Problem 10 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether each system of linear equations has (a) one and only one solution, (b) infinitely many solutions, or (c) no solution. Find all solutions whenever they exist. \(\frac{5}{4} x-\frac{2}{3} y=3\) \(\frac{1}{4} x+\frac{5}{3} y=6\)

Step-by-Step Solution

Answer

The system of linear equations has one unique solution, which is \(x=4\) and \(y=2\).

Step 1: Calculate the determinant

To determine the number of solutions possible, we first need to calculate the determinant of the coefficient matrix. The coefficient matrix for the given system of equations is \[ A = \begin{bmatrix} \frac{5}{4} & -\frac{2}{3} \\ \frac{1}{4} & \frac{5}{3} \end{bmatrix} \] To calculate the determinant, we use the formula \(D = ad - bc\), where \(a = \frac{5}{4}, \: b = -\frac{2}{3}, \: c = \frac{1}{4}, \: d = \frac{5}{3}\) So, \[D = \frac{5}{4} \cdot \frac{5}{3} - \left(-\frac{2}{3}\right) \cdot \frac{1}{4} = \frac{25}{12} + \frac{1}{6} = \frac{27}{12}\]

Step 2: Analyzing the determinant

Since the determinant \(D = \frac{27}{12}\) is not zero, the given system of linear equations has a unique solution. Now we will use Cramer's rule to find the solution.

Step 3: Find the solution using Cramer's rule

Cramer's rule allows us to find the values of \(x\) and \(y\) in a unique solution using determinants. The solution for \(x\) and \(y\) can be calculated using the formulas: \(x = \frac{D_x}{D}\) \(y = \frac{D_y}{D}\) where \(D_x\) is the determinant obtained by substituting the constant term of each equation for its corresponding \(x\)'s coefficient in matrix \(A\): \[ D_x = \begin{vmatrix} 3 & -\frac{2}{3} \\ 6 & \frac{5}{3} \end{vmatrix} \] \(D_y\) is the determinant obtained by substituting the constant term of each equation for its corresponding \(y\)'s coefficient in matrix \(A\): \[ D_y = \begin{vmatrix} \frac{5}{4} & 3 \\ \frac{1}{4} & 6 \end{vmatrix} \] Now, let's calculate \(D_x\) and \(D_y\): \(D_x = 3 \cdot \frac{5}{3} - 6 \cdot \left(-\frac{2}{3}\right) = 5 + 4 = 9\) \(D_y = 6 \cdot \frac{5}{4} - 3 \cdot \frac{1}{4} = \frac{27}{4} - \frac{3}{4} = \frac{24}{4}\) Now, we can find the values of \(x\) and \(y\): \(x = \frac{9}{\frac{27}{12}} = \frac{9}{\frac{27}{12}} \cdot \frac{12}{1} = \frac{108}{27} = 4\) \(y = \frac{\frac{24}{4}}{\frac{27}{12}} = \frac{6}{\frac{9}{4}} \cdot \frac{4}{1} = 6 \cdot \frac{4}{3} = 2 \cdot 4 = 2\)

Step 4: Verify the solution

Now that we have found the values of \(x=4\) and \(y=2\), let's verify if they satisfy both given equations: 1. \(\frac{5}{4} \cdot 4 - \frac{2}{3} \cdot 2 = 5 - \frac{4}{3} = 5 - \frac{4}{3} = 3\): True 2. \(\frac{1}{4} \cdot 4 + \frac{5}{3} \cdot 2 = 1 + \frac{10}{3} = 1 + 3\frac{1}{3} = 6\): True Since both equations are satisfied by the values of \(x=4\) and \(y=2\), we can conclude that the given system of linear equations has one unique solution \((4, 2)\).

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a scalar value that is a function of the entries of a square matrix. It provides important information about the matrix, such as whether the matrix is invertible and the volume scaling factor for the transformation associated with the matrix.

For a 2x2 matrix, the determinant can be calculated using the formula \(D = ad - bc\), where \(a, b, c,\) and \(d\) are the elements of the matrix arranged as follows:
\[A = \begin{bmatrix}a & b \ c & d\end{bmatrix}\]
In the context of linear algebra and systems of equations, the determinant plays a pivotal role in figuring out the nature of the solutions. If the determinant is non-zero, the matrix is said to have an inverse, and the corresponding system of linear equations typically has a unique solution.

Cramer's Rule

Cramer's Rule is a theorem in algebra that provides an explicit expression for the solution of a system of linear equations with as many equations as unknowns, given that the system has a unique solution. It uses determinants to solve for each variable.

For a system of two equations as in the given exercise, Cramer's Rule instructs us to substitute the constants from the equations into their respective columns in the matrix and then evaluate the determinant for each:

\(x = \frac{D_x}{D}\) and \(y = \frac{D_y}{D}\)

where \(D\) is the determinant of the coefficient matrix, and \(D_x\) and \(D_y\) are the determinants of matrices formed by replacing the x and y columns, respectively, with the constant terms from the equations.

This rule provides a straightforward method for solving systems of equations when the determinant of the coefficient matrix is nonzero. However, it is not often used for larger systems due to the computational complexity.

Unique Solution

A unique solution in the context of a system of linear equations means that there is only one set of values for the variables that satisfies all the equations simultaneously. Specifically, for a two-variable system like the one in the exercise, it implies there is exactly one pair (x, y) that makes both equations true.

This occurs when the lines represented by each equation intersect at exactly one point. In terms of the determinant of the system's coefficient matrix, we know we have a unique solution if the determinant is non-zero. This condition guarantees that each equation contributes distinct information and that they are not multiple representations of the same line or parallel lines, which would result in either infinitely many solutions or no solution at all.

Solving Linear Systems

Solving a system of linear equations means finding the values of the variables that make all the equations true at the same time. There are several methods to solve linear systems, including graphing, substitution, elimination, and using matrices such as the augmented matrix method or applying Cramer's Rule.

In the exercise, the determinant and Cramer's Rule are applied to find the unique solution. It's important for students not to just perform the calculations but to understand the properties of linear systems, like how the determinant indicates the nature of solutions. Practical learning also involves verifying solutions by substituting the found values back into the original equations to ensure they hold true, reinforcing the mathematical concepts through application.