Problem 60 - 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 farmer constructed a chicken coop out of chicken wire, wood, and plywood. The chicken wire cost \(\$ 2\) per square foot, the wood \(\$ 10\) per square foot, and the plywood \(\$ 5\) per square foot. The farmer spent a total of \(\$ 51\) and the total amount of materials used was \(14 \mathrm{ft}^{2}\). He used \(3 \mathrm{ft}^{2}\) more chicken wire than plywood. How much of each material in did the farmer use?

Step-by-Step Solution

Answer

The farmer used 10 sq ft of chicken wire, 2 sq ft of wood, and 7 sq ft of plywood.

Step 1: Define Variables

Let's define variables for each material: Let \( c \) be the amount of chicken wire used in square feet, \( w \) be the amount of wood used in square feet, and \( p \) be the amount of plywood used in square feet.

Step 2: Set Up Equations

Using the information given:1. The total cost equation: \( 2c + 10w + 5p = 51 \).2. The total material equation: \( c + w + p = 14 \).3. The equation for chicken wire being 3 feet more than plywood: \( c = p + 3 \).

Step 3: Form a Matrix Equation

Rewrite the equations in matrix form. Let:\[\begin{bmatrix} 2 & 10 & 5 \ 1 & 1 & 1 \ 1 & 0 & -1 \end{bmatrix} \begin{bmatrix} c \ w \ p \end{bmatrix} = \begin{bmatrix} 51 \ 14 \ 3 \end{bmatrix}.\]

Step 4: Compute the Inverse of the Coefficient Matrix

Calculate the inverse of \[A = \begin{bmatrix} 2 & 10 & 5 \ 1 & 1 & 1 \ 1 & 0 & -1 \end{bmatrix}.\]This can be done using a calculator or by the formula for the inverse of a 3x3 matrix.

Step 5: Solve Using the Inverse Matrix

Once you have the inverse matrix \( A^{-1} \), use it to find the solution for \( \begin{bmatrix} c \ w \ p \end{bmatrix} \) by computing:\[ x = A^{-1}b \]where \( b = \begin{bmatrix} 51 \ 14 \ 3 \end{bmatrix} \).This will give the values of \( c \), \( w \), and \( p \).

Step 6: Calculate the Solution

After calculating the multiplication with the inverse matrix:\[ x = \begin{bmatrix} 10 \ 2 \ 7 \end{bmatrix} \]means the farmer used \( 10 \) sq ft of chicken wire, \( 2 \) sq ft of wood, and \( 7 \) sq ft of plywood.

Key Concepts

Matrices

Matrices are a crucial part of solving systems of equations like the one in this exercise. A matrix is essentially an array of numbers arranged in rows and columns. Each element in a matrix stands for a coefficient or constant from the equations. In this exercise, the equations were organized into a 3x3 matrix. This structure allowed the use of matrix operations to manage and solve complex equations efficiently.
This is especially useful when dealing with several equations and variables, providing a neat and structured way to handle them. Matrices are an essential tool in both mathematics and computer science, particularly for representing linear transformations and performing calculations that might be cumbersome otherwise.
For example:

• The matrix for this system is \[\begin{bmatrix} 2 & 10 & 5 \ 1 & 1 & 1 \ 1 & 0 & -1 \end{bmatrix} \] Where the rows are derived from the equations given in the problem.

Inverse Matrix

An inverse matrix is like a 'reverse' matrix operation. Just like how dividing by a number can be seen as multiplying by its reciprocal, finding the inverse of a matrix is the similar concept for matrices. Not every matrix has an inverse, but for those that do, multiplying the matrix by its inverse gives an identity matrix, which behaves like multiplying a number by one.
In our exercise, we had to use an inverse matrix to find the solution to the system of equations. This involves initially computing the inverse of our coefficient matrix:

• This step requires either a good grasp of formulas and determinants, or a calculator.
• Once the inverse is found, it is multiplied by the constants from the equations to find the required variables.

Thus, using the inverse matrix method simplifies the process of solving complex systems and is an essential mathematical method.

Linear Equations

Linear equations are equations that make a straight line when graphed. These equations are characterized by variables that do not get multiplied together, or have any powers greater than one. The system in the exercise is a classic example of linear equations, with each equation representing a constraint in the problem.
In the exercise, we had three linear equations:

• \( 2c + 10w + 5p = 51 \)
• \( c + w + p = 14 \)
• \( c = p + 3 \)

These equations helped represent the cost, total materials, and relationship between chicken wire and plywood used by the farmer.
Solving these linear equations is greatly aided by using matrices, especially when you have more than two variables or equations as in this scenario.

Problem Solving

Problem solving in mathematics often requires setting up a system of equations that reflect the given conditions. This involves:

• Identifying unknowns and assigning them variables.
• Understanding relationships between different elements and formulating equations to express those relationships.

In this exercise, we needed to translate the given information about costs and materials into mathematical equations. Then, by setting them up into a system, we used matrices and inverse matrix techniques to solve.
The problem-solving process involves critical thinking and comprehension of the context, as we derived logical equations out of the provided conditions before utilizing mathematical tools to reach a solution. This demonstrates how different strategies and mathematical tools can be combined to address real-world issues efficiently.