".
• Each pivot is to the right of any pivots in the rows above.
• Rows with all zero elements, if any, are at the bottom.

In the step-by-step solution, Gaussian Elimination was performed on the matrix to transform it into row-echelon form. This makes it easier to identify and solve for the variables.

Augmented matrix

An augmented matrix is a compact way of representing a linear system of equations. It includes the matrix of coefficients along with the constants in a single matrix. When dealing with a homogeneous system like in the given problem, there's no need for an extra column of constants because we are solving for cases where the constant is zero.
For the original system, the augmented matrix started as:

• \[\begin{bmatrix}1 & 1 & 1 & -1 \-1 & 0 & -1 & 2 \1 & 3 & 2 & 2\end{bmatrix}\]

The Gaussian Elimination process was used to modify this matrix, allowing us to visualize how changes affect the overall system.

Solution set

The solution set of a system of equations is the set containing all possible solutions. Here, we find all vectors \( \mathbf{x} \) that satisfy \( A\mathbf{x} = 0 \). When expressed as a solution set, we use known variables in terms of free variables.
For the solved example, only one free variable \(x_4\) was introduced. Using the row-echelon form, we derived expressions for the basic variables in terms of \(x_4\):

• \[x_1 = -2x_4 - x_3\]
• \[x_2 = x_4 + x_3\]
• \[x_3 = -x_4 + 1\]

This provides a comprehensive description of all potential solutions involving \(x_4\).

Linear system

A linear system consists of linear equations involving the same set of variables. The main task is to find the values that satisfy all equations simultaneously.
In the problem, the linear system was formed as:

• \[x_1 + x_2 + x_3 = -x_4\]
• \[x_2 = x_4 + x_3\]
• \[x_3 = -x_4 + 1\]

This system represents interlinked equations where understanding variable dependencies is key. Solving involves expressing some variables in terms of others, often requiring techniques like Gaussian Elimination to simplify and to systematically handle dependencies.