Problem 6 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Given that the augmented matrix in row-reduced form is equivalent to the augmented matrix of a system of linear equations, (a) determine whether the system has a solution and (b) find the solution or solutions to the system, if they exist. \(\left[\begin{array}{rrrr|r}1 & 0 & 0 & 0 & 3 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & 2\end{array}\right]\)

Step-by-Step Solution

Answer

The given system has infinitely many solutions, with the solution expressed as: \(x_1 = 3, x_2 = -1 - x_3, x_3 = x_3, x_4 = 2\), where \(x_3\) can be any real number.

Step 1: Determine if the system has a solution

Review the row-reduced matrix: \(\left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 3 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & 2 \end{array}\right]\) There isn't any row of the form \([0 \: 0 \: 0 \: 0 \: | \: k]\) with \(k \neq 0\), meaning there are no contradictions in the given system. Therefore, the system has a solution.

Step 2: Identify the basic and free variables

The basic variables are those that correspond to the columns with leading 1s: \(x_1, x_2,\) and \(x_4.\) The free variable is the remaining variable, which is \(x_3\).

Step 3: Write down the row-reduced matrix as a system of linear equations

Transform the row-reduced matrix into a system of linear equations: \(x_1 = 3\) \(x_2 + x_3 = -1\) \(x_4 = 2\)

Step 4: Solve the system of linear equations

Solve the given system by expressing the basic variables in terms of the free variables, if any: \(x_1 = 3\) \(x_2 = -1 - x_3\) \(x_3 = x_3\) \(x_4 = 2\) The solution to the system is then: \(x_1 = 3, \: x_2 = -1 - x_3, \: x_3 = x_3, \: x_4 = 2\) As there is only one free variable (which can take on any real value), there are infinitely many solutions to the given system.

Key Concepts

Row-Reduced Echelon Form

When working with systems of linear equations, a common method for finding solutions is to express the augmented matrix of the system in what is known as row-reduced echelon form (RREF). This form is characterized by certain properties. First, the leading entry in each row is 1, and it is referred to as a pivot. This 1 is to the right of the pivot in the row above, if there is one.

Every column that contains a pivot has zeros in every other position, creating a staircase pattern of leading 1s descending from left to right. If a column does not have a leading 1 and is not a column of the augmented part, then the corresponding variable is considered a free variable. The RREF is very handy as it makes it straightforward to see whether the system has a unique solution, no solution, or infinitely many solutions.

Matrix Solutions

The solutions of a system of linear equations can often be deduced directly from the RREF of its augmented matrix. If the matrix leads to a row that translates into an impossible statement (like 0 = 1), the system has no solution. On the other hand, if the RREF does not have such inconsistencies, the system will have at least one solution.

The RREF allows the easy identification of basic and free variables, which in turn helps to express the solution in a parametric form. In the case where the system has unique solutions, each variable will be defined explicitly in terms of the constants from the augmented portion of the matrix. However, if free variables are present, the system will have infinitely many solutions represented by a set of parametric equations.

Free Variables

Free variables are the heartbeats of a system of linear equations that yield infinitely many solutions. They are the variables that do not lead a column in the matrix's row-reduced echelon form. In essence, free variables can take on any real number value, and they provide the freedom needed for the system to have more than one solution.

Identifying free variables is crucial because they are expressed as parameters in the general solution of the system. When a solution involves free variables, it means you can plug in any value for these variables to get a specific solution within the infinite solution set. This concept highlights the idea that not all variables in a linear equation system need to be strictly determined to satisfy the equations.

Infinite Solutions

When a system of linear equations has at least one free variable, it turns out to have an infinite number of solutions. This is because the free variable can assume any value, which in turn generates a different solution for every value it takes. The solutions can typically be written in parametric form, where the free variables serve as the parameters.

As with the example problem, having a free variable (\(x_3\) in this case) in the system implies that there are infinite ways to satisfy the equation set. Since the variable can be any real number, each choice leads to a different set of values for our dependent variables, creating a continuous set of solutions. This is a key concept that differentiates such systems from those with a single solution or no solution at all.