Problem 38 - Algebra for College Students

Question

In Problems 35-42, each matrix is the reduced echelon matrix for a system with variables \(x_{1}, x_{2}, x_{3}\), and \(x_{4}\). Find the solution set of each system. \(\left[\begin{array}{rrrr:r}1 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 2 & -3 \\ 0 & 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\)

Step-by-Step Solution

Answer

The solution set is \((2, -3 - 2t, 4 - 3t, t)\) for any real number \(t\).

Step 1: Analyze the Given Matrix

The matrix provided is a reduced row-echelon form, which makes it easier to identify solutions for systems of linear equations. Examine the entries of the matrix rows to form equations for each variable. The zero row at the bottom indicates a redundant equation, which confirms the system has infinitely many solutions.

Step 2: Write Down Each Equation

From the matrix:- The first row, \([1 \ 0 \ 0 \ 0 \ | \ 2]\), corresponds to the equation: \(x_1 = 2\).- The second row, \([0 \ 1 \ 0 \ 2 \ | -3]\), gives the equation: \(x_2 + 2x_4 = -3\).- The third row, \([0 \ 0 \ 1 \ 3 \ | \ 4]\), translates to: \(x_3 + 3x_4 = 4\).

Step 3: Solve for Basic Variables

Identify basic variables and express all other variables as free variables. From the equations:- Solve for non-free variable directly: \(x_1 = 2\).- Solve for \(x_2\) and \(x_3\): - \(x_2 = -3 - 2x_4\) - \(x_3 = 4 - 3x_4\) Free variable is \(x_4\).

Step 4: Express Solution Set

Express the solution set using the free variable. Since \(x_4\) is free, let it be any real number \(t\):- \(x_1 = 2\)- \(x_2 = -3 - 2t\)- \(x_3 = 4 - 3t\)- \(x_4 = t\)The solution set is: \(\{(2, -3 - 2t, 4 - 3t, t) \mid t \in \mathbb{R}\}\)

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. In our exercise, we have the variables \(x_1, x_2, x_3,\) and \(x_4\). The linear equations in the system can be extracted from the rows of the given matrix in reduced row echelon form (RREF). The system is represented by the equations:

• \(x_1 = 2\)
• \(x_2 + 2x_4 = -3\)
• \(x_3 + 3x_4 = 4\)

The fourth row consists entirely of zeros, which indicates a redundant equation, reflecting that these equations do not contain conflicting information about the variables. Converting the matrix into RREF simplifies finding solutions to the system, as it highlights the dependencies and independent variables clearly. The goal of solving a system is to find values for each variable that make all the equations true simultaneously.

This system of equations can be interpreted to permit infinite values for the free variable(s), showing the versatility and openness of solutions that can be formed under given conditions.

Basic and Free Variables

In the context of solving a system of linear equations using matrices in reduced row echelon form, we often encounter basic and free variables. Basic variables are those that can be directly solved for in terms of other variables. Free variables, in contrast, are not constrained by the leading entries of any row and can take any real number value.

In our given matrix:

• \(x_1\), \(x_2\), and \(x_3\) are basic variables because they can be directly solved from the equations derived from the matrix.
• \(x_4\) is a free variable since it does not appear as a leading variable in any row; instead, it appears freely in the equations \(x_2 + 2x_4 = -3\) and \(x_3 + 3x_4 = 4\).

The freedom of \(x_4\) allows us to express the basic variables \(x_2\) and \(x_3\) in terms of \(x_4\), thus leading us to a solution that incorporates this free variable. Letting \(x_4 = t\), where \(t\) is any real number, we derive:

• \(x_2 = -3 - 2t\)
• \(x_3 = 4 - 3t\)

Here, \(x_4\) being unrestricted illustrates its fundamental role in sculpting a broader range of solutions

Infinitely Many Solutions

The occurrence of a free variable in a system signifies the presence of infinitely many solutions. This stems from the freedom to choose any value for the free variable \(x_4\). For every unique value selected for \(x_4\), a different solution is generated for the system.

In our exercise, setting \(x_4 = t\) (where \(t\) is any real number) helps express other variables in terms of \(t\):

• \(x_1 = 2\)
• \(x_2 = -3 - 2t\)
• \(x_3 = 4 - 3t\)
• \(x_4 = t\)

These expressions translate into the solution set: \(\{(2, -3 - 2t, 4 - 3t, t) \mid t \in \mathbb{R}\}\). The parameter \(t\) effectively counts the degree of freedom in the system, signifying that there are an unlimited number of solutions.

The redundant row of zeros in the matrix supports this by not introducing any new condition to the system, confirming the consistent, yet infinitely varying solutions that satisfy the original set of linear equations.