Problem 56 - College Algebra with Corequisite Support

Question

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at \(\$ 1\) and the chocolate chip cookies at \(\$ 0.75\). They raised \(\$ 700\) and sold 850 items. How many brownies and how many cookies were sold?

Step-by-Step Solution

Answer

250 brownies and 600 cookies were sold.

Step 1: Define Variables

Let's define two variables: \( x \) for the number of brownies and \( y \) for the number of chocolate chip cookies sold.

Step 2: Formulate the System of Equations

According to the problem, the total items sold were brownies plus cookies, equaling 850: \[ x + y = 850 \] The total money raised is the sum of the money from brownies and cookies, equaling $700: \[ 1x + 0.75y = 700 \] So, the system of equations is:\[\begin{cases} x + y = 850 \ x + 0.75y = 700 \end{cases}\]

Step 3: Write in Matrix Form

Write the system of equations in matrix form as \( AX = B \), where:\[A = \begin{bmatrix} 1 & 1 \ 1 & 0.75 \end{bmatrix}, \ X = \begin{bmatrix} x \ y \end{bmatrix}, \ B = \begin{bmatrix} 850 \ 700 \end{bmatrix}\]

Step 4: Find the Inverse of Matrix A

To solve the system using the inverse, we need to find \( A^{-1} \). The formula to find the inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For our matrix:\[ A = \begin{bmatrix} 1 & 1 \ 1 & 0.75 \end{bmatrix} \]\( ad - bc = 1(0.75) - 1(1) = -0.25 \), so:\[ A^{-1} = \frac{1}{-0.25} \begin{bmatrix} 0.75 & -1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 4 \ 4 & -4 \end{bmatrix}\]

Step 5: Solve for X using Matrix Inversion

Using the equation \( X = A^{-1}B \), compute:\[ X = \begin{bmatrix} -3 & 4 \ 4 & -4 \end{bmatrix} \begin{bmatrix} 850 \ 700 \end{bmatrix} \]Executing the multiplication, we get:\[ \begin{bmatrix} -3 \times 850 + 4 \times 700 \ 4 \times 850 - 4 \times 700 \end{bmatrix} = \begin{bmatrix} -2550 + 2800 \ 3400 - 2800 \end{bmatrix} \]Evaluating the results:\[ \begin{bmatrix} 250 \ 600 \end{bmatrix} \]Thus, \( x = 250 \) and \( y = 600 \).

Step 6: Verify the Solution

Verify the solution by substituting \( x = 250 \) and \( y = 600 \) back into the original equations:1. \( x + y = 250 + 600 = 850 \) (satisfied)2. \( x + 0.75y = 250 + 0.75 \times 600 = 250 + 450 = 700 \) (satisfied)Both original equations are satisfied, confirming the solution is correct.

Key Concepts

Matrix Inversion

Matrix inversion is a crucial concept in linear algebra used to solve systems of equations. When we have a system of linear equations like the sorority bake sale problem, we can represent these equations with matrices. The key to matrix inversion is to find a matrix that, when multiplied by the original, results in the identity matrix. This is akin to dividing numbers, where for example, multiplying by the reciprocal undoes a multiplication.
For a 2x2 matrix, the inverse is calculated using the formula:

• Given matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \).
• The inverse \( A^{-1} \) is determined as \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).

In our problem, the matrix \( A \) represents the coefficients of the equations, and by finding \( A^{-1} \), we can compute the solution by using matrix multiplication with the constants matrix \( B \). The result is the values for the unknowns \( x \) and \( y \), which correspond to the number of brownies and cookies sold.

Matrix Formulation

When solving a system of equations, formulating them in terms of matrices is both efficient and insightful. This approach uses a matrix format to represent the equations, capturing the essence of the problem in a neat, compact form. In the sorority bake sale exercise, the system of equations can be written as a matrix equation \( AX = B \).
The breakdown is as follows:

• Matrix \( A \) contains the coefficients of the equations: \( A = \begin{bmatrix} 1 & 1 \ 1 & 0.75 \end{bmatrix} \).
• Matrix \( X \) contains the variables, the unknowns \( x \) and \( y \): \( X = \begin{bmatrix} x \ y \end{bmatrix} \).
• Matrix \( B \) represents the constants from the right-hand side of the equations: \( B = \begin{bmatrix} 850 \ 700 \end{bmatrix} \).

By using the matrix form, students learn to appreciate the structure behind linear equations and how this method provides a systematic way to find solutions, particularly when using advanced techniques like finding the inverse.

Equation Verification

Verification of solutions is always an important step in solving equations. It confirms that the calculated answers satisfy the original problem. In the case of the bake sale, once we've found the number of brownies and cookies sold using matrix inversion, we perform this step to ensure accuracy.The process is straightforward:

• Substitute the values of \( x \) and \( y \) back into the original equations.
• Check if each equation holds true.

For our exercise, we determined \( x = 250 \) and \( y = 600 \). By substituting these values:- First equation: \( 250 + 600 = 850 \) is satisfied.- Second equation: \( 250 + 0.75 \times 600 = 700 \) holds true as well.
If both equations are satisfied, it confirms the correctness of our solution, demonstrating the reliability of matrix methods in solving systems of equations. It's a reassuring step for validating the hard work involved.