Problem 58 - 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. Anna, Ashley, and Andrea weigh a combined \(370 \mathrm{lb}\). If Andrea weighs \(20 \mathrm{lb}\) more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh?

Step-by-Step Solution

Answer

Anna weighs 180 lb, Ashley 120 lb, and Andrea 140 lb.

Step 1: Define the Variables

Let's define the variables for the weights of Anna, Ashley, and Andrea. Let \( x \) be the weight of Anna, \( y \) be the weight of Ashley, and \( z \) be the weight of Andrea.

Step 2: Set Up the Equations

We are given three conditions. First, the combined weight of Anna, Ashley, and Andrea is 370 lb: \( x + y + z = 370 \). Second, Andrea weighs 20 lb more than Ashley: \( z = y + 20 \). Third, Anna weighs 1.5 times as much as Ashley: \( x = 1.5y \).

Step 3: Represent Equations in Matrix Form

Using the equations from Step 2, we represent the system in matrix form: \[ \begin{bmatrix} 1 & 1 & 1 \ 0 & -1 & 1 \ 1.5 & -1 & 0 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 370 \ 20 \ 0 \end{bmatrix} \]

Step 4: Find the Inverse of the Matrix

Calculate the inverse of the coefficient matrix: \( A^{-1} = \begin{bmatrix} 1 & 1 & 1 \ 0 & -1 & 1 \ 1.5 & -1 & 0 \end{bmatrix}^{-1} \). Using a calculator or software, find the inverse.

Step 5: Solve the System Using the Inverse

Multiply the inverse matrix obtained in Step 4 with the constants matrix to find the solution vector: \[ \begin{bmatrix} x \ y \ z \end{bmatrix} = A^{-1} \begin{bmatrix} 370 \ 20 \ 0 \end{bmatrix} \]. Compute the values to find \( x \), \( y \), and \( z \).

Step 6: Interpret the Solution

After calculating the matrix product, we find \( x \,\approx \, 180 \), \( y \,\approx \, 120 \), and \( z \,\approx \, 140 \). Thus, Anna weighs 180 lb, Ashley weighs 120 lb, and Andrea weighs 140 lb.

Key Concepts

Matrix Inverse

When dealing with systems of equations, one powerful method is using an inverse matrix to find solutions. Think of a matrix as a table of numbers arranged in rows and columns that corresponds to a linear transformation. Just like you can reverse numbers with their reciprocals, matrices can be reversed with their inverses.
An inverse matrix is denoted as \( A^{-1} \), where \( A \) is the original matrix. For a matrix to have an inverse, it must be square (same number of rows and columns) and must not be singular (meaning it should have non-zero determinant). If these conditions are met, \( A \times A^{-1} \) results in the identity matrix, which acts like the number 1 in matrix multiplication.
Here’s the general idea to use inverse matrices in solving systems:

• Ensure you have a square coefficient matrix.
• Calculate the inverse of this matrix.
• Multiply this inverse by the constants vector to find your solution vector.

In essence, the inverse matrix helps us transform and solve our complex problem directly.

Linear Algebra

Linear algebra is a branch of mathematics that focuses on systems of linear equations, vectors, and matrices. It provides the tools needed to model and solve problems involving linear relationships. In this context, the system of equations represents a real-world situation with three equations about the weights of Anna, Ashley, and Andrea.
The power of linear algebra in this problem lies in its ability to simplify the messy process of solving these kinds of equations by hand. The interconnected rows and columns in a matrix make it easier to handle multiple linear equations at once.
The matrix form of our system is a compact way to express the relationships between our variables (weights in this case) and their coefficients. Linear algebra allows us to use operations like matrix multiplication and finding determinants to solve such systems efficiently.

• Understanding matrices: They represent a system of linear equations.
• Using operations: Multiply, add, and transform to solve equations.
• Finding solutions: Identify solutions through computations such as finding the inverse.

Overall, the concepts of linear algebra simplify complex systems into manageable and solvable forms using matrices.

Variable Definition

Defining variables is the crucial first step in translating a real-world scenario into a mathematical problem. Variables serve as placeholders for unknowns that we need to solve for. In this specific exercise, the variables represent the weights of the three girls.
Proper definition of variables ensures clarity in setting up the equations. Here, we chose \( x \), \( y \), and \( z \) to represent Anna's, Ashley's, and Andrea's weights respectively. Such clarity helps avoid confusion when creating equations.
To define variables efficiently:

• Identify what the problem is asking for.
• Assign variables to unknown quantities.
• Use consistent symbols throughout the problem to maintain clarity.

By clearly defining variables, you set the stage for framing the system of equations correctly and solving them using appropriate methods, such as the matrix inverse approach described earlier.