Problem 36 - Precalculus Mathematics for Calculus

Question

Use Cramer's Rule to solve the system. $$\left\\{\begin{array}{l} \frac{1}{2} x+\frac{1}{3} y=1 \\ \frac{1}{4} x-\frac{1}{6} y=-\frac{3}{2} \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = -2 \) and \( y = 6 \).

Step 1: Write the System in Matrix Form

First, we need to write the given system of equations in the form of a matrix. The system is:\[\begin{align*}\frac{1}{2}x + \frac{1}{3}y &= 1 \\frac{1}{4}x - \frac{1}{6}y &= -\frac{3}{2}\end{align*}\]In matrix form, this can be expressed as:\[A = \begin{pmatrix} \frac{1}{2} & \frac{1}{3} \ \frac{1}{4} & -\frac{1}{6} \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \ -\frac{3}{2} \end{pmatrix}\]Therefore, the system is \( A\mathbf{x} = \mathbf{b} \).

Step 2: Calculate Determinant of Coefficient Matrix (Det(A))

Next, we find the determinant of matrix \( A \). For a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is given by:\[ \text{det}(A) = ad - bc \]Plugging in the values, we get:\[ \text{det}(A) = \left(\frac{1}{2}\right)\left(-\frac{1}{6}\right) - \left(\frac{1}{3}\right)\left(\frac{1}{4}\right) = -\frac{1}{12} - \frac{1}{12} = -\frac{2}{12} = -\frac{1}{6} \]

Step 3: Calculate Determinants of Modified Matrices

Cramer's Rule requires us to compute determinants of matrices \( A_x \) and \( A_y \), derived from \( A \) by replacing each column with the vector \( \mathbf{b} \).For \( A_x \): Replace the first column with \( \mathbf{b} \):\[ A_x = \begin{pmatrix} 1 & \frac{1}{3} \ -\frac{3}{2} & -\frac{1}{6} \end{pmatrix} \]\[ \text{det}(A_x) = \left(1\right)\left(-\frac{1}{6}\right) - \left(\frac{1}{3}\right)\left(-\frac{3}{2}\right) = -\frac{1}{6} + \frac{1}{2} = -\frac{1}{6} + \frac{3}{6} = \frac{2}{6} = \frac{1}{3} \]For \( A_y \): Replace the second column with \( \mathbf{b} \):\[ A_y = \begin{pmatrix} \frac{1}{2} & 1 \ \frac{1}{4} & -\frac{3}{2} \end{pmatrix} \]\[ \text{det}(A_y) = \left(\frac{1}{2}\right)\left(-\frac{3}{2}\right) - \left(1\right)\left(\frac{1}{4}\right) = -\frac{3}{4} - \frac{1}{4} = -1 \]

Step 4: Apply Cramer's Rule

Cramer's Rule states that the solution for \( x \) can be found as:\[ x = \frac{\text{det}(A_x)}{\text{det}(A)} \]And for \( y \):\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} \]Substitute the determinants:\[ x = \frac{\frac{1}{3}}{-\frac{1}{6}} = \frac{1}{3} \times -6 = -2 \]\[ y = \frac{-1}{-\frac{1}{6}} = \frac{-1}{1} \times -6 = 6 \]

Step 5: Verify Solution

Substitute \( x = -2 \) and \( y = 6 \) back into the original equations to verify:First Equation: \( \frac{1}{2}(-2) + \frac{1}{3}(6) = -1 + 2 = 1 \) ✓Second Equation: \( \frac{1}{4}(-2) - \frac{1}{6}(6) = -\frac{1}{2} - 1 = -\frac{3}{2} \) ✓Both equations are satisfied, confirming the solution is correct.

Key Concepts

Determinant Calculation

Understanding how to calculate the determinant is crucial for applying Cramer's Rule. The determinant provides important information about a matrix, such as whether the matrix is invertible and the solutions of the linear system. For a 2x2 matrix like the one in our exercise, the formula is simple. If you have a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), its determinant, \( \text{det}(A) \), is calculated as:\[\text{det}(A) = ad - bc\]This formula considers the products of the diagonals of the matrix. The determinant impacts the feasibility of using Cramer's Rule because a zero determinant indicates that a unique solution does not exist. Keep in mind, that if \( \text{det}(A) \) is zero, then we can't proceed with this method.

Matrix Algebra

Matrix algebra forms the foundation of solving linear systems with Cramer's Rule. When dealing with matrices, remember that they can compactly represent systems of equations. In our specific problem, we converted the linear equations into the matrix form:

• Coefficient matrix \( A \)
• Variable vector \( \mathbf{x} \)
• Constant vector \( \mathbf{b} \)

This resulted in the equation \( A\mathbf{x} = \mathbf{b} \), which encapsulates our system of equations. Working with matrices allows us to leverage matrix operations and rules, such as multiplication and determinant calculation, to find solutions efficiently. Understanding these operations significantly widens your tools for solving equations, particularly when systems become more complex.

Linear Systems

Linear systems of equations, like the one in our problem, consist of multiple linear equations that we solve together. Cramer's Rule is a powerful way to find solutions to such systems when they are expressed in a certain form. A system is characterized by equations where all terms are either constants or multipliers of variables raised to the exponent of one.When using Cramer's Rule, the system must meet certain conditions:

• The coefficient matrix must be square (same number of equations as variables).
• The determinant of the coefficient matrix must be non-zero.

In our exercise, the original system of linear equations was: \[\left\{\begin{array}{l}\frac{1}{2} x+\frac{1}{3} y=1 \\frac{1}{4} x-\frac{1}{6} y=-\frac{3}{2}\end{array}\right.\]This system was solvable using Cramer's Rule because it met the necessary conditions. Knowing how to convert a problem into a linear system is half the battle, allowing you to apply numerous mathematical strategies for its solution.