Problem 30 - Algebra and Trigonometry

Question

\(29-44\) Use Cramer's Rule to solve the system. $$ \left\\{\begin{array}{l}{6 x+12 y=33} \\ {4 x+7 y=20}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{3}{2}\) and \(y = 2\).

Step 1: Write the System of Equations

Start with the given system of linear equations: \[ \begin{align*} 6x + 12y &= 33, \ 4x + 7y &= 20. \end{align*} \] Here, we need to solve for the variables \(x\) and \(y\) using Cramer's Rule.

Step 2: Set Up the Coefficient Matrix

Identify the coefficients of the variables in the system to form the matrix, called the coefficient matrix \(A\):\[ A = \begin{bmatrix} 6 & 12 \ 4 & 7 \end{bmatrix} \] This matrix is derived by taking the coefficients from each equation.

Step 3: Calculate the Determinant of Coefficient Matrix

Compute the determinant of the coefficient matrix \(A\):\[ \text{det}(A) = \begin{vmatrix} 6 & 12 \ 4 & 7 \end{vmatrix} = (6)(7) - (12)(4) = 42 - 48 = -6. \]The determinant is \(-6\).

Step 4: Form the Matrix for x

Replace the first column of matrix \(A\) with the constants from the right side of the equations to form matrix \(A_x\):\[ A_x = \begin{bmatrix} 33 & 12 \ 20 & 7 \end{bmatrix} \]

Step 5: Calculate the Determinant for Matrix x

Find the determinant of matrix \(A_x\):\[ \text{det}(A_x) = \begin{vmatrix} 33 & 12 \ 20 & 7 \end{vmatrix} = (33)(7) - (12)(20) = 231 - 240 = -9. \]

Step 6: Form the Matrix for y

Replace the second column of matrix \(A\) with the constants from the right side to form matrix \(A_y\):\[ A_y = \begin{bmatrix} 6 & 33 \ 4 & 20 \end{bmatrix} \]

Step 7: Calculate the Determinant for Matrix y

Calculate the determinant of matrix \(A_y\):\[ \text{det}(A_y) = \begin{vmatrix} 6 & 33 \ 4 & 20 \end{vmatrix} = (6)(20) - (33)(4) = 120 - 132 = -12. \]

Step 8: Solve for x and y

Use Cramer's Rule, where \(x = \frac{\text{det}(A_x)}{\text{det}(A)}\) and \(y = \frac{\text{det}(A_y)}{\text{det}(A)}\):\[ x = \frac{-9}{-6} = \frac{3}{2}, \quad y = \frac{-12}{-6} = 2. \]

Step 9: Final Solution

The solution to the system of equations is:\[ x = \frac{3}{2}, \quad y = 2. \]

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations made up of two or more variables that share a common solution. In this exercise, we have a simple system with two equations and two variables, \(x\) and \(y\):

• Equation 1: \(6x + 12y = 33\)
• Equation 2: \(4x + 7y = 20\)

The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Such systems are often presented as matrices, where the coefficients of the variables, the matrix on the left, interact with the variables themselves, producing results shown on the right. To solve these systems, one effective method is using Cramer's Rule, which leverages the properties of determinants to isolate particular variable solutions.

Determinant Computation

Before diving into how Cramer's Rule uses determinants, let's first understand what a determinant is. A determinant is a special number that can be calculated from a square matrix. It provides essential information about the matrix, such as whether it is invertible.

For a 2x2 matrix, the determinant is computed as follows:

• Suppose \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \)
• The determinant of \(A\), denoted as \( \text{det}(A) \), is \(ad - bc\)

In the provided solution, the determinant for the coefficient matrix \(A\) is computed to be \(-6\).

This determinant is crucial because if it is zero, the system of equations either has no solutions or an infinite number of solutions. Here, the non-zero determinant indicates a unique solution exists.

Matrix Manipulation

Matrix manipulation involves altering matrices to form different matrices required for solving equations using methods like Cramer's Rule. In this exercise, matrix manipulation is crucial.

• First, we construct the original coefficient matrix \(A\) from the system of linear equations' coefficients:
• Next, to solve for \(x\), we replace the first column of \(A\) with the constants from the right-hand side of the equations, forming matrix \(A_x\).
• Similarly, to solve for \(y\), we replace the second column of \(A\) to form matrix \(A_y\).

These steps involve careful manipulation of the original matrix to form new ones, \(A_x\) and \(A_y\), whose determinants help isolate solutions for specific variables when used in combination with the original matrix's determinant.

The relationships established through these matrix manipulations underpin the practicality of Cramer's Rule, leading us to the solution \(x = \frac{3}{2}\) and \(y = 2\). This demonstrates how effective matrix manipulation and determinants are in linear algebra.