Problem 32 - Algebra and Trigonometry

Question

\(29-44\) 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 standard form

Rearrange the given system of equations. The original system is: \( \frac{1}{2}x + \frac{1}{3}y = 1 \) and \( \frac{1}{4}x - \frac{1}{6}y = -\frac{3}{2} \). It is already in the standard form, \( ax + by = c \). So nothing needs to be changed here.

Step 2: Identify coefficients and write matrices

From the system, identify the coefficients of \( x \) and \( y \):- Coefficients of the first equation: \( a_1 = \frac{1}{2} \), \( b_1 = \frac{1}{3} \), \( c_1 = 1 \).- Coefficients of the second equation: \( a_2 = \frac{1}{4} \), \( b_2 = -\frac{1}{6} \), \( c_2 = -\frac{3}{2} \).Write the coefficient matrix, constant matrices for solving using Cramer's Rule:- Coefficient matrix \( A = \begin{pmatrix} \frac{1}{2} & \frac{1}{3} \ \frac{1}{4} & -\frac{1}{6} \end{pmatrix} \) - Constant matrix\( B = \begin{pmatrix} 1 \ -\frac{3}{2} \end{pmatrix} \).

Step 3: Find the determinant of the coefficient matrix

Calculate the determinant of \( A \):\[ \det(A) = \left(\frac{1}{2}\right) \left(-\frac{1}{6}\right) - \left(\frac{1}{3}\right) \left(\frac{1}{4}\right) = -\frac{1}{12} - \frac{1}{12} = -\frac{2}{12} = -\frac{1}{6} \]

Step 4: Construct matrices by replacing columns and find determinants

Replace the first column of \( A \) with \( B \) to get matrix \( A_x \), and the second column to get matrix \( A_y \).Matrix \( A_x = \begin{pmatrix} 1 & \frac{1}{3} \ -\frac{3}{2} & -\frac{1}{6} \end{pmatrix} \)\[ \det(A_x) = (1)(-\frac{1}{6}) - (\frac{1}{3})(-\frac{3}{2}) = -\frac{1}{6} + \frac{1}{2} = -\frac{1}{6} + \frac{3}{6} = \frac{2}{6} = \frac{1}{3} \]Matrix \( A_y = \begin{pmatrix} \frac{1}{2} & 1 \ \frac{1}{4} & -\frac{3}{2} \end{pmatrix} \)\[ \det(A_y) = \left(\frac{1}{2}\right)(-\frac{3}{2}) - (1)\left(\frac{1}{4}\right) = -\frac{3}{4} - \frac{1}{4} = -1 \]

Step 5: Calculate values of x and y using Cramer's Rule

Use Cramer's rule: - \( x = \frac{ \det(A_x) }{ \det(A) } = \frac{ \frac{1}{3} }{ -\frac{1}{6} } = -2 \)- \( y = \frac{ \det(A_y) }{ \det(A) } = \frac{-1}{-\frac{1}{6}} = 6 \)

Step 6: Conclude with the solution

The solution to the system of equations is \( x = -2 \) and \( y = 6 \).

Key Concepts

Understanding Systems of Equations

Systems of equations are a set of two or more equations with the same variables. In simple terms, you can think of them as multiple lines on a graph that may intersect at one or more points. Solving these systems means finding the point(s) where the lines meet, which will give the values for the variables that satisfy all equations simultaneously.
In the example provided, we are given a system:

• \( \frac{1}{2}x + \frac{1}{3}y = 1 \)
• \( \frac{1}{4}x - \frac{1}{6}y = -\frac{3}{2} \)

This system has two equations with two variables, \( x \) and \( y \), and we aim to find their values using methods such as Cramer's Rule. Understanding how these equations relate to each other is crucial to solving them effectively.

Determinant Calculation in Cramer's Rule

The determinant is a special number that can be calculated from a square matrix. When using Cramer's Rule to solve systems of equations, finding the determinant is a critical step. The determinant gives us important information about the matrix, such as whether it has an inverse or not. If the determinant is zero, the matrix does not have an inverse, and thus Cramer's Rule cannot be applied.
In the given problem, we calculate the determinant of the coefficient matrix \( A \), which is composed of the coefficients from the system of equations.
The determinant of matrix \( A \) is found using the formula:
\[ \text{det}(A) = a_{11}a_{22} - a_{12}a_{21} \]For our specific example, the determinant \( \det(A) \) is computed as follows:

• \[ \text{det}(A) = \left( \frac{1}{2} \right) \left( -\frac{1}{6} \right) - \left( \frac{1}{3} \right) \left( \frac{1}{4} \right) = -\frac{1}{12} - \frac{1}{12} = -\frac{1}{6} \]

This non-zero determinant indicates that the system of equations has a unique solution.

Matrix Algebra in Solving Systems

Matrix algebra is a fundamental part of solving systems of equations using methods like Cramer's Rule. A matrix is a rectangular array of numbers, which can be manipulated through various operations like addition, multiplication, and finding determinants. In our context, matrices are used to organize coefficients and constants from the system of equations.
First, we construct the **coefficient matrix** \( A \), which consists of the coefficients of the variables \( x \) and \( y \). It looks like this:

• \[ A = \begin{pmatrix} \frac{1}{2} & \frac{1}{3} \ \frac{1}{4} & -\frac{1}{6} \end{pmatrix} \]

We also form a **constant matrix** \( B \) using the right-hand side of the equations, which is:

• \[ B = \begin{pmatrix} 1 \ -\frac{3}{2} \end{pmatrix} \]

To apply Cramer's Rule, we create new matrices \( A_x \) and \( A_y \) by replacing the columns of \( A \) with \( B \) one at a time and compute their determinants. The solutions for \( x \) and \( y \) are then found by dividing these determinants by the original determinant of \( A \). This process highlights the elegance and power of matrix algebra in solving linear systems efficiently.