Problem 55 - 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. Students were asked to bring their favorite fruit to class. \(95 \%\) of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were \(5 \%\) less popular than bananas, what are the percentages of each individual fruit?

Step-by-Step Solution

Answer

Bananas: 20%, Apples: 15%, Oranges: 40%.

Step 1: Define Variables

Let \( x \) represent the percentage of bananas, \( y \) the percentage of apples, and \( z \) the percentage of oranges. We need these to sum to 95%, as stated in the problem.

Step 2: Set Up Equations

We have three conditions:1. \( x + y + z = 95 \)2. Oranges are twice as popular as bananas: \( z = 2x \)3. Apples are 5% less popular than bananas: \( y = x - 5 \)

Step 3: Convert to Standard Form for Matrix

Express the equations in standard linear form.- From \( x + y + z = 95 \), we have: \( x + y + z = 95 \).- For \( z = 2x \), rewrite as: \( -2x + 0y + z = 0 \).- For \( y = x - 5 \), rewrite as: \( -x + y + 0z = -5 \).

Step 4: Write as a Matrix Equation

Formulate the matrix equation from the system:\[\begin{bmatrix} 1 & 1 & 1 \ -2 & 0 & 1 \ -1 & 1 & 0 \end{bmatrix}\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 95 \ 0 \ -5 \end{bmatrix}\]

Step 5: Calculate the Inverse of the Coefficient Matrix

Find the inverse of the matrix:\[A = \begin{bmatrix} 1 & 1 & 1 \ -2 & 0 & 1 \ -1 & 1 & 0 \end{bmatrix}\]The inverse \( A^{-1} \) can be calculated using various methods, like row reduction or a calculator.

Step 6: Solve for the Variables

Multiply both sides of the equation by \( A^{-1} \) to find:\[\begin{bmatrix} x \ y \ z \end{bmatrix} = A^{-1} \begin{bmatrix} 95 \ 0 \ -5 \end{bmatrix}\]Perform the matrix multiplication to solve for \( x \), \( y \), and \( z \).

Step 7: Interpret Result

After performing the operations, you will find:- \( x = 20 \)- \( y = 15 \)- \( z = 40 \)Thus, \( x + y + z = 75 \) are accounted for in the 95% of fruits.

Key Concepts

Matrix Inverse

When solving systems of equations through matrices, one powerful tool is the matrix inverse. Understanding what a matrix inverse is can greatly ease solving complex linear equations.

In simple terms, the inverse of a matrix is like the reciprocal of a number. If a number multiplied by its reciprocal equals one, then a matrix multiplied by its inverse equals the identity matrix. If you have a square matrix \( A \), its inverse (denoted as \( A^{-1} \)) satisfies:

• \( A \cdot A^{-1} = I \)
• \( A^{-1} \cdot A = I \)

Where \( I \) is the identity matrix, a special diagonal matrix with ones on the main diagonal and zeroes elsewhere. Not every matrix has an inverse; a matrix must be square (same number of rows and columns) and have a non-zero determinant to possess an inverse.

Finding the inverse can be done through various methods—row operations, using the adjoint and determinant, or with technology like calculators or software, especially for larger matrices. For our system, finding the inverse of the coefficient matrix allowed us to solve the equations efficiently.

Linear Algebra

Linear algebra is the branch of mathematics that deals with vectors, vector spaces, and matrices. It's a foundational tool in understanding and solving systems of linear equations. The core of linear algebra involves concepts like matrix operations, solving systems using row reduction, and exploring vector spaces.

Let's unpack a few fundamental operations used in solving our system:

• **Matrix Representation**: By converting equations into matrix form, we simplify handling multiple equations at once. Relations become visual and computationally manageable.


• **Linear Equations**: These are equations that can be represented as a linear polynomial, such as \( ax + by + cz = d \).


• **Matrix Equation**: Transforming a structure of equations into a single matrix equation \( A\textbf{x} = \textbf{b} \) where \( extbf{x} \) is the vector of variables.

Linear algebra is behind numerous practical applications, from engineering to computer graphics. In this exercise, linear algebra provided a compact and effective framework to address the fruit percentage problem systematically.

Problem-Solving Steps

Solving problems like the fruit percentage system requires a structured approach. Here is a breakdown of effective problem-solving steps:

• **Define the Variables**: Choose letters to represent unknowns (like percentages of fruits), which helps in setting up equations precisely.


• **Set Up the Equations**: Use the information given (like relationships between bananas, apples, and oranges) to form equations.


• **Convert to Standard Form**: Rearrange equations into a format suitable for matrix representation, aligning all variable coefficients.


• **Matrix Representation**: Write equations as a matrix equation. It simplifies solving by consolidating information into a matrix form: \( A\textbf{x} = \textbf{b} \).


• **Calculate the Matrix Inverse**: Determine \( A^{-1} \) since multiplying it by \( extbf{b} \) allows for solving \( extbf{x} \).


• **Solve for Variables**: Perform matrix multiplication to find the solution vector \( extbf{x} \), giving the percentages of each fruit.


• **Interpret Results**: Finally, check if your solution aligns with the problem's conditions and adjust if necessary.

This systematic approach not only helps in understanding current problems but also serves as a useful template for future problems involving systems of equations.