Problem 42 - Precalculus: Mathematics for Calculus

Question

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

x = \( \frac{3}{2} \), y = 2

Step 1: Write the System as a Matrix Equation

The given system of equations can be represented as a matrix equation of the form \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the vector of variables, and \( \mathbf{b} \) is the constant vector. Given equations:\[\begin{align*}6x + 12y &= 33 \4x + 7y &= 20\end{align*}\]This is equivalent to:\[A = \begin{pmatrix} 6 & 12 \ 4 & 7 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 33 \ 20 \end{pmatrix}\]

Step 2: Calculate the Determinant of Matrix A

The determinant of matrix \( A \) needs to be calculated as it is used in Cramer's Rule to check if the system has a unique solution.The determinant of \( A \), \( \det(A) \), is given by:\[\det(A) = \begin{vmatrix} 6 & 12 \ 4 & 7 \end{vmatrix} = 6 \cdot 7 - 12 \cdot 4 = 42 - 48 = -6\]Since \( \det(A) eq 0 \), the system has a unique solution.

Step 3: Form the Matrices A_x and A_y

Using Cramer's Rule, replace the respective columns of \( A \) with the vector \( \mathbf{b} \) to find matrices \( A_x \) and \( A_y \).Matrix \( A_x \) is formed by replacing the first column of \( A \) with \( \mathbf{b} \):\[A_x = \begin{pmatrix} 33 & 12 \ 20 & 7 \end{pmatrix} \]Matrix \( A_y \) is formed by replacing the second column of \( A \) with \( \mathbf{b} \):\[A_y = \begin{pmatrix} 6 & 33 \ 4 & 20 \end{pmatrix} \]

Step 4: Calculate the Determinants of A_x and A_y

Now, compute the determinants of \( A_x \) and \( A_y \) to apply Cramer's rule.The determinant of \( A_x \):\[\det(A_x) = \begin{vmatrix} 33 & 12 \ 20 & 7 \end{vmatrix} = 33 \cdot 7 - 12 \cdot 20 = 231 - 240 = -9\]The determinant of \( A_y \):\[\det(A_y) = \begin{vmatrix} 6 & 33 \ 4 & 20 \end{vmatrix} = 6 \cdot 20 - 33 \cdot 4 = 120 - 132 = -12\]

Step 5: Solve for x and y Using Cramer's Rule

Cramer's Rule states that the solutions for \( x \) and \( y \) are given by:\[x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}\]Substituting the determinants found:\[x = \frac{-9}{-6} = \frac{3}{2}, \quad y = \frac{-12}{-6} = 2\]Thus, the solution to the system is \( x = \frac{3}{2} \) and \( y = 2 \).

Key Concepts

Determinant

The determinant is a special number that can be calculated from a square matrix, like a 2x2 or 3x3 matrix. It is an essential part of Cramer's Rule, which is used to solve systems of linear equations.

For a 2x2 matrix \( A \), which looks like \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is given by the formula \( ad - bc \). This value helps to determine whether the system has a unique solution.

Why is the determinant important?

• If \( \det(A) eq 0 \), the matrix is invertible, which means there is a unique solution to the system of equations.
• If \( \det(A) = 0 \), the matrix does not have an inverse, and there could be no solution or infinitely many solutions.

In the case of our original problem, we calculated the determinant of matrix \( A \) as \( -6 \). This non-zero determinant confirms that a unique solution exists for the equations given.

Matrix Equation

Matrix equations are a straightforward way to represent and solve linear equations systematically. In our given exercise, the two equations:\[ 6x + 12y = 33 \]\[ 4x + 7y = 20 \]naturally fit into a matrix form.

The idea is to express these equations as \( A \mathbf{x} = \mathbf{b} \), where:

• \( A \) is the coefficient matrix \( \begin{pmatrix} 6 & 12 \ 4 & 7 \end{pmatrix} \)
• \( \mathbf{x} \) is the vector of variables \( \begin{pmatrix} x \ y \end{pmatrix} \)
• \( \mathbf{b} \) is the constant vector \( \begin{pmatrix} 33 \ 20 \end{pmatrix} \)

Using matrices simplifies handling and understanding of complex equations. It also makes operations like finding solutions using Cramer's Rule feasible through straightforward determinant calculations.

Unique Solution

A unique solution to a system of equations means there exists exactly one set of values for the variables that satisfies all the equations simultaneously.

This condition in a system can be confirmed when the determinant of the coefficient matrix \( A \) is not zero. In linear algebra, when solving for variables \( \begin{pmatrix} x & y \end{pmatrix} \) using Cramer's Rule, having a non-zero determinant of \( A \) ensures that we can obtain distinct values for \( x \) and \( y \).

With Cramer's Rule, the solutions are given by:

• \( x = \frac{\det(A_x)}{\det(A)} \)
• \( y = \frac{\det(A_y)}{\det(A)} \)

In this exercise, the determinants were calculated to give a result where \( x = \frac{3}{2} \) and \( y = 2 \). Thus, this demonstrates that the equations intersect at one specific point, meaning they have a unique solution. Determining this unique solution is essential for understanding how different variables interact with each other in any context.