Problem 25 - Precalculus

Question

Use Cramer's rule to solve each system of equations, if possible. $$\begin{aligned} &3 x+\frac{1}{2} y=1\\\ &4 x+\frac{1}{3} y=\frac{5}{3} \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{2} \) and \( y = -1 \).

Step 1: Write matrix equation

Represent the system of equations in matrix form as \( A \mathbf{x} = \mathbf{b} \), where \( A = \begin{pmatrix} 3 & \frac{1}{2} \ 4 & \frac{1}{3} \end{pmatrix} \), \( \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix} \), and \( \mathbf{b} = \begin{pmatrix} 1 \ \frac{5}{3} \end{pmatrix} \).

Step 2: Calculate determinant of matrix A

To use Cramer's Rule, first find the determinant of the matrix \( A \). For a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is given by \( \text{det}(A) = ad - bc \). Calculate: \[ \text{det}(A) = 3 \cdot \frac{1}{3} - 4 \cdot \frac{1}{2} = 1 - 2 = -1. \]

Step 3: Determine if Cramer's Rule can be applied

We can apply Cramer's Rule only if the determinant of matrix \( A \) is not equal to zero. From Step 2, we found that \( \text{det}(A) = -1 \), so Cramer's Rule is applicable.

Step 4: Set up matrices for Cramer's Rule

To find \( x \), replace the first column of \( A \) with \( \mathbf{b} \) to form matrix \( A_x = \begin{pmatrix} 1 & \frac{1}{2} \ \frac{5}{3} & \frac{1}{3} \end{pmatrix} \). To find \( y \), replace the second column of \( A \) with \( \mathbf{b} \) to form matrix \( A_y = \begin{pmatrix} 3 & 1 \ 4 & \frac{5}{3} \end{pmatrix} \).

Step 5: Calculate determinants for modified matrices

Calculate the determinant of \( A_x \): \[ \text{det}(A_x) = 1 \cdot \frac{1}{3} - \frac{5}{3} \cdot \frac{1}{2} = \frac{1}{3} - \frac{5}{6} = \frac{-3}{6} = -\frac{1}{2}. \] Calculate the determinant of \( A_y \): \[ \text{det}(A_y) = 3 \cdot \frac{5}{3} - 4 \cdot 1 = 5 - 4 = 1. \]

Step 6: Solve for variables using Cramer's Rule

Cramer's Rule states that \( x = \frac{\text{det}(A_x)}{\text{det}(A)} \) and \( y = \frac{\text{det}(A_y)}{\text{det}(A)} \). Using this, we find \[ x = \frac{-\frac{1}{2}}{-1} = \frac{1}{2}, \] and \[ y = \frac{1}{-1} = -1. \]

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements. It's essential for various applications, including solving systems of equations with Cramer's Rule. For a 2x2 matrix like \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated using the formula:

• \( \text{det}(A) = ad - bc \)

This means you multiply the diagonal elements \( a \) and \( d \), subtract the product of the off-diagonal elements \( b \) and \( c \). In our example, the matrix \( A \) is \( \begin{pmatrix} 3 & \frac{1}{2} \ 4 & \frac{1}{3} \end{pmatrix} \). Using the formula gives us:

• \( \text{det}(A) = 3 \cdot \frac{1}{3} - 4 \cdot \frac{1}{2} = 1 - 2 = -1 \)

Since the determinant is not zero, Cramer's Rule can be applied. The fact that it's non-zero ensures that the system of equations has a unique solution.
This is crucial for solving linear equations using matrices.

System of Equations

A system of equations is a collection of two or more equations that share variables. Solving them means finding values for the variables that make all equations true at the same time. Consider the given equations:

• \( 3x + \frac{1}{2}y = 1 \)
• \( 4x + \frac{1}{3}y = \frac{5}{3} \)

These equations form a system with two unknowns \( x \) and \( y \). Various methods exist to solve these systems, including substitution, elimination, and using matrices with Cramer's Rule. Using Cramer's Rule is particularly effective when the determinant of the coefficient matrix is non-zero.
When using this method, you calculate determinants of matrices modified by substituting columns, providing the solutions directly for each variable. This mathematical tool provides a direct and structured way to tackle potentially complicated systems of equations.

Matrix Representation

Matrix representation simplifies working with systems of equations by organizing coefficients and constants into matrices. For the equations provided:

• \( 3x + \frac{1}{2}y = 1 \)
• \( 4x + \frac{1}{3}y = \frac{5}{3} \)

We set up the system in the format \( A\mathbf{x} = \mathbf{b} \). The matrix \( A \) comprises coefficients of \( x \) and \( y \):

• \( A = \begin{pmatrix} 3 & \frac{1}{2} \ 4 & \frac{1}{3} \end{pmatrix} \)

\( \mathbf{x} \) is the column matrix of the variables:

• \( \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix} \)

And \( \mathbf{b} \) is the constant matrix:

• \( \mathbf{b} = \begin{pmatrix} 1 \ \frac{5}{3} \end{pmatrix} \)

This matrix setup allows us to use Cramer's Rule efficiently. This format not only organizes the problem but lays the groundwork for applying linear algebra techniques to find solutions.