Problem 28 - Intermediate Algebra

Question

Suppose that the augmented matrix of a system of three linear equations in three variables can be changed to the following matrix. $$ \left[\begin{array}{rrr:r} 1 & 0 & 1 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ What can be said about the solution set of the system?

Step-by-Step Solution

Answer

The system has infinitely many solutions.

Step 1: Identify the Matrix Form

The given matrix is an augmented matrix resulting from a system of equations. Let's rewrite it comprehendibly: \[ \begin{bmatrix} 1 & 0 & 1 & | & 1 \ 0 & 1 & -1 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \]. It represents three equations in three variables, where the rightmost column represents constants.

Step 2: Translate Rows into Equations

Convert each row of the augmented matrix into a linear equation. The first row gives the equation \( x + z = 1 \), the second row gives \( y - z = 0 \), and the third row \( 0 = 0 \) indicates no new information.

Step 3: Analyze the Third Row

The third row consists entirely of zeros on the left side and zero on the right side, \( 0 = 0 \). This suggests there is no contradiction and the system is consistent, meaning there is at least one solution.

Step 4: Solve for Each Variable

Using the obtained equations, solve for the variables. From the second equation, \( y - z = 0 \), we find \( y = z \). Substitute \( z = y \) into the first equation \( x + z = 1 \) to get \( x + y = 1 \).

Step 5: Express the Solution Set

The system has two leading variables, \( x \) and \( y \), which means \( z \) is free. Express \( x \) and \( y \) in terms of \( z \): \( x = 1 - z \) and \( y = z \). Therefore, the solution can be expressed as \((1 - z, z, z)\), where \( z \) is any real number.

Key Concepts

Understanding Systems of Linear Equations

A system of linear equations is a set of equations with multiple variables. Each equation describes a line, plane, or hyperplane in multidimensional space. The goal is often to find points where these lines or planes intersect. This is called finding the solution set.
\[\]When dealing with systems of linear equations, one common method of finding solutions is using augmented matrices, which simplify the process. An augmented matrix compresses the system’s equations by focusing on coefficients and constants.
\[\]In our exercise, the system derived from the augmented matrix can be represented as:

• \(x + z = 1\)
• \(y - z = 0\)
• \(0 = 0\)

Here, the first two rows translate into equations in three variables, forming the system of linear equations. The third equation does not affect the solution due to its trivial nature, \(0 = 0\). This means the system doesn’t have contradictions and is consistent.

Role of a Free Variable

In systems of linear equations, a free variable occurs when the number of variables exceeds the number of independent equations. It means not all variables are directly solvable from the system equations.
\[\]In our example, the variable \(z\) acts as a free variable. This is because the system has only two independent equations for three variables:\([x, y, z]\). When a variable is free, it can take any real number, allowing for infinitely many solutions.
\[\]Given this flexibility, we can express the dependent variables, \(x\) and \(y\), in terms of \(z\):

• \(x = 1 - z\)
• \(y = z\)

This assignment of values captures all potential solutions, constrained only by the system equations, as \(z\) varies freely across all real numbers.

Describing the Solution Set

Once a free variable is identified, it helps in outlining the complete solution set for the equations. The solution set can be visualized as a line or plane depending on the dimensions affected by the free variable. Here, the solutions are expressed parametrically using the free variable \(z\).
\[\]For the given system:

• \((x, y, z) = (1 - z, z, z)\)

This shows that as \(z\) changes, new solutions are generated, forming a line of possible solutions in 3D space.
\[\]The expression involves parameterizing the variables \(x\) and \(y\) using \(z\), indicating an infinite number of solutions. The concept of a free variable is fundamental in defining solution sets for such systems, allowing us to easily identify and express all possible solutions.