Problem 11 - Precalculus

Question

The following matrices are in reduced row echelon form. Determine the solution of the corresponding system of linear equations or state that the system is inconsistent. $$ \left[\begin{array}{rrrr|r} 1 & 0 & -8 & 1 & 7 \\ 0 & 1 & 4 & -3 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$

Step-by-Step Solution

Answer

The system has infinitely many solutions: \( x_1 = 8x_3 - x_4 + 7 \), \( x_2 = -4x_3 + 3x_4 + 2 \) where \( x_3, x_4 \) are free.

Step 1: Identify the Matrix Structure

The given matrix is already in reduced row echelon form, where leading ones are at the start of each row that contains a pivot, and each leading one is the only non-zero entry in its column. The system corresponds to the following equations: \( x_1 - 8x_3 + x_4 = 7 \) and \( x_2 + 4x_3 - 3x_4 = 2 \). The remaining rows represent the equation \( 0 = 0 \), indicating no additional constraints.

Step 2: Recognize Free and Basic Variables

Identify the variables corresponding to leading ones (basic variables) and those that are not leading in any row (free variables). Here, \( x_1 \) and \( x_2 \) are basic variables, while \( x_3 \) and \( x_4 \) are free variables since they do not lead any row.

Step 3: Express Basic Variables in Terms of Free Variables

Using the transformed system of equations, express the basic variables \( x_1 \) and \( x_2 \) in terms of the free variables \( x_3 \) and \( x_4 \). From the equations: \( x_1 = 8x_3 - x_4 + 7 \) and \( x_2 = -4x_3 + 3x_4 + 2 \).

Step 4: Formulate the General Solution

The general solution of the system is constructed by expressing all variables in terms of the free variables. The solution set, in parametric form, can be written as: \( x_1 = 8x_3 - x_4 + 7 \), \( x_2 = -4x_3 + 3x_4 + 2 \), \( x_3 = x_3 \), and \( x_4 = x_4 \).

Key Concepts

Reduced Row Echelon Form

In the study of systems of linear equations, the reduced row echelon form (RREF) of a matrix is a simplified version where solving becomes easier.
This form has specific characteristics to help quickly identify basic features of the solution:

• Each leading entry in a row is 1.
• Each leading one is the only non-zero entry in its column.
• Any rows containing all zeros are placed at the bottom.

In the matrix provided, each leading 1 appears at the start of each row, setting a clear path for interpreting solutions.
This structure simplifies back substitution when finding variable values.

Free Variables

Free variables are a crucial part of determining the general solution to a system of linear equations.
These are the variables that do not have a leading 1 in any of the rows of the matrix.
Instead, free variables remain unspecified and can take any real number value.

• Free variables provide flexibility in the solution, representing unlimited possibilities for values.
• They are necessary for formulating solutions in parametric form, allowing the expression of certain variables and equations in terms of others.

In our system, the free variables are identified as those without leading coefficients: here, they are the third and fourth variables, denoted by \(x_3\) and \(x_4\).

Basic Variables

Basic variables are the anchors of the system of equations, corresponding to the leading 1s in the reduced row echelon form.
They can be expressed in terms of the free variables, leading to a concise solution.

• Basic variables have fixed expressions in terms of the free variables, offering more determinate values in connection with the parameterization of the solutions.
• Each basic variable corresponds to a row in the matrix with a leading 1, and they are usually solved directly from equations.

In this particular exercise, the basic variables are \(x_1\) and \(x_2\), which appear as the first variable in each pivotal equation. Their solutions are critically dependent on the values assumed by the free variables.

Parametric Solution

A parametric solution is a way to express the infinite solution set of a linear system that includes free variables.
Instead of a single solution, you get a family of solutions, expressing each variable in terms of free parameters, demonstrating how solutions vary.

• Each free variable is treated as a parameter, such as \(t\) or \(s\), to describe how other variables depend on them.
• This approach captures all possible solutions by altering the values of the free variables.

For example, our solution is formulated as:- \(x_1 = 8x_3 - x_4 + 7\)- \(x_2 = -4x_3 + 3x_4 + 2\)- \(x_3 = x_3\), and- \(x_4 = x_4\)Thus, changing \(x_3\) and \(x_4\) results in different solutions, illustrating the system's infinite nature.