Problem 61 - College Algebra

Question

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. Jay has lemon, orange, and pomegranate trees in his backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of \(70 \mathrm{lb}, 10\) oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick?

Step-by-Step Solution

Answer

Jay picked 70 lemons, 62 oranges, and 10 pomegranates.

Step 1: Define the Variables

Let \( x \) be the number of lemons, \( y \) be the number of oranges, and \( z \) be the number of pomegranates. We need to create equations based on these variables.

Step 2: Create the First Equation (Total Fruits)

The total number of fruits picked is 142, so we write the equation: \( x + y + z = 142 \).

Step 3: Create the Second Equation (Total Weight)

We have to convert the total weight into ounces. Given: 70 lb 10 oz = 1130 oz (since 1 lb = 16 oz). The weight equation is: \( 5x + 8y + 11z = 1130 \).

Step 4: Create the Third Equation (Relation between Oranges and Pomegranates)

Jay picked 15.5 times more oranges than pomegranates, hence: \( y = 15.5z \).

Step 5: Form the System of Equations

We now have the system of equations: \[ \begin{align*} x + y + z &= 142 \ 5x + 8y + 11z &= 1130 \ y &= 15.5z \end{align*} \]

Step 6: Substitute and Simplify

Substitute \( y = 15.5z \) into the other two equations: \[ x + 15.5z + z = 142 \ 5x + 8(15.5z) + 11z = 1130\] Simplify these equations: \[\begin{align*}x + 16.5z &= 142 \5x + 123z &= 1130\end{align*}\]

Step 7: Express Simplified System in Matrix Form

We express the system as a matrix equation: \( A\mathbf{x} = \mathbf{b} \). Where \( A = \begin{bmatrix} 1 & 16.5 \ 5 & 123 \end{bmatrix} \), \( \mathbf{x} = \begin{bmatrix} x \ z \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} 142 \ 1130 \end{bmatrix} \).

Step 8: Calculate the Inverse of Matrix A

The inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \( A^{-1} = \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \cdot \frac{1}{ad-bc} \). Calculate for: \[ A^{-1} = \begin{bmatrix} 123 & -16.5 \ -5 & 1 \end{bmatrix} \cdot \frac{1}{123 - (16.5 \times 5)} \].

Step 9: Multiply Inverse Matrix by b

Calculate \( \mathbf{x} = A^{-1} \cdot \mathbf{b} \) to find values of \( x \) and \( z \). Use matrix multiplication to get \( x \) and \( z \).

Step 10: Solve for y

Utilize \( y = 15.5z \) to calculate \( y \) once \( z \) is known from previous step by substituting \( z \) into the equation.

Key Concepts

System of Equations

A system of equations consists of multiple equations that are solved together to find common solutions to all of them. In our exercise, we constructed a system based on Jay's fruit-picking scenario. We needed to figure out how many lemons, oranges, and pomegranates he collected. Each equation in the system represents a different aspect of the scenario:

1. **Total Fruits equation**: This tells us how many pieces of fruit were picked in total. It is given by: \( x + y + z = 142 \), where \( x, y, \) and \( z \) represent the number of lemons, oranges, and pomegranates, respectively.
2. **Total Weight equation**: Here, we take into account the weight of each piece of fruit, expressed in ounces. This equation is: \( 5x + 8y + 11z = 1130 \), knowing conversions from pounds and ounces.
3. **Relationship equation**: Jay picked 15.5 times more oranges than pomegranates, which gives us: \( y = 15.5z \).

By solving this system of equations, we can determine the number of each type of fruit that Jay picked.

Inverse Matrix

To solve a system of linear equations efficiently, especially when represented in matrix form, the inverse matrix method can be applied. A matrix is a rectangular array of numbers, and an inverse matrix is analogous to the reciprocal of a number. When multiplied by the original matrix, it yields the identity matrix, essentially undoing the transformations represented by the original matrix.

In our exercise, the system of equations was represented as a matrix equation \( A\mathbf{x} = \mathbf{b} \). Here \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constants matrix. To solve for \( \mathbf{x} \), we calculated the inverse of matrix \( A \), denoted as \( A^{-1} \). The key process here is:
\[ \mathbf{x} = A^{-1} \cdot \mathbf{b} \]
Obtaining \( A^{-1} \) involved calculating determinants and adjugates of matrix \( A \). This method is particularly useful in handling systems with consistent and independent solutions.

Algebraic Problem Solving

Algebraic problem solving involves using algebraic methods to find unknown values within a given problem. In our scenario, we started by defining variables and breaking down the problem statement into workable components. Key steps in this process include:

• **Defining Variables**: Identifying what each variable stands for — in this case, \( x \) for lemons, \( y \) for oranges, and \( z \) for pomegranates.
• **Creating Equations**: Translating the real-world situation into mathematical equations based on the problem’s conditions, such as total weight and counts.
• **Substituting and Simplifying**: Using given relations, such as \( y = 15.5z \), to simplify the equations and reduce the number of variables.

These methods help frame the problem within a mathematical context, making it easier to apply algebraic and matrix methods for solutions.

Weights and Measures Conversion

Dealing with different units of measurement is a common need in problem-solving, especially when they mix systems like pounds and ounces. For Jay's fruit, the total weight given was a combination of pounds and ounces, requiring conversion to a single unit to make calculations easier.

To convert measurements:

• Recognize that 1 pound (lb) is equal to 16 ounces (oz).
• Convert pounds into ounces by multiplying the number of pounds by 16.
• Add the remaining ounces to this total to get the complete weight in ounces.

In the original problem, Jay's fruit weighed 70 lb and 10 oz. By converting this entirely to ounces, we get 70 multiplied by 16, plus 10, leading to 1130 ounces. Having a unified measure ensures calculations remain consistent and simplifies the algebra needed to solve the equations.